Explain and discuss the most common negligent error by any allied health professional of your crossing.

Explain and discuss the most common negligent error by any allied health professional of your crossing.

Health care law and legislation discussion board week 3

Hospital Department and Allied Health Professionals Options Menu: Forum

Describe and discuss the most common negligent error by any allied health professional of your crossing.

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Health Care Law And Legislation DB Week three

Health Care Law And Legislation DB Week three

Health care law and legislation discussion board week 3

Hospital Department and Allied Health Professionals Options Menu: Forum

Describe and discuss the most common negligent error by any allied health professional of your crossing.

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Analysis Anova: Single Factor

Analysis Anova: Single Factor

Analysis
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Column 1 55 27.7395 0.5043545455 0.2961900025
Column 2 49 24.945 0.5090816327 0.4145881186
Column 3 53 11.6171 0.219190566 0.1736197093
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 2.9003011651 2 1.4501505826 4.9712754701 0.0080878949 3.0547708302
Within Groups 44.9227147151 154 0.2917059397
Total 47.8230158803 156
Data
Wet Moderate Dry
0 0 0
0 0.63 0
0.56 0 0
0 1.04 0
1.3863 1.426 0
1.08 2.164 0
1.48 1.323 0
0 0 0.562
0 0 0
1.2437 1.2729 0
0.608 0 0.397
0.7357 0 0
0 0 0
0 0 0
0.322 0 0
0 0.598 0
0.6931 0 0
1.5581 0 0
0 0 0
0 0 0
0 0 0.6931
0 0 0
0.56 1.69 0
1.048 0.69 0
1.004 0.7 0
0.95 1.02 1.1
0 0 0
0 0.69 0.56
1.11 1.72 0
0 0 0
0 0.7 0
0.68 1.09 0
0 1.81 1.535
0 1.7 0
1.1 0.6931 0
0 0 0
0 1.33 0
0 0 0
0.7 0 0
0.7 0.6 1.06
1.08 0.84 0.7
0 0 0
0 0 0
0 1.098 0.69
0.535 0 0.69
0.694 0 0
0.694 0.12 0.287
1.0986 0 1.55
1.32 0 0
1.5 0.693
0 0
0 0
0.64 1.1
1.05
1.609
Graph
Average Variance
Wet 0.5043545455 0.2961900025
Moderate 0.5090816327 0.4145881186
Dry 0.219190566 0.1736197093
Figure 1. Invertebrate soil diversity in wet, moderate and dry habitats at cal poly pomona in spring of 2019
Graph
0.2961900025
0.4145881186
0.1736197093
Wet Moderate Dry
Habitats
Mean Invertebrate Diversity (H’)

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Soil Invertebrate Diversity

Soil Invertebrate Diversity

1

Formal Lab

Foundations of Biology: Evolution, Ecology, and Biodiversity Lab

Soil Invertebrate Diversity

LEARNING OUTCOMES During this three-day laboratory exercise, you will:

  1. Sample the invertebrates in soil and leaf litter using field and laboratory techniques. 2. Use a dichotomous key to identify the soil invertebrates. 3. Calculate and compare the invertebrate diversity within soil samples taken from habitats

with different water availabilities. 4. Statistically analyze the collective data from all lab sections to serve as the basis for a

formal lab report. BACKGROUND Invertebrates are one of the dominant groups of organisms in soil food webs. Probably the most common invertebrate groups in the soil are arthropods and nematodes, but other phyla are represented as well. Soil invertebrates may be predators, herbivores, or detritivores, or may feed specifically on fungi and bacteria that are abundant in the soil. Of course, soil invertebrates live where they do not only for the food, but also for the shelter—they require the shade, moisture, and stable temperatures that soil and leaf litter provide. The diversity of soil invertebrates can affect the properties of the very soil in which they reside, due to the fact that they influence key soil processes such as nutrient cycling, nutrient retention, the formation of soil structure, and decomposition rates. To sample soil invertebrates, you first need to obtain a small quantity of soil or leaf litter. The tricky part is then extracting all of the small invertebrates from the soil. To get them to come out of their otherwise optimal habitat, you need to make the soil unpleasantly hot and dry. An apparatus called a Berlese funnel will help with this job. The Berlese funnel does not extract all of the organisms from the soil. Fungi and bacteria are not motile enough to escape, and some invertebrates are also connected to the soil particles. However, this technique does give a good estimate of the true soil invertebrate diversity. EXERCISES Activity 1 (Week 1) – Sampling a Soil Habitat for Invertebrates

  1. You will work with your lab partner to obtain a soil and leaf litter sample from one of three local campus habitats. Your lab instructor will lead you to all three locations so that everyone is able to observe the conditions encountered and methods employed at each.
  2. The three locations sampled will be habitats that differ in water availability. The first site

2

will be a manicured grove that receives an optimal amount of water on a regular basis. The second site will be a natural woodland that on occasion has a small stream trickling through, so there is some (but not an abundant amount of) water present much of the time. Therefore, even when water is not present, the soil remains somewhat damp. The third and final site will be a very dry grassland / shrubland location that gets minimal water throughout the year. Note that you may not sample from these three locations in this particular order.

  1. All lab sections will sample from these same three habitats—data will be combined for

all sections. As each lab section has twelve groups of two, four groups from each section will sample from one particular location. Specific sampling spots within each habitat will be chosen randomly by each group. Following the directions given by your instructor, obtain a leaf litter sample from your sampling spot. Fill about three-fourths of a plastic bag with soil and leaf litter scooped from the ground surface.

  1. Spend a few minutes at each location examining the habitat that was just sampled. Do

you see any invertebrates crawling on the surface of the soil or in the leaf litter? How wet or dry does the site appear? Begin thinking about which habitat may contain the greatest invertebrate diversity.

Activity 2 (Week 1) – Constructing and Using a Berlese Funnel A Berlese funnel is a simple device used for extracting small invertebrate animals from soil and leaf litter samples. The Berlese funnel creates a very hot and dry environment at one end of the funnel, driving the invertebrates to the other end. Here they will fall out into a container of alcohol that is set up to collect them. Once you have extracted the invertebrates from your sample, you can observe them using a dissecting microscope and, with the aid of a dichotomous key, identify what kinds of invertebrates are present in the sample. This process, along with calculating the Shannon index, will allow you to determine your soil invertebrate diversity. First, though, the Berlese funnel must be set up—follow the procedure below to construct your own funnel.

  1. Cut an empty and clean 2-liter bottle 3 cm below the top of the label using a sharp scalpel. Then cut off the neck by cutting along the middle of the curvature. The bottom of the bottle will serve as the jar. The top of the bottle with the neck removed will serve as the funnel. The neck of the bottle can be discarded.
  2. Fill approximately one-fourth of the jar with 70% isopropyl alcohol.
  3. On a strip of tape placed on the jar, write a label for your sample. Include your names,

lab section, and the habitat from which you sampled.

  1. Invert the funnel end of the bottle into the jar (see Figure 1).
  2. Place an approximately 30 cm × 30 cm piece of square screening into the funnel so that it lines the entire inside. The screening has mesh large enough for invertebrate animals to

3

fall through but will keep much of the soil and leaves from falling into the alcohol. Use masking tape to secure the square screening onto the outside of the funnel by taping down all four corners.

  1. Transfer your leaf litter sample from your plastic bag into the lined funnel.
  2. Place the entire funnel apparatus underneath a heat lamp. This heat source will drive the

invertebrates to the lower end of the soil sample, causing them to fall through the screening and the mouth of the funnel into the alcohol, where they will be collected as your sample. Adjust the lamp so that the bulb is about three to four inches away from the top of your leaf litter.

  1. Let the Berlese funnel sit under the heat lamp for one week. Most of the invertebrates

should be extracted in that amount of time.

Figure 1: Diagram of a Berlese Funnel Activity 3 (Week 2) – Identifying the Invertebrates in Your Soil Sample

  1. Remove the jar containing the extracted soil invertebrates from your Berlese funnel.
  2. Carefully pour a small amount of alcohol from the jar into a Petri dish. Only pour enough to fill half of the Petri dish.
  3. With your lab partner, observe the Petri dish under a dissecting microscope. Be sure to

adjust the light source, magnification, and focus for optimal viewing.

  1. As you find the invertebrates in your sample, make a sketch or detailed notes of each one so that you are very familiar with the organisms you are observing. Use a small paintbrush to move around or pick up the small animals in the Petri dish. Using forceps

4

is not likely to be effective and may actually crush some of the individuals in your sample.

  1. Use the dichotomous key provided by your lab instructor to identify the invertebrates in

your sample. As most of your sample will consist of arthropods, the key is specifically for this phylum. Key out the arthropods to the class level. As the class Insecta will be the most abundant, the key enables you to identify insects to the order level. All other phyla can simply be recorded as such. Record the name of each phylum, class, or order that you identify in Table 1. Keep a tally of the number of individuals in each taxonomic group that you encounter in your sample. You will probably want to remove the organisms from your Petri dish once you have recorded them so that you can keep an accurate tally, but do not discard them completely in case you need to recheck any of your work.

  1. Repeat steps 2 through 5 until all of the alcohol is drained from your jar. If necessary,

you may have to add additional alcohol to “rinse” out any invertebrates or small particles that stick to the walls of the jar.

  1. Once completed, count your tallies and record the total number of individuals from each

taxon in Table 1. Sum these counts to determine the total number of individuals in your entire sample.

Activity 4 (Week 2) – Calculating the Soil Invertebrate Diversity You will use the Shannon index (H′) to calculate the diversity of your soil / leaf litter sample. Recall the equation for this diversity index: abundance of a given taxon (i) where pi = H′ = -ΣS (pi ln pi) total abundance of all taxa and S = the # of different taxa Complete Table 1 by calculating pi, ln pi, and pi ln pi for each invertebrate taxon. Determine the total diversity by summing all of the pi ln pi values and reversing the negative sign. This will yield the diversity for your sample. Your lab instructor will indicate the proper procedures for cleaning up / discarding / putting away funnels, mesh, soil, leaf litter, alcohol, invertebrates, and lamps. Have your work checked by your lab instructor prior to disassembling your Berlese funnel and cleaning up your lab bench. Be sure to give your diversity value, as well as the habitat from which the value came, to your instructor. All of the diversity values from all lab sections will be combined to produce one large set of data.

5

Table 1: Shannon Diversity Index Calculations Lab Partner Names Habitat Sampled

Invertebrate Taxon (Phylum, Class, or

Order)

of Individuals of

this Taxon pi ln pi pi ln pi

Total # of individuals

(add the second column)

H′ = – ΣS (pi ln pi) (add the last column and reverse the – sign)

pi = # of individuals of a particular taxon / total # of individuals in the entire sample

6

Activity 5 (Week 3) – Analyzing Differences in Diversity Among the Habitats A statistical technique known as ANOVA will be used to determine if there is a difference in the diversity of invertebrates among the three habitats (which do differ in the amount of water that they receive). Let us explore the topic of statistics further using a simple example so that you feel comfortable with the analysis that you will be performing. Suppose you want to know if there is a difference between the mean (average) height of all of the male students on campus and the mean (average) height of all of the female students on campus. Here, the heights of all of the male students on campus and the heights of all of the female students on campus are termed populations, and the means of these populations are termed parameters. A parameter is a numerical summary of a population. You can probably imagine that it might be challenging to obtain the height of every single male student and every single female student on campus. Because populations are often very large (as in this example), population parameters will likely not be able to be determined. To resolve this, a sample (random subset) needs to be taken from each of the populations. You could sample 30 male students and measure their heights and then sample 30 female students and measure their heights. Here, the heights of the random subsets from the populations are termed samples, and the means of these samples are termed statistics. A statistic is a numerical summary of a sample. Sample statistics are therefore estimates of population parameters. Statistics will rarely if ever equal the parameters exactly. The difference between the statistic and the parameter (which a researcher would not know, as parameters are not known) is termed sampling error. Not only will statistics estimate the parameters of interest, they will allow for an objective method of dealing with this sampling error. Statistics are therefore very important tools, as they enable scientists to make conclusions about entire populations, though they are working with just samples. Statistics will reveal how reliable these conclusions are. How exactly are these conclusions determined? All hypotheses formulated in biology are set up as null and alternative hypotheses. A null hypothesis is a hypothesis of no difference or no relationship. In the student height example, the null hypothesis would be that there is no difference between the mean male height and the mean female height of all students on campus. An alternative hypothesis must accompany each null hypothesis, and together these two hypotheses must account for all possibilities. In this example, the alternative hypothesis would be that there is a difference in the mean heights of male and female students on campus. As you can see, there are no other possibilities—there either is a difference or there is not a difference between male and female heights. Based on the type of data, the number of samples collected, and the question of interest, a particular statistical test will be performed. Remember, for our soil invertebrate study you will be utilizing a test known as ANOVA. For the student height example, a different statistical test (known as a t-test) would be employed. Understanding when to use which type of statistical test need not concern you here—you will deal with that when you take a biostatistics class. What is important now is to understand the results that the tests give you.

7

Each statistical test will yield a result known as a p-value. A p-value is a probability value and therefore always ranges between 0 and 1. Again, the calculation of the p-value is beyond the scope of this course (we will let Microsoft Excel calculate it for us). However, you do need to be able to interpret it. A p-value is the probability (percent chance) that a sample could have been taken from a population in which the null hypothesis is true. Remember what the null hypothesis is—the hypothesis of no difference. If the p-value in the height example turned out to be 0.85, then there would be an 85% chance that the samples were taken from populations in which there is no difference in the mean heights of male and female students. This is a very high probability, so we would accept the null hypothesis and conclude that there is no difference between the mean height of males and the mean height of females on campus. Suppose the p-value was instead 0.01. This would mean that there is only a 1% chance that the samples came from populations in which the null hypothesis (no difference in the mean heights) is true. This is a very small probability, so we would therefore reject the null hypothesis, concluding instead that there is in fact a difference between the mean height of males and the mean height of females on campus (the alternative hypothesis). Where is the line drawn between accepting and rejecting the null hypothesis? It is standard practice in biology to only reject the null hypothesis if the p-value is less than or equal to 0.05. The reason for 0.05 will become apparent in a biostatistics course. For now, rest assured that there is a solid mathematical basis for this value. Once it is known that there is a difference between the populations (if in fact there is one), how can it be determined what that difference is? Go back and look at the mean values from the samples. If the males have a higher mean value, then it can be concluded that males on campus are significantly taller than females. If the females have a higher mean value, then it can be concluded that the females on campus are significantly taller. Again, remember that this is the goal of statistics—making overall conclusions about populations using data obtained only from samples. It is a very important tool, as we need to make conclusions about entire populations, but they are just too large to work with. Practice your understanding of statistics using another example: Is there a difference in photosynthesis rates among plants grown in low, medium, and high light levels? The sample mean photosynthesis rate from plants in low light is 20 units, the sample mean from medium light is 25 units, the sample mean from high light is 30 units, and the p-value is 0.03. (Units of photosynthesis = μmol/m2/sec.) What is the null hypothesis?

What is the alternative hypothesis?

8

Is the null hypothesis accepted or rejected? What conclusion can you make based on these results?

Now for some more details as to how ANOVA in particular works. ANOVA is used when you are analyzing differences in means among three or more samples. As we have data from three different samples (the three different habitats), ANOVA is the appropriate statistical test for our study. ANOVA stands for analysis of variance, which indicates what the test does—it analyzes the variances of the samples. Variances are measures of variability. If there is much variability among all of the samples, then it should make sense that the samples are not the same—there is a difference in the diversity of invertebrates, for example, among three sites that differ in water availability. Take a look at the following example, which shows hypothetical values for invertebrate diversity at three locations (only five values are shown for each in this example, but remember that our data set will be much larger as we are combining data from all lab sections): High H2O Low H2O No H2O 7 3 2 7 3 2 7 3 2 7 3 2 7 3 2 Mean: 7 3 2 In the above example, there is variation among the three habitats. The habitat with much water has the highest mean invertebrate diversity, the habitat with no water has the lowest mean invertebrate diversity, and the habitat with intermediate water availability have intermediate invertebrate diversity. In this case, the null hypothesis of no difference among the habitats would be rejected and we would instead conclude that there is a difference in diversity. Now consider another example, which also shows variation, but variation of a different kind: High H2O Low H2O No H2O 7 4 3 6 3 2 4 2 7 3 7 6 2 6 4 Mean: 4.4 4.4 4.4 In the above example, there is variation within each sample rather than among the samples. Notice that within each habitat, various different diversity values are obtained, but the mean

9

diversity value among all three habitats is identical. In this case, the null hypothesis of no difference among the samples would be accepted—there is no difference in invertebrate diversity among habitats with differing water availabilities. What causes the variation within each sample in this example? Perhaps it is different availabilities of other nutrients. Perhaps it is differing temperatures. However, as these factors are not being tested for in our study, we cannot say with certainty what is causing the variation. What can be said with certainty is that what we are testing for, a difference in water availability, has no effect on invertebrate diversity. In the first example, there was variation among the samples but no variation within each sample, and we rejected the null hypothesis of no difference. In the second example, there was variation within each sample but no variation among the samples, and we accepted the null hypothesis of no difference. A much more likely scenario is that there will be some variation among the samples and some variation within each sample: High H2O Low H2O No H2O 6 4 3 7 2 3 6 5 2 8 3 2 7 4 2 Mean: 6.8 3.6 2.4 You should convince yourself that, in order to reject the null hypothesis and conclude that there is a difference among the three habitats, there must be greater variation among the samples rather than within the samples. How is it determined which is greater? This is where the calculation of a p-value by Microsoft Excel comes into play. Enter the raw data into an Excel spreadsheet and then select Tools → Data Analysis → Anova: Single Factor. When you are asked to input the range, highlight your data. Click OK and find the p-value in the output table. Practice using the data analysis feature in Excel by entering the above example (the one with some variation among the samples and some variation within the samples). You should find a p- value of 1.18×10-5 (0.0000118) in the output table. Will you accept or reject the null hypothesis of no difference in this example? Why?

Is there more variation among the habitats or within each habitat in this example?

What conclusion can you make about invertebrate diversity in habitats with different water

availabilities? Note that the means for each sample are given in the output table as well.

10

You are now ready to input the actual invertebrate diversity data from the three different habitats sampled by the entire class. Don’t be overwhelmed by all of this background information—after all, Excel is doing the work for you. You just need to have a big picture understanding of the results that Excel gives you. Ask your instructor about your results to make sure that you are interpreting them correctly. Activity 6 (Week 3) – Searching for Sources as Background Information You will be writing your lab report in the form of a formal scientific paper. Your lab instructor will discuss the required components of a scientific paper. One requirement is to find other papers that have performed studies similar to yours, citing them in your own paper as background information or to compare results. Cited sources must be peer-reviewed scientific journal articles. Textbooks and websites are not permitted. While in lab, you will start to look up journal articles and have your lab instructor confirm that they are indeed relevant sources for your paper. Two good locations to begin your search for sources are:

  1. The Cal Poly Pomona website. From the homepage, search for the University Library, then begin searching for journal articles by typing in some keywords (for example, invertebrate diversity and water availability). The search results will indicate which sources are peer-reviewed journal articles.
  2. Google Scholar. Go to https://scholar.google.com/, type in some keywords, and examine

the peer-reviewed articles that result to determine which may be appropriate for inclusion in your paper.

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Foundations of Biology: Evolution, Ecology, and Biodiversity Lab

Foundations of Biology: Evolution, Ecology, and Biodiversity Lab

1

Formal Lab

Foundations of Biology: Evolution, Ecology, and Biodiversity Lab

Soil Invertebrate Diversity

LEARNING OUTCOMES During this three-day laboratory exercise, you will:

  1. Sample the invertebrates in soil and leaf litter using field and laboratory techniques. 2. Use a dichotomous key to identify the soil invertebrates. 3. Calculate and compare the invertebrate diversity within soil samples taken from habitats

with different water availabilities. 4. Statistically analyze the collective data from all lab sections to serve as the basis for a

formal lab report. BACKGROUND Invertebrates are one of the dominant groups of organisms in soil food webs. Probably the most common invertebrate groups in the soil are arthropods and nematodes, but other phyla are represented as well. Soil invertebrates may be predators, herbivores, or detritivores, or may feed specifically on fungi and bacteria that are abundant in the soil. Of course, soil invertebrates live where they do not only for the food, but also for the shelter—they require the shade, moisture, and stable temperatures that soil and leaf litter provide. The diversity of soil invertebrates can affect the properties of the very soil in which they reside, due to the fact that they influence key soil processes such as nutrient cycling, nutrient retention, the formation of soil structure, and decomposition rates. To sample soil invertebrates, you first need to obtain a small quantity of soil or leaf litter. The tricky part is then extracting all of the small invertebrates from the soil. To get them to come out of their otherwise optimal habitat, you need to make the soil unpleasantly hot and dry. An apparatus called a Berlese funnel will help with this job. The Berlese funnel does not extract all of the organisms from the soil. Fungi and bacteria are not motile enough to escape, and some invertebrates are also connected to the soil particles. However, this technique does give a good estimate of the true soil invertebrate diversity. EXERCISES Activity 1 (Week 1) – Sampling a Soil Habitat for Invertebrates

  1. You will work with your lab partner to obtain a soil and leaf litter sample from one of three local campus habitats. Your lab instructor will lead you to all three locations so that everyone is able to observe the conditions encountered and methods employed at each.
  2. The three locations sampled will be habitats that differ in water availability. The first site

2

will be a manicured grove that receives an optimal amount of water on a regular basis. The second site will be a natural woodland that on occasion has a small stream trickling through, so there is some (but not an abundant amount of) water present much of the time. Therefore, even when water is not present, the soil remains somewhat damp. The third and final site will be a very dry grassland / shrubland location that gets minimal water throughout the year. Note that you may not sample from these three locations in this particular order.

  1. All lab sections will sample from these same three habitats—data will be combined for

all sections. As each lab section has twelve groups of two, four groups from each section will sample from one particular location. Specific sampling spots within each habitat will be chosen randomly by each group. Following the directions given by your instructor, obtain a leaf litter sample from your sampling spot. Fill about three-fourths of a plastic bag with soil and leaf litter scooped from the ground surface.

  1. Spend a few minutes at each location examining the habitat that was just sampled. Do

you see any invertebrates crawling on the surface of the soil or in the leaf litter? How wet or dry does the site appear? Begin thinking about which habitat may contain the greatest invertebrate diversity.

Activity 2 (Week 1) – Constructing and Using a Berlese Funnel A Berlese funnel is a simple device used for extracting small invertebrate animals from soil and leaf litter samples. The Berlese funnel creates a very hot and dry environment at one end of the funnel, driving the invertebrates to the other end. Here they will fall out into a container of alcohol that is set up to collect them. Once you have extracted the invertebrates from your sample, you can observe them using a dissecting microscope and, with the aid of a dichotomous key, identify what kinds of invertebrates are present in the sample. This process, along with calculating the Shannon index, will allow you to determine your soil invertebrate diversity. First, though, the Berlese funnel must be set up—follow the procedure below to construct your own funnel.

  1. Cut an empty and clean 2-liter bottle 3 cm below the top of the label using a sharp scalpel. Then cut off the neck by cutting along the middle of the curvature. The bottom of the bottle will serve as the jar. The top of the bottle with the neck removed will serve as the funnel. The neck of the bottle can be discarded.
  2. Fill approximately one-fourth of the jar with 70% isopropyl alcohol.
  3. On a strip of tape placed on the jar, write a label for your sample. Include your names,

lab section, and the habitat from which you sampled.

  1. Invert the funnel end of the bottle into the jar (see Figure 1).
  2. Place an approximately 30 cm × 30 cm piece of square screening into the funnel so that it lines the entire inside. The screening has mesh large enough for invertebrate animals to

3

fall through but will keep much of the soil and leaves from falling into the alcohol. Use masking tape to secure the square screening onto the outside of the funnel by taping down all four corners.

  1. Transfer your leaf litter sample from your plastic bag into the lined funnel.
  2. Place the entire funnel apparatus underneath a heat lamp. This heat source will drive the

invertebrates to the lower end of the soil sample, causing them to fall through the screening and the mouth of the funnel into the alcohol, where they will be collected as your sample. Adjust the lamp so that the bulb is about three to four inches away from the top of your leaf litter.

  1. Let the Berlese funnel sit under the heat lamp for one week. Most of the invertebrates

should be extracted in that amount of time.

Figure 1: Diagram of a Berlese Funnel Activity 3 (Week 2) – Identifying the Invertebrates in Your Soil Sample

  1. Remove the jar containing the extracted soil invertebrates from your Berlese funnel.
  2. Carefully pour a small amount of alcohol from the jar into a Petri dish. Only pour enough to fill half of the Petri dish.
  3. With your lab partner, observe the Petri dish under a dissecting microscope. Be sure to

adjust the light source, magnification, and focus for optimal viewing.

  1. As you find the invertebrates in your sample, make a sketch or detailed notes of each one so that you are very familiar with the organisms you are observing. Use a small paintbrush to move around or pick up the small animals in the Petri dish. Using forceps

4

is not likely to be effective and may actually crush some of the individuals in your sample.

  1. Use the dichotomous key provided by your lab instructor to identify the invertebrates in

your sample. As most of your sample will consist of arthropods, the key is specifically for this phylum. Key out the arthropods to the class level. As the class Insecta will be the most abundant, the key enables you to identify insects to the order level. All other phyla can simply be recorded as such. Record the name of each phylum, class, or order that you identify in Table 1. Keep a tally of the number of individuals in each taxonomic group that you encounter in your sample. You will probably want to remove the organisms from your Petri dish once you have recorded them so that you can keep an accurate tally, but do not discard them completely in case you need to recheck any of your work.

  1. Repeat steps 2 through 5 until all of the alcohol is drained from your jar. If necessary,

you may have to add additional alcohol to “rinse” out any invertebrates or small particles that stick to the walls of the jar.

  1. Once completed, count your tallies and record the total number of individuals from each

taxon in Table 1. Sum these counts to determine the total number of individuals in your entire sample.

Activity 4 (Week 2) – Calculating the Soil Invertebrate Diversity You will use the Shannon index (H′) to calculate the diversity of your soil / leaf litter sample. Recall the equation for this diversity index: abundance of a given taxon (i) where pi = H′ = -ΣS (pi ln pi) total abundance of all taxa and S = the # of different taxa Complete Table 1 by calculating pi, ln pi, and pi ln pi for each invertebrate taxon. Determine the total diversity by summing all of the pi ln pi values and reversing the negative sign. This will yield the diversity for your sample. Your lab instructor will indicate the proper procedures for cleaning up / discarding / putting away funnels, mesh, soil, leaf litter, alcohol, invertebrates, and lamps. Have your work checked by your lab instructor prior to disassembling your Berlese funnel and cleaning up your lab bench. Be sure to give your diversity value, as well as the habitat from which the value came, to your instructor. All of the diversity values from all lab sections will be combined to produce one large set of data.

5

Table 1: Shannon Diversity Index Calculations Lab Partner Names Habitat Sampled

Invertebrate Taxon (Phylum, Class, or

Order)

of Individuals of

this Taxon pi ln pi pi ln pi

Total # of individuals

(add the second column)

H′ = – ΣS (pi ln pi) (add the last column and reverse the – sign)

pi = # of individuals of a particular taxon / total # of individuals in the entire sample

6

Activity 5 (Week 3) – Analyzing Differences in Diversity Among the Habitats A statistical technique known as ANOVA will be used to determine if there is a difference in the diversity of invertebrates among the three habitats (which do differ in the amount of water that they receive). Let us explore the topic of statistics further using a simple example so that you feel comfortable with the analysis that you will be performing. Suppose you want to know if there is a difference between the mean (average) height of all of the male students on campus and the mean (average) height of all of the female students on campus. Here, the heights of all of the male students on campus and the heights of all of the female students on campus are termed populations, and the means of these populations are termed parameters. A parameter is a numerical summary of a population. You can probably imagine that it might be challenging to obtain the height of every single male student and every single female student on campus. Because populations are often very large (as in this example), population parameters will likely not be able to be determined. To resolve this, a sample (random subset) needs to be taken from each of the populations. You could sample 30 male students and measure their heights and then sample 30 female students and measure their heights. Here, the heights of the random subsets from the populations are termed samples, and the means of these samples are termed statistics. A statistic is a numerical summary of a sample. Sample statistics are therefore estimates of population parameters. Statistics will rarely if ever equal the parameters exactly. The difference between the statistic and the parameter (which a researcher would not know, as parameters are not known) is termed sampling error. Not only will statistics estimate the parameters of interest, they will allow for an objective method of dealing with this sampling error. Statistics are therefore very important tools, as they enable scientists to make conclusions about entire populations, though they are working with just samples. Statistics will reveal how reliable these conclusions are. How exactly are these conclusions determined? All hypotheses formulated in biology are set up as null and alternative hypotheses. A null hypothesis is a hypothesis of no difference or no relationship. In the student height example, the null hypothesis would be that there is no difference between the mean male height and the mean female height of all students on campus. An alternative hypothesis must accompany each null hypothesis, and together these two hypotheses must account for all possibilities. In this example, the alternative hypothesis would be that there is a difference in the mean heights of male and female students on campus. As you can see, there are no other possibilities—there either is a difference or there is not a difference between male and female heights. Based on the type of data, the number of samples collected, and the question of interest, a particular statistical test will be performed. Remember, for our soil invertebrate study you will be utilizing a test known as ANOVA. For the student height example, a different statistical test (known as a t-test) would be employed. Understanding when to use which type of statistical test need not concern you here—you will deal with that when you take a biostatistics class. What is important now is to understand the results that the tests give you.

7

Each statistical test will yield a result known as a p-value. A p-value is a probability value and therefore always ranges between 0 and 1. Again, the calculation of the p-value is beyond the scope of this course (we will let Microsoft Excel calculate it for us). However, you do need to be able to interpret it. A p-value is the probability (percent chance) that a sample could have been taken from a population in which the null hypothesis is true. Remember what the null hypothesis is—the hypothesis of no difference. If the p-value in the height example turned out to be 0.85, then there would be an 85% chance that the samples were taken from populations in which there is no difference in the mean heights of male and female students. This is a very high probability, so we would accept the null hypothesis and conclude that there is no difference between the mean height of males and the mean height of females on campus. Suppose the p-value was instead 0.01. This would mean that there is only a 1% chance that the samples came from populations in which the null hypothesis (no difference in the mean heights) is true. This is a very small probability, so we would therefore reject the null hypothesis, concluding instead that there is in fact a difference between the mean height of males and the mean height of females on campus (the alternative hypothesis). Where is the line drawn between accepting and rejecting the null hypothesis? It is standard practice in biology to only reject the null hypothesis if the p-value is less than or equal to 0.05. The reason for 0.05 will become apparent in a biostatistics course. For now, rest assured that there is a solid mathematical basis for this value. Once it is known that there is a difference between the populations (if in fact there is one), how can it be determined what that difference is? Go back and look at the mean values from the samples. If the males have a higher mean value, then it can be concluded that males on campus are significantly taller than females. If the females have a higher mean value, then it can be concluded that the females on campus are significantly taller. Again, remember that this is the goal of statistics—making overall conclusions about populations using data obtained only from samples. It is a very important tool, as we need to make conclusions about entire populations, but they are just too large to work with. Practice your understanding of statistics using another example: Is there a difference in photosynthesis rates among plants grown in low, medium, and high light levels? The sample mean photosynthesis rate from plants in low light is 20 units, the sample mean from medium light is 25 units, the sample mean from high light is 30 units, and the p-value is 0.03. (Units of photosynthesis = μmol/m2/sec.) What is the null hypothesis?

What is the alternative hypothesis?

8

Is the null hypothesis accepted or rejected? What conclusion can you make based on these results?

Now for some more details as to how ANOVA in particular works. ANOVA is used when you are analyzing differences in means among three or more samples. As we have data from three different samples (the three different habitats), ANOVA is the appropriate statistical test for our study. ANOVA stands for analysis of variance, which indicates what the test does—it analyzes the variances of the samples. Variances are measures of variability. If there is much variability among all of the samples, then it should make sense that the samples are not the same—there is a difference in the diversity of invertebrates, for example, among three sites that differ in water availability. Take a look at the following example, which shows hypothetical values for invertebrate diversity at three locations (only five values are shown for each in this example, but remember that our data set will be much larger as we are combining data from all lab sections): High H2O Low H2O No H2O 7 3 2 7 3 2 7 3 2 7 3 2 7 3 2 Mean: 7 3 2 In the above example, there is variation among the three habitats. The habitat with much water has the highest mean invertebrate diversity, the habitat with no water has the lowest mean invertebrate diversity, and the habitat with intermediate water availability have intermediate invertebrate diversity. In this case, the null hypothesis of no difference among the habitats would be rejected and we would instead conclude that there is a difference in diversity. Now consider another example, which also shows variation, but variation of a different kind: High H2O Low H2O No H2O 7 4 3 6 3 2 4 2 7 3 7 6 2 6 4 Mean: 4.4 4.4 4.4 In the above example, there is variation within each sample rather than among the samples. Notice that within each habitat, various different diversity values are obtained, but the mean

9

diversity value among all three habitats is identical. In this case, the null hypothesis of no difference among the samples would be accepted—there is no difference in invertebrate diversity among habitats with differing water availabilities. What causes the variation within each sample in this example? Perhaps it is different availabilities of other nutrients. Perhaps it is differing temperatures. However, as these factors are not being tested for in our study, we cannot say with certainty what is causing the variation. What can be said with certainty is that what we are testing for, a difference in water availability, has no effect on invertebrate diversity. In the first example, there was variation among the samples but no variation within each sample, and we rejected the null hypothesis of no difference. In the second example, there was variation within each sample but no variation among the samples, and we accepted the null hypothesis of no difference. A much more likely scenario is that there will be some variation among the samples and some variation within each sample: High H2O Low H2O No H2O 6 4 3 7 2 3 6 5 2 8 3 2 7 4 2 Mean: 6.8 3.6 2.4 You should convince yourself that, in order to reject the null hypothesis and conclude that there is a difference among the three habitats, there must be greater variation among the samples rather than within the samples. How is it determined which is greater? This is where the calculation of a p-value by Microsoft Excel comes into play. Enter the raw data into an Excel spreadsheet and then select Tools → Data Analysis → Anova: Single Factor. When you are asked to input the range, highlight your data. Click OK and find the p-value in the output table. Practice using the data analysis feature in Excel by entering the above example (the one with some variation among the samples and some variation within the samples). You should find a p- value of 1.18×10-5 (0.0000118) in the output table. Will you accept or reject the null hypothesis of no difference in this example? Why?

Is there more variation among the habitats or within each habitat in this example?

What conclusion can you make about invertebrate diversity in habitats with different water

availabilities? Note that the means for each sample are given in the output table as well.

10

You are now ready to input the actual invertebrate diversity data from the three different habitats sampled by the entire class. Don’t be overwhelmed by all of this background information—after all, Excel is doing the work for you. You just need to have a big picture understanding of the results that Excel gives you. Ask your instructor about your results to make sure that you are interpreting them correctly. Activity 6 (Week 3) – Searching for Sources as Background Information You will be writing your lab report in the form of a formal scientific paper. Your lab instructor will discuss the required components of a scientific paper. One requirement is to find other papers that have performed studies similar to yours, citing them in your own paper as background information or to compare results. Cited sources must be peer-reviewed scientific journal articles. Textbooks and websites are not permitted. While in lab, you will start to look up journal articles and have your lab instructor confirm that they are indeed relevant sources for your paper. Two good locations to begin your search for sources are:

  1. The Cal Poly Pomona website. From the homepage, search for the University Library, then begin searching for journal articles by typing in some keywords (for example, invertebrate diversity and water availability). The search results will indicate which sources are peer-reviewed journal articles.
  2. Google Scholar. Go to https://scholar.google.com/, type in some keywords, and examine

the peer-reviewed articles that result to determine which may be appropriate for inclusion in your paper.

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What is the alternative hypothesis?

What is the alternative hypothesis?

1

Formal Lab

Foundations of Biology: Evolution, Ecology, and Biodiversity Lab

Soil Invertebrate Diversity

LEARNING OUTCOMES During this three-day laboratory exercise, you will:

  1. Sample the invertebrates in soil and leaf litter using field and laboratory techniques. 2. Use a dichotomous key to identify the soil invertebrates. 3. Calculate and compare the invertebrate diversity within soil samples taken from habitats

with different water availabilities. 4. Statistically analyze the collective data from all lab sections to serve as the basis for a

formal lab report. BACKGROUND Invertebrates are one of the dominant groups of organisms in soil food webs. Probably the most common invertebrate groups in the soil are arthropods and nematodes, but other phyla are represented as well. Soil invertebrates may be predators, herbivores, or detritivores, or may feed specifically on fungi and bacteria that are abundant in the soil. Of course, soil invertebrates live where they do not only for the food, but also for the shelter—they require the shade, moisture, and stable temperatures that soil and leaf litter provide. The diversity of soil invertebrates can affect the properties of the very soil in which they reside, due to the fact that they influence key soil processes such as nutrient cycling, nutrient retention, the formation of soil structure, and decomposition rates. To sample soil invertebrates, you first need to obtain a small quantity of soil or leaf litter. The tricky part is then extracting all of the small invertebrates from the soil. To get them to come out of their otherwise optimal habitat, you need to make the soil unpleasantly hot and dry. An apparatus called a Berlese funnel will help with this job. The Berlese funnel does not extract all of the organisms from the soil. Fungi and bacteria are not motile enough to escape, and some invertebrates are also connected to the soil particles. However, this technique does give a good estimate of the true soil invertebrate diversity. EXERCISES Activity 1 (Week 1) – Sampling a Soil Habitat for Invertebrates

  1. You will work with your lab partner to obtain a soil and leaf litter sample from one of three local campus habitats. Your lab instructor will lead you to all three locations so that everyone is able to observe the conditions encountered and methods employed at each.
  2. The three locations sampled will be habitats that differ in water availability. The first site

2

will be a manicured grove that receives an optimal amount of water on a regular basis. The second site will be a natural woodland that on occasion has a small stream trickling through, so there is some (but not an abundant amount of) water present much of the time. Therefore, even when water is not present, the soil remains somewhat damp. The third and final site will be a very dry grassland / shrubland location that gets minimal water throughout the year. Note that you may not sample from these three locations in this particular order.

  1. All lab sections will sample from these same three habitats—data will be combined for

all sections. As each lab section has twelve groups of two, four groups from each section will sample from one particular location. Specific sampling spots within each habitat will be chosen randomly by each group. Following the directions given by your instructor, obtain a leaf litter sample from your sampling spot. Fill about three-fourths of a plastic bag with soil and leaf litter scooped from the ground surface.

  1. Spend a few minutes at each location examining the habitat that was just sampled. Do

you see any invertebrates crawling on the surface of the soil or in the leaf litter? How wet or dry does the site appear? Begin thinking about which habitat may contain the greatest invertebrate diversity.

Activity 2 (Week 1) – Constructing and Using a Berlese Funnel A Berlese funnel is a simple device used for extracting small invertebrate animals from soil and leaf litter samples. The Berlese funnel creates a very hot and dry environment at one end of the funnel, driving the invertebrates to the other end. Here they will fall out into a container of alcohol that is set up to collect them. Once you have extracted the invertebrates from your sample, you can observe them using a dissecting microscope and, with the aid of a dichotomous key, identify what kinds of invertebrates are present in the sample. This process, along with calculating the Shannon index, will allow you to determine your soil invertebrate diversity. First, though, the Berlese funnel must be set up—follow the procedure below to construct your own funnel.

  1. Cut an empty and clean 2-liter bottle 3 cm below the top of the label using a sharp scalpel. Then cut off the neck by cutting along the middle of the curvature. The bottom of the bottle will serve as the jar. The top of the bottle with the neck removed will serve as the funnel. The neck of the bottle can be discarded.
  2. Fill approximately one-fourth of the jar with 70% isopropyl alcohol.
  3. On a strip of tape placed on the jar, write a label for your sample. Include your names,

lab section, and the habitat from which you sampled.

  1. Invert the funnel end of the bottle into the jar (see Figure 1).
  2. Place an approximately 30 cm × 30 cm piece of square screening into the funnel so that it lines the entire inside. The screening has mesh large enough for invertebrate animals to

3

fall through but will keep much of the soil and leaves from falling into the alcohol. Use masking tape to secure the square screening onto the outside of the funnel by taping down all four corners.

  1. Transfer your leaf litter sample from your plastic bag into the lined funnel.
  2. Place the entire funnel apparatus underneath a heat lamp. This heat source will drive the

invertebrates to the lower end of the soil sample, causing them to fall through the screening and the mouth of the funnel into the alcohol, where they will be collected as your sample. Adjust the lamp so that the bulb is about three to four inches away from the top of your leaf litter.

  1. Let the Berlese funnel sit under the heat lamp for one week. Most of the invertebrates

should be extracted in that amount of time.

Figure 1: Diagram of a Berlese Funnel Activity 3 (Week 2) – Identifying the Invertebrates in Your Soil Sample

  1. Remove the jar containing the extracted soil invertebrates from your Berlese funnel.
  2. Carefully pour a small amount of alcohol from the jar into a Petri dish. Only pour enough to fill half of the Petri dish.
  3. With your lab partner, observe the Petri dish under a dissecting microscope. Be sure to

adjust the light source, magnification, and focus for optimal viewing.

  1. As you find the invertebrates in your sample, make a sketch or detailed notes of each one so that you are very familiar with the organisms you are observing. Use a small paintbrush to move around or pick up the small animals in the Petri dish. Using forceps

4

is not likely to be effective and may actually crush some of the individuals in your sample.

  1. Use the dichotomous key provided by your lab instructor to identify the invertebrates in

your sample. As most of your sample will consist of arthropods, the key is specifically for this phylum. Key out the arthropods to the class level. As the class Insecta will be the most abundant, the key enables you to identify insects to the order level. All other phyla can simply be recorded as such. Record the name of each phylum, class, or order that you identify in Table 1. Keep a tally of the number of individuals in each taxonomic group that you encounter in your sample. You will probably want to remove the organisms from your Petri dish once you have recorded them so that you can keep an accurate tally, but do not discard them completely in case you need to recheck any of your work.

  1. Repeat steps 2 through 5 until all of the alcohol is drained from your jar. If necessary,

you may have to add additional alcohol to “rinse” out any invertebrates or small particles that stick to the walls of the jar.

  1. Once completed, count your tallies and record the total number of individuals from each

taxon in Table 1. Sum these counts to determine the total number of individuals in your entire sample.

Activity 4 (Week 2) – Calculating the Soil Invertebrate Diversity You will use the Shannon index (H′) to calculate the diversity of your soil / leaf litter sample. Recall the equation for this diversity index: abundance of a given taxon (i) where pi = H′ = -ΣS (pi ln pi) total abundance of all taxa and S = the # of different taxa Complete Table 1 by calculating pi, ln pi, and pi ln pi for each invertebrate taxon. Determine the total diversity by summing all of the pi ln pi values and reversing the negative sign. This will yield the diversity for your sample. Your lab instructor will indicate the proper procedures for cleaning up / discarding / putting away funnels, mesh, soil, leaf litter, alcohol, invertebrates, and lamps. Have your work checked by your lab instructor prior to disassembling your Berlese funnel and cleaning up your lab bench. Be sure to give your diversity value, as well as the habitat from which the value came, to your instructor. All of the diversity values from all lab sections will be combined to produce one large set of data.

5

Table 1: Shannon Diversity Index Calculations Lab Partner Names Habitat Sampled

Invertebrate Taxon (Phylum, Class, or

Order)

of Individuals of

this Taxon pi ln pi pi ln pi

Total # of individuals

(add the second column)

H′ = – ΣS (pi ln pi) (add the last column and reverse the – sign)

pi = # of individuals of a particular taxon / total # of individuals in the entire sample

6

Activity 5 (Week 3) – Analyzing Differences in Diversity Among the Habitats A statistical technique known as ANOVA will be used to determine if there is a difference in the diversity of invertebrates among the three habitats (which do differ in the amount of water that they receive). Let us explore the topic of statistics further using a simple example so that you feel comfortable with the analysis that you will be performing. Suppose you want to know if there is a difference between the mean (average) height of all of the male students on campus and the mean (average) height of all of the female students on campus. Here, the heights of all of the male students on campus and the heights of all of the female students on campus are termed populations, and the means of these populations are termed parameters. A parameter is a numerical summary of a population. You can probably imagine that it might be challenging to obtain the height of every single male student and every single female student on campus. Because populations are often very large (as in this example), population parameters will likely not be able to be determined. To resolve this, a sample (random subset) needs to be taken from each of the populations. You could sample 30 male students and measure their heights and then sample 30 female students and measure their heights. Here, the heights of the random subsets from the populations are termed samples, and the means of these samples are termed statistics. A statistic is a numerical summary of a sample. Sample statistics are therefore estimates of population parameters. Statistics will rarely if ever equal the parameters exactly. The difference between the statistic and the parameter (which a researcher would not know, as parameters are not known) is termed sampling error. Not only will statistics estimate the parameters of interest, they will allow for an objective method of dealing with this sampling error. Statistics are therefore very important tools, as they enable scientists to make conclusions about entire populations, though they are working with just samples. Statistics will reveal how reliable these conclusions are. How exactly are these conclusions determined? All hypotheses formulated in biology are set up as null and alternative hypotheses. A null hypothesis is a hypothesis of no difference or no relationship. In the student height example, the null hypothesis would be that there is no difference between the mean male height and the mean female height of all students on campus. An alternative hypothesis must accompany each null hypothesis, and together these two hypotheses must account for all possibilities. In this example, the alternative hypothesis would be that there is a difference in the mean heights of male and female students on campus. As you can see, there are no other possibilities—there either is a difference or there is not a difference between male and female heights. Based on the type of data, the number of samples collected, and the question of interest, a particular statistical test will be performed. Remember, for our soil invertebrate study you will be utilizing a test known as ANOVA. For the student height example, a different statistical test (known as a t-test) would be employed. Understanding when to use which type of statistical test need not concern you here—you will deal with that when you take a biostatistics class. What is important now is to understand the results that the tests give you.

7

Each statistical test will yield a result known as a p-value. A p-value is a probability value and therefore always ranges between 0 and 1. Again, the calculation of the p-value is beyond the scope of this course (we will let Microsoft Excel calculate it for us). However, you do need to be able to interpret it. A p-value is the probability (percent chance) that a sample could have been taken from a population in which the null hypothesis is true. Remember what the null hypothesis is—the hypothesis of no difference. If the p-value in the height example turned out to be 0.85, then there would be an 85% chance that the samples were taken from populations in which there is no difference in the mean heights of male and female students. This is a very high probability, so we would accept the null hypothesis and conclude that there is no difference between the mean height of males and the mean height of females on campus. Suppose the p-value was instead 0.01. This would mean that there is only a 1% chance that the samples came from populations in which the null hypothesis (no difference in the mean heights) is true. This is a very small probability, so we would therefore reject the null hypothesis, concluding instead that there is in fact a difference between the mean height of males and the mean height of females on campus (the alternative hypothesis). Where is the line drawn between accepting and rejecting the null hypothesis? It is standard practice in biology to only reject the null hypothesis if the p-value is less than or equal to 0.05. The reason for 0.05 will become apparent in a biostatistics course. For now, rest assured that there is a solid mathematical basis for this value. Once it is known that there is a difference between the populations (if in fact there is one), how can it be determined what that difference is? Go back and look at the mean values from the samples. If the males have a higher mean value, then it can be concluded that males on campus are significantly taller than females. If the females have a higher mean value, then it can be concluded that the females on campus are significantly taller. Again, remember that this is the goal of statistics—making overall conclusions about populations using data obtained only from samples. It is a very important tool, as we need to make conclusions about entire populations, but they are just too large to work with. Practice your understanding of statistics using another example: Is there a difference in photosynthesis rates among plants grown in low, medium, and high light levels? The sample mean photosynthesis rate from plants in low light is 20 units, the sample mean from medium light is 25 units, the sample mean from high light is 30 units, and the p-value is 0.03. (Units of photosynthesis = μmol/m2/sec.) What is the null hypothesis?

What is the alternative hypothesis?

8

Is the null hypothesis accepted or rejected? What conclusion can you make based on these results?

Now for some more details as to how ANOVA in particular works. ANOVA is used when you are analyzing differences in means among three or more samples. As we have data from three different samples (the three different habitats), ANOVA is the appropriate statistical test for our study. ANOVA stands for analysis of variance, which indicates what the test does—it analyzes the variances of the samples. Variances are measures of variability. If there is much variability among all of the samples, then it should make sense that the samples are not the same—there is a difference in the diversity of invertebrates, for example, among three sites that differ in water availability. Take a look at the following example, which shows hypothetical values for invertebrate diversity at three locations (only five values are shown for each in this example, but remember that our data set will be much larger as we are combining data from all lab sections): High H2O Low H2O No H2O 7 3 2 7 3 2 7 3 2 7 3 2 7 3 2 Mean: 7 3 2 In the above example, there is variation among the three habitats. The habitat with much water has the highest mean invertebrate diversity, the habitat with no water has the lowest mean invertebrate diversity, and the habitat with intermediate water availability have intermediate invertebrate diversity. In this case, the null hypothesis of no difference among the habitats would be rejected and we would instead conclude that there is a difference in diversity. Now consider another example, which also shows variation, but variation of a different kind: High H2O Low H2O No H2O 7 4 3 6 3 2 4 2 7 3 7 6 2 6 4 Mean: 4.4 4.4 4.4 In the above example, there is variation within each sample rather than among the samples. Notice that within each habitat, various different diversity values are obtained, but the mean

9

diversity value among all three habitats is identical. In this case, the null hypothesis of no difference among the samples would be accepted—there is no difference in invertebrate diversity among habitats with differing water availabilities. What causes the variation within each sample in this example? Perhaps it is different availabilities of other nutrients. Perhaps it is differing temperatures. However, as these factors are not being tested for in our study, we cannot say with certainty what is causing the variation. What can be said with certainty is that what we are testing for, a difference in water availability, has no effect on invertebrate diversity. In the first example, there was variation among the samples but no variation within each sample, and we rejected the null hypothesis of no difference. In the second example, there was variation within each sample but no variation among the samples, and we accepted the null hypothesis of no difference. A much more likely scenario is that there will be some variation among the samples and some variation within each sample: High H2O Low H2O No H2O 6 4 3 7 2 3 6 5 2 8 3 2 7 4 2 Mean: 6.8 3.6 2.4 You should convince yourself that, in order to reject the null hypothesis and conclude that there is a difference among the three habitats, there must be greater variation among the samples rather than within the samples. How is it determined which is greater? This is where the calculation of a p-value by Microsoft Excel comes into play. Enter the raw data into an Excel spreadsheet and then select Tools → Data Analysis → Anova: Single Factor. When you are asked to input the range, highlight your data. Click OK and find the p-value in the output table. Practice using the data analysis feature in Excel by entering the above example (the one with some variation among the samples and some variation within the samples). You should find a p- value of 1.18×10-5 (0.0000118) in the output table. Will you accept or reject the null hypothesis of no difference in this example? Why?

Is there more variation among the habitats or within each habitat in this example?

What conclusion can you make about invertebrate diversity in habitats with different water

availabilities? Note that the means for each sample are given in the output table as well.

10

You are now ready to input the actual invertebrate diversity data from the three different habitats sampled by the entire class. Don’t be overwhelmed by all of this background information—after all, Excel is doing the work for you. You just need to have a big picture understanding of the results that Excel gives you. Ask your instructor about your results to make sure that you are interpreting them correctly. Activity 6 (Week 3) – Searching for Sources as Background Information You will be writing your lab report in the form of a formal scientific paper. Your lab instructor will discuss the required components of a scientific paper. One requirement is to find other papers that have performed studies similar to yours, citing them in your own paper as background information or to compare results. Cited sources must be peer-reviewed scientific journal articles. Textbooks and websites are not permitted. While in lab, you will start to look up journal articles and have your lab instructor confirm that they are indeed relevant sources for your paper. Two good locations to begin your search for sources are:

  1. The Cal Poly Pomona website. From the homepage, search for the University Library, then begin searching for journal articles by typing in some keywords (for example, invertebrate diversity and water availability). The search results will indicate which sources are peer-reviewed journal articles.
  2. Google Scholar. Go to https://scholar.google.com/, type in some keywords, and examine

the peer-reviewed articles that result to determine which may be appropriate for inclusion in your paper.

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Biology Lab Report/3 Pages/24 Hrs

Biology Lab Report/3 Pages/24 Hrs

The title is reasonably lengthy in order to be descriptive and

is centered across the top of the first page

Name California State Polytechnic University, Pomona

This is where you will place your abstract. The abstract is a complete, yet succinct, synopsis of the paper. This is basically like a Cliff’s Notes version of your entire paper where you summarize your entire report in one paragraph. It should be italicized and in 10 font. This will most likely be the part you write last so that you have a complete understanding of your entire paper. Journal articles have many different formats for the abstract, but this is what I will expect from you (horizontal lines optional). In the papers you cite, note how they go about summarizing their paper, as this will be helpful in guiding you towards writing a good abstract.

Introduction

In this part of the paper you need to include at least one source (the entire

paper needs three, so if you only have one here you will need two in the discussion

portion or vice versa). The rest of the report will continue in twelve font. Please use

a standard font like Times New Roman. Your sources should be properly cited. Say

for example you want to include a journal article written by Joe Schmoe in 1997. The

end of the sentence containing his information would look exactly like this (Schmoe,

1997). Notice the period is after the close parenthesis. If there are two authors it

looks like this (Schmoe and Toe, 1997). If there are more than two you do this

(Schmoe et al., 1997). Et al. means “and his whole posse”. Or you could include a

citation like this: Schmoe, in his 1997 study of riparian forests, found that more

arthropods than non-arthropods exist in the leaf litter of such a community. Either

way is acceptable. Please do not quote directly from the source. You must

paraphrase (put in your own words). Be sure to examine the literature cited section at

the end of this guide for proper format there.

Now that we have finished the technical stuff for the intro, let’s talk about the

information you need to include in this portion. We have already discussed citations

• Title bold and 16 font • Name and affiliation normal, 12 font • Only capitalize first word of title • NO COVER SHEET!!!! • No period after title

but let me stress that they need to be relevant. You should probably not be using a sea

star article as a reference for this paper. What you should write about is what other

people have found in their research (this is the background stuff, first paragraph).

Then you discuss what you specifically are looking into. You could explain that you

are interested in seeing if there is a difference between two sites, two seasons, etc.

You do not state your hypotheses like this: H0: There is no difference between site

one and site two. HA: There is a significant difference between site one and site two.

This is a no-no. Your words should clearly indicate what your HA is and from there

the reader can infer the H0. Let’s pretend you were comparing a riparian site to the

parking lot. A good way to get your alternative hypothesis across is like this: In this

study, arthropod samples from a riparian community were compared with those

collected from a parking lot to see if the communities exhibited similar or different

levels of diversity and evenness. From this the reader can infer what your hypotheses

are. The introduction should be written in the present tense since this is the most

current information available.

Methods

This portion of your paper is where you lay out exactly all the details of your

project. Since you have already completed this portion, it should be written in the

past tense. You need enough details to allow for reproducibility. This means that

anyone who reads your paper should be able to conduct the exact same experiment to

see if they get the same results. For materials, there should not be a list of items used.

You will incorporate them into your explanation (real words and sentences) of how

you conducted your experiment. You can see how this is done when you read your

journal articles that you have selected to cite. Remember to include exact

measurements of amounts (when possible), time of day, or any other factors that

could influence the outcome. Normally this portion is long and tedious, but our

Figure 1. Brief description of graph

experiment was rather simple so it should not be too bad. You also need to state what

statistical test was applied.

Results

This portion is where you report your findings. You will talk about the

arthropods you collected, etc. I am requiring that you also have graphs or tables to

illustrate your results and you must refer to them in your writing!!! For example,

diversity is greatest in the intermediate group (Figure 1). Make sure you also include

a legend, meaning that at the bottom (for figures) or at the top (for tables) you write

(in 10 font):

YOU MUST DESCRIBE YOUR DATA; ONLY HAVING TABLES AND/OR

FIGURES IS UNACCEPTABLE. We will discuss in class what type of statistics you

will use, but most likely it will be the Shannon index and/or some other statistical test

that gives you a p-value. You do not put all of your math work in this part. You only

need to state some descriptive statistics (means, standard deviations) and the result

(for example, p = 0.0012 or H′ = 1.034) of the statistical test/index. I will talk about

what the p-value means and how you should interpret it for your discussion in class so

pay close attention! Interpreting what your data are telling you will be saved for the

discussion portion. You only describe what you found and report what your numbers

are here. This section should also be written in the past tense.

Discussion

This is the conclusion, where you tie it all together. NOW you can say stuff

about what you should infer from what the data have told you (by the way, data is

plural, datum singular). This is the absolute most important part of the whole paper,

and the part on which I will grade you the toughest. A general rule of thumb is that it

should be at least as long as your introduction. You should have citations here as

well. A good way to do this is to compare your findings to those of another article or

use the conclusions of another article to explain your results. It is important to be

clear whether or not you reject or accept your null hypothesis without actually saying,

“We reject/accept our null hypothesis” (just like the rules in the introduction).

You should also allude to further possible research that could be done to

expand on your project or ways in which your project could have been improved

upon. The most important part of your paper is inferring why you got the results you

did. If there was a significant difference, you should be able to discuss why you think

there is a difference between the sites. Your entire paper must be at least 5 pages,

double spaced.

Literature cited

Polis, G. A. 1981. The evolution and dynamics of intraspecific predation. Annual Review of Ecology and Systematics 12:225-251.

Shaffer, H. B. and M. L. McKnight. 1996. The polytypic species revisited: genetic

differentiation and molecular phylogenetics of the Tiger Salamander Ambystoma tigrinum (Amphibia: Caudata) complex. Evolution 50:417-433.

Pfennig, D. W., J. P. Collins, and R. E. Ziemba. 1999. A test of alternative

hypotheses for kin recognition in cannibalistic Tiger Salamanders. Behavioral Ecology 10:436-443.

One author:

Two authors: Notice how 2nd author’s name is arranged! 3 + authors:

This section should continue right under your discussion. Do not start a separate page for this portion. There are different acceptable formats, but for our purposes we will use the following format (alphabetical by author’s last name). Notice that each citation is single-spaced.

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Attention And Conscious Thought

Attention And Conscious Thought

Select at least one of the research articles listed in this week’s Learning Resources or referenced by the University of Utah, Department of Psychology/American Psychological Association.

Imagine you were a legislator or inventor in the future. You are familiar with the research on attention and distracted driving, but people still need to be able to communicate with others while they travel. What law would you propose or technology would you invent to enable drivers to communicate with those outside the car while reducing the dangers of distracted driving? (No, you cannot just invent a driverless car or hands free cell phone.) What research would inform your proposed law or invention?

Post a 3- to 4-paragraph response to the following:

For your chosen article, summarize how the researchers conducted the study and what they found.

Based on the research article you summarized and on your reading from the text, critique the law that drivers can only use hands-free cell phones while driving.

For the given scenario, what law would you propose or technology would you invent to enable communication with people outside the car while reducing the dangers of distracted driving? What research would inform your proposed law?

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Laboratory Analysis of Signals and Systems (LASS) University of M’sila

Laboratory Analysis of Signals and Systems (LASS) University of M’sila

SAI Intelligent Systems Conference 2015 November 10-11, 2015 | London, UK

773 | P a g e 978-1-4673-7606-8/15/$31.00 ©2015 IEEE

Images Segmentation based on Interval Type-2 Fuzzy C-Means

ASSAS Ouarda* Department of Computer Science,

Laboratory Analysis of Signals and Systems (LASS) University of M’sila

M’sila, Algeria e-mail: assas_warda@yahoo.fr

Abstract—Segmentation process helps to find region of interest in a particular image. The main goal is to make image more simple and meaningful. This work is an improvement of an existing method which is Fuzzy C-Means (FCM) to partitioning an image into several constituent components – type 2 Fuzzy C- Means-. First, membership function defined by Hamid R Tizhoosh is used to measure the image fuzziness. Second, new membership functions are proposed. The evaluation of adopted approaches was compared using the validity functions: Partition Coefficient Vpc, Partition Entropy Vpe and Peak Signal and Noise Ratio PSNR. The experimental results on real images prove that the proposed approaches are more accurate and robust than the standard FCM approach.

Keywords—component; Segmentation; Fuzzy Logic; Type-2 Fuzzy Sets; Fuzzy C-Means

I. INTRODUCTION Image segmentation plays an important role in vision and

image processing applications. It is a widely employed technique in many fields like: document image analysis, scene or map processing, Signature Identification, Biomedical Imaging, and Target Identification [1]. The images segmentation has remained a challenge. Many approaches have been studied, including Methods based edge, methods based region, methods based on thresholding, methods based artificial neural networks, data fusion methods, Markov random field methods and hybrid Methods.

In fuzzy segmentation, the image pixel values can belong to more than one segment, and associated with each of the points are membership grades that indicate the degree to which the data points belong to different segments.

Segmentation process also helps to find region of interest in a particular image. The main goal is to make image more simple and meaningful. Fuzzy C-Means (FCM) is a unsupervised fuzzy classification algorithm. Resulting from the C-means algorithm (C-means), it introduces the concept of fuzzy set in the class definition: each point in the data set for each cluster with a certain degree, and all clusters are characterized by their center of gravity.

In this paper, the use of fuzzy logic is extended to a higher order, which called type-2 fuzzy logic. The fuzzy C-Means using type-2 fuzzy logic with new membership functions is proposed to segment human MR Brain images.

The organisation of the paper is as follows. In section 2 the fuzzy C-Means technique of segmentation is reviewed and in section 3 describes briefly the type-2 fuzzy sets. Section 4 present a complete description of proposed segmenting approach using tye-2fuzzy logic, where each step of the algorithm is developed in detail. Ssection 5 illustrates the obtained experimental results and discussions and section 6 concludes this paper.

II. FUZZY C-MEANS TECHNIQUE Modeling inaccuracy is done by considering gradual

boundaries instead of clear borders between classes. The uncertainty is expressed by the fact that a pixel has attributes that assign a class than another. So, Fuzzy clustering assigns not a pixel a label on a single class, but its degree of membership in each class. These values indicate the uncertainty of a pixel belonging to a region and are called membership degrees. The membership degree s in the interval [0, 1] and the obtained classes are not necessarily disjoint. In this case, the data Xj are not assigned to a single class, but many through degrees of membership Uij of the vector Xj to class i. The purpose of classification algorithms is not only calculating cluster centers bi but all degrees of membership vectors to classes. If Uij is the membership degree of Xj to class i, the matrix U(CxN, C number of cluster and N is the data size) is called fuzzy C-partitions matrix if and only if it satisfies the conditions (1) and (2):

[ ] [ ] [ ]∈

∈∀∈∀ =

N

j ij

ij

Nu

u NjCi

1

0

1,0 ,1,,1 (1)

[ ] =

=∈∀ C

i ijuCi

1

1 ,1 (2)

The objective function to minimize J and the solutions bi, Uij, of the problem of the FCM are described by the following formulas:

= =

= C

i

N

j ij

m ij bxdUXUBJ

1 1

2 ),()(),,( (3)

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=

== N

j

m ij

N

j j

m ij

u

Xu bi

1

1

)(

.)( (4)

1

)1( 2

1 2

2

),( ),(

=

= mC

k kj

ij ij bXd

bXd u (5)

With the variable m is the fuzzification coefficient which takes values in the interval [0, + [. The FCM algorithm stops when the partition becomes stable

Like other unsupervised classification algorithms, it uses a criterion minimization of intra-class distances and maximizing inter-class distances, but gives a degree of membership of each class for each pixel. This algorithm requires prior knowledge of the number of clusters and generates classes through an iterative process by minimizing an objective function. Thus, it allows to obtaining a fuzzy partition of the image by providing each pixel with a membership degree (between 0 and 1) to a given class. The cluster which is associated with a pixel is one whose degree of membership is the highest.

The main steps of the Fuzzy C-means algorithm are:

1) Input the image Xj: j=1..N, N: size of image. 2) Set the parameters of the algorithm: C: number of

cluster, m: fuzzy coefficient, : convergence error. 3) Initialize the membership matrix U with random values

in the range [0,1]. 4) Update the centers bi using the equation (4) and

evaluation of the objective function Jold using the formula (3). 5) Update the membership matrix U using the equation

(5) and evaluation of the objective function Jnew using the formula (3).

6) Repeat steps 4 and 5 until satisfaction of the stopping criterion which is written: || Jold-Jnew:||

7) The outputs are the membership matrix U and the centers bi.

III. TYPE-2 FUZZY SETS [2] A new area in fuzzy logic is introduced in this section,

which called type-2 fuzzy logic. Type-2 fuzzy set theory was introduced by Zadeh in [3] to solve the problem in defining the complex uncertainty which the problem is unable to define the existing type-1 fuzzy set theory. A type-2 fuzzy set is a set in which we also have uncertainty about the membership function. Type-2 fuzzy logic is a generalisation of conventional fuzzy logic (type-1) in the sense that uncertainty is not only limited to the linguistic variables but also is present in the definition of the membership functions.[4]

Speaking of uncertainty, there are two main types of uncertainty, linguistic and random. The first is associated with the word and the fact that it can mean different things to different people while the second is associated with unpredictability. Probability theory is used to treat the random uncertainty and fuzzy set is used to treat the linguistic

uncertainty. As the variance provides a measure of dispersion around the average of probabilistic uncertainty, a fuzzy set needs a dispersion measurement of linguistic uncertainty. A type-2 fuzzy set provides precisely this measure of dispersion.

A Type-2 fuzzy set is an extension of the type-1 fuzzy set. It has the degrees of membership which are themselves fuzzy. For each value of the primary value (pressure, temperature …etc.), membership is a function and not a value (secondary membership function), whose domain, is the primary membership in the interval [0, 1] and whose row (second degrees) must also be in the interval [0, 1]. The membership function of a type-2 fuzzy set is three-dimensional, and it is this new third dimension that provides new degrees of freedom in the design to treat uncertainty. These sets are very useful in situations where it is difficult to determine the exact membership function for a fuzzy set.

The term footprint of uncertainty (FOU) is used in the literature to verbalize the shape of type-2 fuzzy sets(shaded area in Fig.1)[5][6]. The FOU implies that there is a distribution that sits on top of that shaded area. When they all equal one, the resulting type-2 fuzzy sets are called interval type-2 fuzzy sets. For which, the membership function provides an interval [7].

(a) (b)

Fig. 1. (a)Type-1 membership function and (b) FOU for an interval type-2 fuzzy set [6]

Definition: A type-2 fuzzy set à is defined by a type-2 membership function μÃ(x, u), where x X and u Jx [0,1] [6].

Ã={((x, u), μÃ(x, u)) ∀ x X, ∀ u Jx [0,1]} (6)

in which 0 μÃ(x, u) 1. à can also be expressed in the usual notation of fuzzy sets as:

(7) where the double integral denotes the union over all x and

u. In order to define a type-2 fuzzy set, one can define a type-1 fuzzy set and assign upper and lower membership degrees to each element to (re)construct the footprint of uncertainty (Fig.1). A more practical definition for a type-2 fuzzy set can be given as follows:

Ã={(x, μU(x), μL(x)) ∀ x X, μL(x)) μ(x) μU(x), μ [0,1]} (8)

SAI Intelligent Systems Conference 2015 November 10-11, 2015 | London, UK

775 | P a g e 978-1-4673-7606-8/15/$31.00 ©2015 IEEE

Tizhoosh [7] has defined the upper and lower membership degrees μU(x) and μL(x) of initial (skeleton) membership function μ by means of linguistic hedges:

μU(x) = [μ(x)]1/ , (9)

μL(x) = μ(x)

Where ]1,+ [. In the conducted experiments, ]1, 2] has been used because >> is usually not meaningful for image data.

For =2, the upper and lower membership degrees represent dilatation and concentration:

μU(x) = [μ(x)]0.5, (11)

μL(x) = [μ(x)]2 (12)

Of course, other linguistic hedges such as de-accentuation and accentuation can also be employed:

μU(x) = [μ(x)]0.75, (13)

μL(x) = [μ(x)]1.25. (14)

IV. FUZZY C-MEANS USING TYPE-2 FUZZY SETS

In the proposed approaches, first, μL=μ and μU= μ 1/ are

taken. The membership functions can be calculated by three ways as follows:

μ(x) = (μL+ μU)/2 (15)

Or μ(x) = (μL * μU) 1/2 (16)

Or μ(x) = (μL + μL*μU) (17)

Second, new upper and lower membership degrees μU(x) and μL(x) of initial (skeleton) membership function μ are proposed by the flowing expressions:

μU(x) = μ(x)*( μ(x)+1), (18) μL(x) = μ(x)/( μ(x)+1). (19)

And

μU(x) = μ(x)2*( μ(x)2+1)/2, (20) μL(x) = μ(x)1/2/2( μ(x)1/2+1). (21)

The general algorithm for the type-2 fuzzy C-means approach can be formulated as follows:

1) Input the image Xj: j=1..N, which N is the image size. 2) Set the parameters of the algorithm: C: number of

cluster, m: fuzzy coefficient, : convergence error. 3) Initialize the membership matrix U with random values

in the range [0, 1]. 4) Update the centers bi using the equation (4) and

evaluation of the objective function Jold using the formula (3). 5) Update the membership matrix U using the equation

(5) and Compute the upper and lower membership using the equations couple (9) and (10) or (18) and (19) or (20) and (21).

6) Calculate the membership functions using one of the formula (15), (16) or (17) then Evaluate the objective function Jnew using the formula (3).

7) Repeat steps 4 and 5 until satisfaction of the stopping criterion which is written: || Jold-Jnew:||

8) The outputs are the membership matrix U and the centers bi.

9) Assign all pixels to clusters by using the maximum membership value of every pixel.

V. EXPERIMENTAL RESULTS The proposed algorithm for a fuzzy 2-partition thresholding

has been tested on many images with various histogram distributions to ensuring its efficiency. Each image is presented by eight bits, that is, grey levels are ranging from 0 (the darkest) to 255 (the brightest). The experimental results of the proposed method are presented and discussed through the dataset of standard 512×512 grayscale test images (Fig 2).

Validation functions of resulting classes of fuzzy partitioning are often used to evaluate the performance of different methods of classification. These functions are: partition coefficient Vpc and partition entropy Vpe. They are defined as follows:

(22)

(23)

The idea of the validity of these functions is that the partition with less fuzzy means better performance. Therefore, the best partition is reached when the Vpc value is maximum and Vpe is minimum.

Also, Peak signal to noise ratio (PSNR) is used to determine the quality of the segmented image. The PSNR give the similarity of an image against a reference image based on the mean square error (MSE) of each pixel:

(24) )255(log20 10 RMSE PSNR =

Where, RMSE is the root mean-squared error, defines as:

( ) ( ) ,,1 2

−= M N

jiÎjiI MN

RMSE

Here I and Î are the original and segmented images of size MxN, respectively.

Partition coefficient Vpc, partition entropy Vpe and PSNR are used to compare the performance of the adopted techniques for segmentation. Performance of the proposed methods is compared with fuzzy c-means method using type-1 fuzzy sets. The numerical results obtained using type-1 Fuzzy c-Means and type-2 Fuzzy c-Means with c=4 are presented in Table 1.

Using the three proposed membership functions and upper and lower membership degrees, performance value tends to increase. Tables II and III show the obtained values of

97

partition coefficient Vpc, partition entropy Vp signal to noise ratio (PSNR) using the propose the test images. A mean (μ) and standard d calculated on efficiency in order to show the the proposed and other method as in table I , I table I, as is apparent, for Vpc a biggest m standard deviation 0,1327 and for Vpe a low and standard deviation 0,1253 are obtained proposed membership function which confirm improvement. Figure 3 shows segmented imag

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mean 3,0320 and west mean 0,0538 d from the third ms the qualitative ge for test images

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TABLE I. EXPERIMENTAL RESULTS USING TYPE-1FCM AND TYPE-2 FCM (TIZHOOSH)

Image Type-1 FCM

Type-2 FCM (Tizhoosh)

μ(x)=(μL+μU)/2 μ(x)=(μL * μU) ½ μ(x)=μL+μL*μU

pvc Pve PSNR Pvc pve PSNR pvc pve PSNR pvc pve PSNR

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

0,7666 0,7501 0,7659 0,7229 0,7604 0,7446 0,7784 0,8406 0,7427 0,4899 0,8233 0,7995 0,7939 0,7466 0,4898 0,7720 0,7599 0,8053 0,7819 0,7528 0,8425 0,7321 0,7723 0,7984 0,7630 0,7365 0,8028 0,7590 0,7453 0,8098 0,7704 0,8128 0,7643 0,7656 0,7663 0,7703 0,7372 0,7782 0,8380 0,7434 0,7455 0,7588 0,8046 0,7458 0,7720 0,7420 0,7601 0,7787 0,8335

0,8931 0,9174 0,8876 0,9812 0,8913 0,9186 0,8685 0,6957 0,8942 1,4696 0,7628 0,8256 0,8174 0,9238 1,4696 0,8442 0,9028 0,7860 0,8554 0,9073 0,6527 0,9475 0,8296 0,8003 0,8507 0,9581 0,7752 0,9043 0,9465 0,8008 0,8757 0,7870 0,8886 0,8984 0,9014 0,8885 0,9323 0,8703 0,7386 0,9347 0,9222 0,8907 0,7855 0,8939 0,8372 0,9362 0,9052 0,8203 0,7424

59,9219 59,8711 54,6581 55,6092 53,4672 54,3089 57,0702 62,8419 56,3880 54,1540 56,2574 55,6175 58,7398 56,6427 52,4359 52,3838 51,7095 55,6994 55,2753 56,7824 57,2201 54,6739 57,4169 55,9750 55,4341 56,6102 53,7848 53,7657 52,9606 56,6207 55,6399 51,8087 54,2598 54,7323 53,4955 56,5148 55,3828 55,0082 56,0596 57,3124 55,9885 56,8288 56,7750 57,9561 53,5282 55,5564 59,2656 54,0085 56,2768

0,7666 0,7501 0,7659 0,7230 0,7604 0,7446 0,7784 0,8406 0,7427 0,7852 0,8233 0,7995 0,7939 0,7466 0,7062 0,7720 0,7599 0,8053 0,7819 0,7528 0,8425 0,7321 0,7723 0,7984 0,7630 0,7365 0,8028 0,7590 0,7453 0,8098 0,7704 0,8128 0,7643 0,7656 0,7663 0,7703 0,7372 0,7782 0,8380 0,7434 0,7455 0,7588 0,8046 0,7458 0,7720 0,7420 0,7602 0,4901 0,8335

0,8931 0,9174 0,8876 0,9800 0,8913 0,9186 0,8685 0,6958 0,8943 0,7835 0,7628 0,8256 0,8174 0,9238 0,9921 0,8442 0,9028 0,7860 0,8554 0,9073 0,6527 0,9475 0,8296 0,8003 0,8507 0,9581 0,7752 0,9042 0,9465 0,8008 0,8757 0,7870 0,8886 0,8984 0,9014 0,8885 0,9323 0,8703 0,7386 0,9347 0,9222 0,8907 0,7855 0,8939 0,8372 0,9362 0,9051 1,4686 0,7424

56,1735 53,9295 53,4129 54,5694 54,3653 53,8159 50,8998 60,6100 58,9479 56,2691 57,0273 54,1700 58,8416 58,0339 59,6437 59,2650 59,0343 65,8172 59,1305 55,9023 54,6723 56,6540 54,3464 54,3215 53,8422 54,8598 58,1743 58,3142 54,6892 57,8771 52,7551 56,8707 54,7562 58,3036 59,2990 57,5287 59,0841 54,3958 61,8339 57,8382 61,1035 58,9056 57,6529 59,4435 56,1949 54,2426 53,7222 56,2187 58,4327

0,6945 0,6486 0,6745 0,6499 0,6524 0,6714 0,6680 0,7818 0,7345 0,6899 0,7426 0,7084 0,7206 0,6482 0,6681 0,6699 0,6638 0,7286 0,6843 0,6441 0,7597 0,6320 0,7112 0,7368 0,6826 0,6152 0,7100 0,6876 0,6462 0,7462 0,6875 0,7301 0,7002 0,6728 0,6725 0,7116 0,6596 0,6950 0,7777 0,6398 0,6335 0,7182 0,7328 0,6525 0,7309 0,6365 0,6458 0,6768 0,7727

0,3303 0,3792 0,3521 0,3795 0,3764 0,3554 0,3559 0,2356 0,2880 0,3277 0,2735 0,3138 0,3012 0,3802 0,3586 0,3536 0,3614 0,2908 0,3404 0,3821 0,2586 0,3951 0,3102 0,2825 0,3393 0,4175 0,3106 0,3386 0,3823 0,2732 0,3365 0,2914 0,3236 0,3531 0,3537 0,3099 0,3681 0,3293 0,2398 0,3899 0,3969 0,3025 0,2851 0,3752 0,2883 0,3931 0,3834 0,3442 0,2469

53,6640 54,6443 54,0172 54,0910 59,1329 53,8223 64,1287 52,4717 56,2632 62,2692 50,7993 58,9390 58,7882 54,2536 52,5250 56,7600 54,1686 62,2923 55,3751 53,8923 59,6786 54,5118 55,1036 53,4889 55,4598 54,3100 55,6546 59,4443 56,5714 52,4261 54,3140 57,9523 52,2208 54,8520 57,0213 54,3461 56,2751 53,9393 62,2455 58,4859 59,6311 59,9732 60,1866 62,0558 54,2636 57,7441 57,7412 56,8627 51,9249

3,0043 2,9353 3,0002 2,8275 2,9816 2,9128 3,0564 3,3228 2,9282 3,0853 3,2432 3,1459 3,1216 2,9201 2,7614 3,0293 2,9789 3,1696 3,0711 2,9488 3,3301 2,8648 3,0312 3,1465 2,9912 2,8801 3,1596 2,9766 2,9176 3,1884 3,0188 3,2042 2,9933 3,0021 3,0036 3,0261 2,8832 3,0548 3,3100 2,9092 2,9176 2,9744 3,1673 2,9175 3,0263 2,9025 2,9794 3,0575 3,2915

0,1019 0,1480 0,0968 0,2541 0,1046 0,1492 0,0598 -0,2270 0,1032 -0,0587 -0,1205 -0,0170 -0,0267 0,1569 0,2830 0,0307 0,1212 -0,0789 0,0365 0,1344 -0,3084 0,1960 0,0077 -0,0598 0,0503 0,2095 -0,0871 0,1179 0,1862 -0,0595 0,0755 -0,0822 0,0993 0,1091 0,1132 0,0903 0,1789 0,0608 -0,1672 0,1747 0,1589 0,1027 -0,0758 0,1276 0,0281 0,1832 0,1214 -0,0067 -0,1606

57,0920 54,2034 53,7383 56,0729 53,4672 53,8986 51,8257 58,5402 56,6325 54,3553 54,9643 52,5391 53,7258 55,6121 58,4502 63,7951 56,6779 55,0349 52,0729 60,0354 57,2684 53,9845 53,6453 55,9060 54,7053 54,4536 55,7208 58,8689 57,2568 54,0072 54,8326 54,6798 54,0912 54,5045 55,4136 57,9124 58,6629 59,0623 59,3004 57,8382 59,4168 54,2835 52,0865 54,9039 57,1192 53,4033 58,2266 52,4531 52,4697

Mean(μ) 0,7620 0,8863 55,7285 0,7665 0,8757 56,8611 0,6902 0,3338 56,3466 3,0320 0,0538 55,6982

STD(σ) 0,0642 0,1403 2,1728 0,0512 0,1132 2,7741 0,0418 0,0456 3,1775 0,1327 0,1253 2,5130

SAI Intelligent Systems Conference 2015 November 10-11, 2015 | London, UK

779 | P a g e 978-1-4673-7606-8/15/$31.00 ©2015 IEEE

TABLE II. EXPERIMENTAL RESULTS USING TYPE-2 FCM (FIRST SET OF LOWER AND UPPER MEMBERSHIP FUNCTION)

Image

Type-2 FCM (first set of lower and upper membership function)

μ(x)=(μL+μU)/2 μ(x)=(μL * μU) ½ μ(x)=μL+μL*μU

pvc pve PSNR pvc pve PSNR pvc Pve PSNR

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

0,7866 0,7580 0,7732 0,7480 0,7575 0,7635 0,7737 0,8463 0,8092 0,7879 0,8274 0,7989 0,8041 0,7551 0,7601 0,7736 0,7648 0,8128 0,7798 0,7501 0,8334 0,7392 0,7935 0,7629 0,7794 0,7297 0,8000 0,7890 0,7551 0,8237 0,7831 0,8113 0,7880 0,7699 0,7730 0,7980 0,7587 0,7875 0,8452 0,7426 0,7432 0,7926 0,8134 0,7574 0,8108 0,7454 0,7550 0,7788 0,8403

0,4435 0,4908 0,4639 0,5126 0,4943 0,4837 0,4528 0,3174 0,4053 0,4131 0,3580 0,4159 0,4099 0,4978 0,4841 0,4579 0,4779 0,3891 0,4551 0,4949 0,3480 0,5164 0,4157 0,4677 0,4428 0,5396 0,4032 0,4342 0,5043 0,3724 0,4418 0,3895 0,4370 0,4708 0,4669 0,4182 0,4932 0,4416 0,3304 0,5200 0,5182 0,4222 0,3826 0,4922 0,3905 0,5142 0,4983 0,4380 0,3355

58,0114 54,1158 54,1943 56,2675 55,8964 52,9515 54,4158 60,4551 54,0658 57,6955 56,9141 56,3290 52,9021 55,6785 56,7441 53,1870 55,4249 55,2785 53,4788 54,5379 54,4867 53,7188 57,1086 56,5554 56,7251 55,7708 52,0194 56,5601 52,0970 54,0422 52,4676 56,6268 57,1359 52,9191 56,7614 52,9470 58,6128 56,3935 60,4722 54,5518 56,3790 55,4692 56,0689 56,6179 60,7960 55,0983 56,9498 54,1289 51,6725

0,7829 0,7532 0,7686 0,7444 0,7530 0,7601 0,7733 0,8435 0,8066 0,7829 0,8240 0,7948 0,8004 0,7502 0,7563 0,7690 0,7604 0,8093 0,7755 0,7449 0,8296 0,7351 0,7906 0,7575 0,7752 0,7241 0,7958 0,7853 0,7505 0,8205 0,7787 0,8076 0,7843 0,7655 0,7685 0,7946 0,7546 0,7835 0,8424 0,7379 0,7380 0,7895 0,8098 0,7526 0,8078 0,7404 0,7499 0,7741 0,8375

0,3984 0,4458 0,4206 0,4644 0,4480 0,4366 0,4078 0,2838 0,3599 0,3776 0,3210 0,3736 0,3677 0,4525 0,4394 0,4152 0,4326 0,3485 0,4104 0,4523 0,3119 0,4701 0,3750 0,4295 0,4012 0,4953 0,3650 0,3899 0,4561 0,3317 0,4000 0,3496 0,3938 0,4258 0,4222 0,3751 0,4467 0,3968 0,2920 0,4735 0,4729 0,3796 0,3435 0,4473 0,3490 0,4685 0,4535 0,3989 0,2990

56,1009 54,1158 52,8344 54,8775 55,0472 54,7517 51,1537 60,7640 60,7482 55,7561 56,9141 54,1326 58,8055 57,6285 62,4823 58,2009 59,0757 67,4584 59,1707 55,0355 54,5979 57,0306 55,0569 53,4375 53,3964 54,1858 57,8649 56,9489 54,9167 57,4828 52,3924 56,5985 54,9776 59,5421 59,2929 58,8171 60,3189 53,9600 62,1836 56,8584 63,0499 58,0777 58,1505 59,2349 55,9368 54,1975 54,2658 55,4020 58,0624

2,7408 2,5738 2,6509 2,5579 2,5695 2,6438 2,6234 3,0973 2,9051 2,7190 2,9326 2,7899 2,8463 2,5384 2,6378 2,6299 2,6095 2,8770 2,6934 2,5283 3,0071 2,4739 2,8095 2,8612 2,6862 2,4029 2,7966 2,7362 2,5189 2,9508 2,7045 2,8818 2,7627 2,6468 2,6417 2,8072 2,5949 2,7372 3,0798 2,5174 2,4834 2,8474 2,8972 2,5628 2,8882 2,4942 2,5281 2,6570 3,0584

-0,5971 -0,4917 -0,5400 -0,4803 -0,4884 -0,5354 -0,5222 -0,8202 -0,6994 -0,5922 -0,7262 -0,6293 -0,6663 -0,4691 -0,5307 -0,5339 -0,5166 -0,6882 -0,5697 -0,4637 -0,7662 -0,4299 -0,6419 -0,6782 -0,5676 -0,3798 -0,6339 -0,6023 -0,4580 -0,7329 -0,5754 -0,6842 -0,6122 -0,5375 -0,5349 -0,6417 -0,5064 -0,5951 -0,8108 -0,4546 -0,4328 -0,6677 -0,7011 -0,4840 -0,6941 -0,4396 -0,4611 -0,5503 -0,7927

64,3974 63,6257 54,2071 53,6361 54,3348 55,8766 55,2204 64,7776 57,6872 57,9310 56,2191 55,0321 58,4994 57,4087 55,5968 52,1124 52,1464 55,2456 54,3341 56,2381 57,0828 53,7559 54,6017 56,4853 56,4605 55,7733 53,8747 56,0625 52,7059 54,0088 59,4106 51,8460 54,5465 56,1485 53,8518 56,6778 55,5568 55,5481 56,0457 56,5197 55,5099 58,5497 58,1479 56,6943 53,3449 54,4854 59,3397 53,9827 55,7968

Mean(μ) 0,7823 0,4442 55,5040 0,7783 0,4014 56,9651 2,7183 -0,5842 56,0682

STD(σ) 0,0295 0,0554 2,1449 0,0300 0,0525 3,1282 0,1723 0,1095 2,7697

SAI Intelligent Systems Conference 2015 November 10-11, 2015 | London, UK

780 | P a g e 978-1-4673-7606-8/15/$31.00 ©2015 IEEE

TABLE III. EXPERIMENTAL RESULTS USING TYPE-2 FCM (SECOND SET OF LOWER AND UPPER MEMBERSHIP FUNCTION)

Image

Type-2 FCM (Second set of lower and upper membership function)

μ(x)=(μL+μU)/2 μ(x)=(μL * μU) ½ μ(x)=μL+μL*μU

Pvc pve PSNR pvc pve PSNR pvc Pve PSNR

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

0,7990 0,7726 0,7867 0,7632 0,7722 0,7776 0,7868 0,8555 0,8189 0,8017 0,8373 0,8112 0,8155 0,7699 0,7746 0,7876 0,7790 0,8241 0,7826 0,7656 0,8431 0,7542 0,8048 0,7785 0,7926 0,7463 0,8123 0,8013 0,7696 0,8339 0,7963 0,8237 0,8001 0,7837 0,7866 0,8095 0,7724 0,8001 0,8542 0,7580 0,7595 0,8031 0,8244 0,7718 0,8208 0,7609 0,7699 0,7933 0,8498

0,4395 0,4874 0,4601 0,5104 0,4897 0,4809 0,4475 0,3117 0,4028 0,4065 0,3533 0,4097 0,4051 0,4946 0,4815 0,4517 0,4737 0,3823 0,4695 0,4912 0,3422 0,5154 0,4139 0,4626 0,4382 0,5387 0,3978 0,4289 0,5008 0,3664 0,4372 0,3816 0,4337 0,4672 0,4630 0,4142 0,4914 0,4365 0,3233 0,5179 0,5146 0,4205 0,3770 0,4895 0,3865 0,5117 0,4950 0,4312 0,3301

56,3236 56,0583 54,7588 57,7972 55,5695 55,7860 52,7795 64,5674 52,2592 55,7561 61,7700 54,4593 52,8105 56,0758 59,7442 56,7740 60,2183 54,7455 60,2128 54,2830 53,0893 56,9982 54,6687 55,2864 56,2108 55,4325 53,8458 55,2230 54,7296 56,5287 51,7657 52,5377 52,4483 55,5154 55,2493 55,2819 55,4197 58,1013 53,1381 56,7719 62,8947 56,6196 56,5632 54,1339 57,0435 54,0836 53,7265 60,2106 52,8832

0,7885 0,7599 0,7748 0,7508 0,7595 0,7659 0,7807 0,8477 0,8114 0,7897 0,8290 0,8004 0,8058 0,7570 0,7633 0,7755 0,7669 0,8145 0,7818 0,7521 0,8343 0,7417 0,7956 0,7635 0,7812 0,7320 0,8014 0,7911 0,7576 0,8253 0,7849 0,8133 0,7900 0,7718 0,7748 0,8001 0,7613 0,7892 0,8466 0,7449 0,7452 0,7953 0,8152 0,7592 0,8123 0,7475 0,7567 0,7807 0,8416

0,4095 0,4566 0,4312 0,4765 0,4593 0,4488 0,4171 0,2914 0,3710 0,3848 0,3297 0,3835 0,3776 0,4636 0,4493 0,4252 0,4434 0,3578 0,4207 0,4620 0,3202 0,4819 0,3853 0,4388 0,4110 0,5059 0,3738 0,4002 0,4673 0,3411 0,4097 0,3577 0,4042 0,4367 0,4331 0,3853 0,4579 0,4075 0,3006 0,4850 0,4840 0,3891 0,3522 0,4584 0,3595 0,4797 0,4644 0,4075 0,3077

52,8140 55,8919 52,6392 56,9631 61,6082 56,2946 52,5695 60,4551 55,2960 55,6238 56,2191 52,7332 55,6251 54,5281 62,4823 61,0080 56,4124 54,3773 57,5570 53,8276 52,7048 54,8569 56,6002 54,4250 54,6339 54,1858 55,6894 53,7690 52,0970 56,9665 52,7867 53,1172 53,2713 54,3618 54,9815 63,0127 57,6230 54,6672 55,1217 56,3772 56,7796 55,4070 56,0440 56,6179 59,8215 56,0687 61,5534 54,3100 51,8908

3,0222 2,8917 2,9589 2,8504 2,8900 2,9229 2,9610 3,2970 3,1312 3,0125 3,1938 3,0733 3,0977 2,8742 2,9080 2,9548 2,9203 3,1354 2,9865 2,8526 3,2322 2,8118 3,0566 3,0880 2,9848 2,7619 3,0777 2,9941 2,8582 3,1881 2,9995 3,1358 3,0298 2,9439 2,9555 3,0729 2,8957 3,0242 3,2881 2,8207 2,8209 3,0594 3,1374 2,8873 3,1308 2,8328 2,8730 2,9787 3,2693

-0,5123 -0,4110 -0,4646 -0,3689 -0,4038 -0,4318 -0,4841 -0,7639 -0,5996 -0,5521 -0,6777 -0,5645 -0,5779 -0,3932 -0,4275 -0,4717 -0,4376 -0,6191 -0,4853 -0,3945 -0,7028 -0,3587 -0,5642 -0,5534 -0,5049 -0,3029 -0,5848 -0,4843 -0,3754 -0,6567 -0,5063 -0,6203 -0,5238 -0,4525 -0,4592 -0,5632 -0,4078 -0,5153 -0,7436 -0,3481 -0,3492 -0,5598 -0,6290 -0,4071 -0,6190 -0,3600 -0,3911 -0,5090 -0,7311

53,2956 59,5215 54,5085 59,5984 56,9342 53,3775 60,9337 51,2168 56,2447 60,1496 56,6787 55,1438 52,3275 56,6076 58,0196 58,2009 60,1757 52,0311 56,8419 56,3215 57,8601 53,1958 53,6429 58,8791 54,2159 58,9863 56,7557 57,5123 52,1138 60,1435 56,2712 53,1566 56,9685 53,2451 55,2860 52,9466 58,6236 56,4354 54,5059 56,5060 55,0264 54,9645 55,4124 59,1778 56,0283 56,6617 55,9110 54,2513 53,9538

Mean(μ) 0,7950 0,4403 55,9010 0,7843 0,4115 55,8095 3,0029 -0,5066 56,0565

STD(σ) 0,0276 0,0566 2,7716 0,0292 0,0534 2,7640 0,1341 0,1144 2,4731

SAI Intelligent Systems Conference 2015 November 10-11, 2015 | London, UK

781 | P a g e 978-1-4673-7606-8/15/$31.00 ©2015 IEEE

VI. CONCLUSION In this work, an approach is proposed to segment images

based on the concept of type-2 fuzzy c-means. The image can carry on as much information as possible when the image is transformed to the fuzzy domain. The main idea of this work was to use of type-2 fuzzy sets into fuzzy c-means. For this purpose, new membership functions, upper and lower membership degrees are proposed. The experimental results have shown the effectiveness and usefulness of the proposed methods for image segmentation. The type-2 fuzzy c-means approach can deliver satisfactory performance to segmenting images.

REFERENCES [1] O. Assas, Fuzzy C-Partition Using Particle Swarm Optimization

Algorithm (ICCS’12 International Conférence en Complex Systems

Agadir, Moroco. 978-1-4673-4766-2/12/$31.00 ©2012 IEEE. http://ieeexplore.ieee.org/xpl/mostRecentIssue.jsp?punumber=6422825

[2] O. Assas, Threshold Selection Based On Type-2 Fuzzy 2-Partition Entropy Approach (WCCS’14 2nd World Conférence en Complex Systems Agadir, Moroco. 978-1-4799-4647-1/14/$31.00©2014 IEEE.

[3] L. A. Zadeh, The concept of linguistic variable and its application to approximate reasoning-Part I-II-III Information Science , 8,8,9 (1979), 199-249,301-357,143-180.

[4] O. Castillo, Type-2 Fuzzy Logic: Theory and Applications, 2008 Springer-Verlag Berlin Heidelberg. ISBN 978-3-540-76283-6]

[5] J. M. Mendel , R.I Bob John, “Type-2 fuzzy sets made simple”, IEEE Trans. Fuzzy Syst. 10(2) (2002) 117-127.

[6] J.M. Mendel, Uncertain Rule based Fuzzy Logic Systems, Prentice- Hall, Englwood Cliffs, NJ,2001.

[7] H. R. Tizhoosh, “Image thresholding using type II fuzzy sets”,Pattern recognition, 38(2005), 2363-2373.

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