CarolinaTM Quantitative Analysis and Statistics for AP BiologyQuantitative Analysis and Statistics Kit for AP Biology Student Guide The chi-square test is a useful method to analyze

Student GuideNAME

DATE

CarolinaTM Quantitative Analysis and Statistics for AP Biology

S-1

Investigation 1. Chi-square Analysis, with Corn Genetics

In this investigation, you will analyze the physical traits, or phenotypes, of F2 generation ears of corn produced from monohybrid and dihybrid crosses. Using the ratio of contrasting phenotypic traits in an ear of corn, you will model genetic crosses to predict the likely parental genotypes responsible for the inheritance of alleles that control the expression of traits such as kernel color and texture. To determine how closely your observed data match your predicted results, you will employ a statistical method known as chi-square analysis. Simply stated, the chi-square (c2) test is a statistical test used to determine if the difference between an observed and an expected result may be attributed to random chance alone, or if the difference is so great that it is unlikely to be due to random chance. Ultimately, you will be able to use the results of the chi-square test to accept or reject your hypothesis about the inheritance of genetic traits in corn based on how closely the collected data match the results predicted by your initial Punnet square analysis.

At the completion of this laboratory, you should be able to

• collect and organize data from genetic crosses.

• use chi-square analysis to determine parental genotypes.

• predict patterns of inheritance given relevant data.

BackgroundBecause corn (Zea mays) is one of the world’s most important food crops, corn genetics has been studied extensively. Four easily recognizable phenotypes may be observed in the corn samples you will observe in this investigation. Demonstrating patterns of Mendelian inheritance, the alleles responsible for the inheritance of these kernel traits segregate independently to produce predictable phenotypic ratios involving kernel color and texture characteristics. For this investigation, we will assume that the allele responsible for normal corn kernel color, yellow, is recessive to the dominant purple allele.

In addition to kernel color, you will be observing a second trait known as kernel texture. The genes that control these two traits are found on separate chromosomes. The corn kernels observed in today’s investigation may exhibit either a smooth or wrinkled phenotype. We may assume that the allele responsible for expressing the smooth texture phenotype is completely dominant to the allele responsible for expressing the wrinkled texture phenotype in kernels. Table 1 shows the symbols used for the traits examined in this investigation.

Table 1. Symbols used to denote dominant and recessive alleles for corn kernel color and texture

Dominant Allelles Recessive Allelles

P = purple kernel color p = yellow kernel color

S = smooth texture s = wrinkled texture

In these activities, you will investigate several phenotypes of corn that are expressed in the kernel. You will be given ears of corn for your investigation. Each kernel on an ear results from a separate fertilization event. The kernels on the ear are the F2 from a cross that began with two parental varieties of corn with contrasting phenotypes. Take a moment to familiarize yourself with the observable phenotypic traits in your corn sample. (Refer to Figure 1.) Find an example of each observable phenotype on your dihybrid cross ear of corn.

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Quantitative Analysis and Statistics Kit for AP Biology Student Guide

Figure 1. Observable phenotypic traits in an F² Zea mays dihybrid cross

Pre-laboratory Questions

Refer to Figure 1 to answer the following questions.

1. Which letter represents a kernel exhibiting a purple, wrinkled phenotype? What are the possible genotype(s), that may produce such a phenotype?

2. Which letter represents a kernel exhibiting a yellow, smooth phenotype? What are the possible genotype(s) that may produce such a phenotype?

3. Which letter represents a kernel exhibiting a yellow, wrinkled phenotype? What are the possible genotype(s) that may produce such a phenotype?

4. Which letter represents a kernel exhibiting a purple, smooth phenotype? What are the possible genotype(s) that may produce such a phenotype?

The correct identification of each kernel phenotype will be critical to your ability to accept or reject your initial hypothesis concerning the genetics of corn during today’s lab investigation. Please take a moment to find each possible phenotype on your corn ear labeled 176600.

To successfully complete these activities, you must have a good background knowledge of genetics. Use your knowledge of Mendelian genetics and Punnett squares to answer the following questions.

5. A geneticist crosses a heterozygous Purple Zea mays with a Yellow Zea mays. What are the genotypes of the parents?

6. Create a Punnett square depicting a monohybrid cross between Pp and pp corn plants. What are the expected phenotypic and genotypic ratios of the resulting F1 offspring?

A B C D

Student Guide Quantitative Analysis and Statistics Kit for AP Biology

7. A geneticist crosses two Purple, Smooth corn plants, each heterozygous for both the color and texture trait.

a. What are the genotypes of the parents?

b. List the genotypes of all potential gametes (color allele and texture allele) that one of these parental plants may produce by meiosis.

8. a. Create a Punnett square that depicts the dihybrid cross of two Purple, Smooth corn plants, each heterozygous for both the color and texture traits.

b. Analyze the Punnett square results from 8a. What is the expected phenotypic ratio of the offspring?

c. If 1600 offspring were produced by these parents, how many offspring would be expected to express the homozygous recessive phenotype for both traits?

The Chi-square (c2) TestCorn is a diploid organism—its cells contain two copies of each chromosome. You should expect to find two alleles for a gene responsible for the expression of a phenotypic trait observed in every kernel, or seed, of corn. This combination of alleles, and thus the inheritance of any observable trait, is the result of random chance.

During sexual reproduction, two identical copies of each chromosome separate before being randomly assorted into haploid gametes. When parental gametes combine in fertilization, new combinations of alleles are created that may be observed in the phenotype of each kernel. Simple chance accounts for the assortment and recombination of alleles that gave rise to the phenotypes observed in F2 generation corn kernels.

Genetics, like gambling, deals with chance and probabilities. When you flip a coin, you have the same chance of getting a head as a tail: a one-to-one ratio. That does not mean that if you flip a coin 100 times you will get 50 heads and 50 tails. You might get 53 heads and 47 tails. That is probably close enough to a 1:1 ratio that we could accept it without a second thought. But what if you got 61 heads and 39 tails? At what point do you begin to suspect that something other than chance is at work in determining the landing of your coin?

Look back at your data. Based on your hypothesis about the inheritance of genetic traits in corn plants, you were expecting a 3:1 or 9:3:3:1 phenotypic ratio in the observed F2 generation. As mentioned before, this assumes that chance alone has been operating in the assortment and recombination of alleles that gave rise to the F2 generation observed by your class. Thus, any variation between the observed and the expected results are due to chance rather than to some other factor that may be influencing your results. The null hypothesis, denoted by H0, is usually the hypothesis that sample observations result purely from chance. But do your data actually support the null hypothesis, or should you instead consider an alternative mechanism of inheritance to explain the results obtained by your class?

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The chi-square test is a useful method to analyze this variability to determine how well observed ratios fit expected ratios. Ultimately, the results of chi-square analysis allow a researcher to determine if the difference between observed and expected results are due to random chance alone, or if there may be a factor other than chance, such as a trick coin or a sneaky epistatic gene, influencing the outcome of an experiment. The difference between the number observed and the number expected for a phenotype is squared and then divided by the number expected. This is repeated for each phenotype class. The chi-square value consists of the summation of these values for all classes.

The chi-square (c2) value is calculated as follows:

c2 = ∑

(observed – expected)2

expected

for all cases

Note that the symbol “∑” stands for “sum.”

The calculated value for c2 is then compared to the values given in a statistical table, such as Table 2.

Table 2. Chi-square table, with probability (P) values in shaded row

Degrees of freedom (df) ACCEPT NULL HYPOTHESIS REJECT

0.95 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.005

1 0.004 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 7.88

2 0.10 0.45 0.71 1.30 2.41 3.22 4.60 5.99 9.21 10.59

3 0.35 1.00 1.42 2.37 3.67 4.64 6.25 7.82 11.34 12.38

4 0.71 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.28 14.86

5 1.14 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 16.75

6 1.64 3.07 3.38 5.35 7.23 8.56 10.65 12.59 16.81 18.55

7 2.17 3.84 4.67 6.35 8.38 9.80 12.02 14.07 18.48 20.28

In this table, any c2 value falling in a column under a P value of 0.05 or less would cause an experimenter to reject the null hypothesis, meaning that the variation between observed and expected data is significantly different and that a factor other than chance is probably influencing the results. Note the column titled “Degrees of freedom.” The degree of freedom is always one less than the number of different outcomes possible. For a monohybrid F2 experiment we have two possible phenotypic outcomes, so there is 1 degree of freedom (2 – 1 = 1). For a dihybrid F2, there are four possible phenotypic outcomes and 3 degrees of freedom.

The numbers to the right of the Degrees of freedom column in the table are c2 values. The probability (P) values given at the top of each column represent the probability that the variation of the observed results from the expected results is due to chance. If the probability value is greater than 5% (P value of 0.05), we accept the null hypothesis; that is, our data fit the expected ratios. However, if our chi-square value corrected for the appropriate degrees of freedom falls within a P value less than or equal to 0.05 then we may feel confident in concluding that a factor other than chance is responsible for creating a significant difference between observed and expected values.

Following are two scenarios involving fruit fies, one for a monohybrid cross and one for a dihybrid cross.

In an F2 population of 100 Drosophila (fruit flies), there are 60 with normal wings and 40 with vestigial wings (expected ratio would be 75 normal wings:25 vestigial wings). Therefore,

c2 =

∑ (observed – expected)2

or c2 =

(60 – 75)2 +

(40 – 25)2 =

3 + 9 = 12

expected 75 25



Look at the chi-square table. In the row for 1 degree of freedom, for a c2 value of 12, the probability is less than 0.05, or 5%. Therefore, these results do not support the expectation (or null hypothesis) of a 3:1 ratio, since the probability is significant (less than 5%) that deviation from the expected ratio is due to chance.

Note: Always consider the possibility of error when the data do not fit the expected ratio(s). Possible sources of error include the following: sample size too small, phenotypes scored incorrectly, crosses made incorrectly.

Now consider the following data for F2 Drosophila of a dihybrid cross of F1 flies having normal wings and red eyes with F1 flies having vestigial wings and sepia eyes. The alleles for normal wings and red eyes are dominant. The expected phenotype ratio is 9 normal wings, red eyes:3 normal wings, sepia eyes:3 vestigial wings, red eyes:1 vestigial wings, sepia eyes.

Table 3. Sample F2 counts for a dyhybrid cross in Drosophila

1 2 3 4 Total

PhenotypeNormal wings,

Red eyesNormal wings

Sepia eyes,Vestigial wings,

Red eyesVestigial wings,

Sepia eyes4

Phenotype Count

577 204 176 59 1016

Expected Numbers

571.5 190.5 190.5 63.5 1016

c2 =

∑ (observed – expected)2

or (577 – 571.5)2

+ (204 – 190.5)2

+ (176 – 190.5)2

+ (59 – 63.5)2

= 2.43

expected

571.5 190.5 190.5 63.5

The probability of 2.43 from the table for 3 degrees of freedom is greater than 30% (P = 0.30) but less than 50% (P = 0.50). This means that a deviation as large or larger would be expected to occur purely by chance more often than 30 percent but less often than 50% of the time. Such a deviation is not significant (because the probability is greater than 5%), so we accept the null hypothesis in favor of the 9:3:3:1 ratio. Note that this acceptance is provisional. Additional data collection could always cause us to reject the null hypothesis at a later date.

Investigation 1 Activity A: Analyzing Results of a Known Cross

Materials

Carolina segregating corn ear, item 176600

transparency marker

Procedure

Formulating a Null Hypothesis

Your teacher will give you an F2 ear of corn that resulted from crossing F1 generation parent plants that were heterozygous for both kernel color and texture.

Observe your Punnett square results from Pre-laboratory Question 8. Essentially, your predicted Punnett square results serve as a null hypothesis for the transmission of the color phenotype in Zea mays. You are assuming that the corn kernel color trait is controlled by a single gene with two alleles, and that the purple allele is dominant to the yellow allele. You are also assuming that the corn kernel texture trait is controlled by a single gene with two alleles, and that the smooth allele is dominant to the wrinkled allele.

What would happen if you were to carefully record the color and texture of every kernel of corn observed? As long as the two kernel phenotype genes are found on separate chromosomes and are inherited independently,



then you would expect the phenotypic ratio to be very close to the ratio predicted by your Punnett square, given your null hypothesis about corn color genetics.

Take a look at the corn ear provided by your teacher. Does your corn ear appear to demonstrate the phenotypic ratio between the four phenotypic categories predicted by your Punnett square in question 8? If we were to count every single kernel, would the observed phenotypic ratio between the four possible phenotypes be close enough to your expected results that you would feel confident stating that any deviation from your expected ratio, or null hypothesis, could be attributed to chance alone? What if the observed results deviate drastically from your expected ratio? Would you then need to consider an alternative hypothesis for the mechanisms that govern the inheritance of one or more traits (for example, maybe the purple allele is not completely dominant to the yellow allele) to explain your observed results?

The chi-square statistical test helps us compare the differences between expected results and data collected from experimental observation in order to determine whether our results support our null hypothesis. Therefore, you should not be alarmed today if your chi-square value suggests that you should reject your null hypothesis in favor of an alternate hypothesis. In fact, a rejection of a null hypothesis in favor of an alternate hypothesis may be the first step toward a possible improvement in your experimental design or even in a novel scientific discovery.

Testing Your Hypothesis

Working in pairs, count and record in Table 4 the number of kernels of each phenotype. You will need to fill out both the parental genotypes and the four possible phenotypes in the table. One person should call out the phenotypes in five consecutive rows of corn while the other tallies them in the table. Mark the beginning of one row of kernels with a transparency marker and count and record the phenotypes of each kernel in that row. Continue counting, marking the beginning of each row as you count. When finished counting five rows, total your results. Then obtain and record the class totals for the same cross.

Table 4. F2 Phenotype count for __________________ × ___________________

Phenotype

Team Count

Total: Total: Total: Total:

Team Total for All Phenotypes Counted:

Class Count Total: Total: Total: Total:

Class Total for All Phenotypes Counted:

Refer to the Punnett square that you made previously for the dihybrid cross.

Calculate the expected counts for the F2 generation and record them below. For example, if your Punnett square results indicated that 9 16 of your F2 generation kernels are expected to be both yellow and smooth, and you counted 300 total kernels within five rows, then we would expect

(9 16)300 = 168.8 kernels to be both yellow and smooth

Phenotype ______________ Expected Count ___________________ Actual Count _______________






Your expected count for each phenotypic combination of kernel color and texture serves as a tentative hypothesis, or null hypothesis, for the transmission of these traits (that purple is completely dominant to yellow kernel color and that smooth is completely dominant to wrinkled texture).

By how much may we deviate from our expected values before we must admit that we were incorrect about our inheritance hypothesis, and thus reject our null hypothesis in favor of an alternate explanation for the transmission of these phenotypic traits? We will need to use the chi-square test to determine if our observed results deviate significantly from our expected results, or if the differences between observed and expected values may be attributed to chance alone.

Your Dihybrid Cross Chi-square (c2) Value

Using your class observed data from Table 4 and the expected phenotypic results from your calculations, fill out the chi-square table below. The following terminology key will help.

Table 5. Chi-square analysis for class F2 data: Dihybrid Cross

O E O – E (O – E)2 (O – E)2

E

Purple, Smooth

Purple, Wrinkled

Yellow, Smooth

Yellow, Wrinkled

Calculated chi-square value (c2): Sum of all answers in column (O – E)2

E∑ =

Degrees of freedom: Number of possible phenotype outcomes – 1 d.f. =

Chi-square table terminology:

O = Observed

E = Expected

Degrees of Freedom (d.f.) = # possible outcomes – 1

∑ = sum of (O – E)2

E

for all categories

Essentially, your expected count for each phenotype serves as your null hypothesis for the mode of inheritance of this particular trait. In other words, we may use the results of our Punnett square analysis to create an expected F2 phenotypic ratio of potential offspring produced from a cross between RrSs and RrSs plants. If the observed ratio is drastically different from our expected ratio, then we may have to reject our null hypothesis in favor of an alternate hypothesis, one that does not assume simple Mendelian inheritance in which the purple allele is dominant to the yellow allele, and that the smooth allele is completely dominant to the wrinkled allele.



Investigation 1 Activity B: Analyzing an Ear of Corn with Unknown Parental Genotypes

Materials

Carolina segregating corn ear of unknown parental genotypes

transparency marker

science notebook or paper for students’ data tables

Your teacher will give you an F2 ear of corn that resulted from crossing F1 generation parent plants that are of unknown genotype. Your lab group will be given only 30 seconds to observe this F2 ear of corn. You must use your observations of the phenotypic ratios presented by the F2 ear of corn to create a hypothesis as to the F1 parental genotypes that may have produced this F2 corn ear.

Procedure

1. From your instructor, obtain an F2 ear of corn that resulted from crossing F1 generation parent plants of unknown genotype.

2. You will observe the F2 ear of corn for 30 seconds.

• Pay close attention to any differences in kernel color (purple or yellow) and texture (smooth or wrinkled) that may be presented.

• In a notebook or on a sheet of paper, record the frequency of observed phenotypes. These phenotypic frequencies will be critical to your formation of an inheritance hypothesis.

3. After observing the ear of corn for 30 seconds, form a prediction as to the F1 genotypes that may have produced this F2 ear of corn.

4. On a separate sheet of paper, create a Punnett square (monohybrid or dihybrid) depicting a cross between the two F1 genotypes that you predicted as parents of the mystery ear.

5. Consider the Punnett square and form a null and an alternate hypothesis for the inheritance mechanism that produced the phenotypic ratio(s) observed in the F2 ear.

For example, If a student group recorded roughly one-half of all kernels as purple and roughly one-half of all kernels as yellow after 30 seconds of observing an F2 ear of corn with unknown parental genotypes, then they may predict the following parental genotypes in the F1 generation: Pp × pp.

F1 generation genotypes: Pp × pp

This prediction results in the following null and alternate hypotheses:

H0: If corn kernel color is controlled by a single gene with two alleles, and the purple kernel color trait is completely dominant to the yellow kernel color trait, then the resulting cross of Pp × pp plants in the F1 generation will yield an F2 phenotypic ratio not significantly different from 1 purple:1 yellow.

HA: The resulting cross of Pp × pp plants in the F1 generation will yield an F2 phenotypic ratio that is significantly different from 1 purple:1 yellow.

6. Once every student group has formed a null hypothesis, follow your instructor’s directions for performing the actual phenotypic counts for your assigned corn ear of unknown parental genotype. Each student group will be responsible for counting two rows of kernels and recording the phenotypic data.

7. Create a table and record the phenotypic data collected by all student groups who were assigned to the same mystery ear.

P

p

p

p

pp

pp

Pp

Pp



c.Chi-square calculation

Null Hypothesis:

Observed (O) Expected (E)(O – E)2

E

Green Leaf

Red Leaf

Total

d. Explain whether your null hypothesis is supported by the chi-square test and justify your explanation. Be sure to include your chi-square value, your degrees of freedom and the P value associated with your test (less than or greater than the P = 0.05 rejection value).

8. Using the expected values from your Punnett square and the collected data from the groups’ counts, calculate the chi-square value and degrees of freedom for your results.

9. Create a statement of acceptance or rejection of your null hypothesis. Include the degrees of freedom used and the resulting P value when stating your conclusion.

Student Questions

1. Describe the importance of using chi-square analysis in genetics.

2. Suppose that a classmate calculated a c2 value of 4.14 from the results of flipping a coin 100 times and recording the observed results.

a. How many degrees of freedom would be used to interpret this statistic?

b. What would the null hypothesis be for this experiment?

c. What would these results indicate about the need to accept or reject the null hypothesis?

3. Refer to class data collected in Table 4 (the dihybrid cross) and calculate the chi-square value. What does the value tell you about the class data? When stating your conclusion, include a statement of acceptance or rejection of your null hypothesis, the degree of freedom used, and the resulting P value.

4. When designing an experiment, why might scientists choose a large sample size rather than a small one if they intend to analyze the data using a chi-square test?

5. Female painted lady butterflies, Vanessa cardui, lay green eggs on the substrate that the larvae will eat after hatching. In an investigation of the influence of color preference on butterfly egg-laying behavior, a scientist prepares two leaf-shaped pieces of plastic foam, one red and one green, and soaks both of them in a nectar solution. The two leaf models are then placed on opposite ends of a covered terrarium. To test the butterflies’ preference for the color of their egg-laying site, five gravid female painted lady butterflies are released into the midpoint of the covered terrarium. After 1 week, the number of eggs laid on each of the model leaves is observed and recorded. The total number of eggs counted on the leaves is 1624. Assume that captive painted lady butterflies lay eggs only on leaf-shaped objects and that no eggs are observed on any other surface in the terrarium.

a. Predict the distribution of eggs on the leaves after one week and justify your prediction.

b. Perform a chi-square test on the data for the butterfly experiment. Specify the null hypothesis that you are testing and enter the values from your calculations in the Chi-square Calculation table below.

Student GuideNAME

DATE



Investigation 2. Analyzing Data Sets, with Hominid Height

Introduction In this investigation, you will analyze height data collected from the skeletal remains of two extinct human hominid species and determine whether any difference in mean height between the two sampled populations is statistically significant. You will be introduced to the powerful statistical analysis tool known as the t-test.

At the completion of this laboratory, you should be able to

• collect and organize data from your own observations.

• analyze and interpret the statistical significance of data sets using the t-test.

• determine the mean, median, mode, and standard deviation of a data set.

Background

Data analysis skills are essential to any scientific investigation. Data collected from investigations are often quantitative, and may range from direct measurements to behavioral observations. After collecting data, scientists often use descriptive statistics and detailed graphs to evaluate their original hypothesis as they begin to answer a research question. Effective data analysis provides an experimenter with powerful tools to interpret and communicate the results of a well-designed investigation.

A challenge faced by students when designing an original investigation is to determine the most practical data collection methodology to apply in answering a testable question. During the design process, think about the type of data needed to answer your question. Will you be comparing two means? Will you be comparing rates? Will you need to determine if the observed values in an experiment are significantly different from the values that you were expecting? Remember that good investigations are built on proper data collection and supported by solid data analysis.

The following formulas and definitions will assist in today’s data collection and analysis investigations.

Formulas:

sampled mean ( x ): sum of all data values

number of data values

standard deviation (s): s = where x = mean and n = size of the sample

standard error of the mean (SEM): standard deviation of the sample

√number of observations

= s

Definitions:

Mode. the value that occurs most frequently in a data set

Median. the middle number in a given sequence of numbers, taken as the average of the two middle numbers when the sequence has an even number of numbers

Σ (x – x 2

n – 1)

√n



Mean. the sum of all data points divided by the number of data points

Range. the value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample maximum)

Standard Deviation. the measure of variation in a distribution, or how spread out the values are

Standard Error of the Mean. a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn

A hypothetical example will be used to introduce the data analysis terms and methods needed to complete today’s investigations. Imagine that a student collected the following data for the heights of 10 radish plants labeled A–J and recorded their data in Table 6, below.

Table 6. Height data collected from 10 radish plants, labeled A–J

Plant A B C D E F G H I J

Height (cm)

5.4 7.2 4.9 9.3 7.2 8.1 8.5 5.4 7.8 10.2

To calculate the mean of this data set, simply take the sum of all values (∑ = 74) and divide it by the number of data values collected.

mean = 74

= 7.4 cm10

To calculate the median, the student would order the data from the smallest to largest value and then take the mean of the middle two values. Using data from Table 6, we would calculate the median as follows:

4.9 5.4 5.4 7.2 7.2 7.8 8.1 8.5 9.3 10.2

median = (7.2 + 7.8)

= 7.5 cm

2

The mode for a data set of 1, 2, 2, 2, 2, 3 would be the number 2, as it is the number that occurs the most frequently. It may be argued that the radish plant data set in Table 6 has two modes: 7.2 and 5.4 occur at an equal frequency.

To calculate the range of a set of data values, simply subtract the smallest observation observed value (sample minimum) from the greatest observation (sample maximum). The range of the radish height data from Table 6 would be calculated as follows:

range = 10.2 – 4.9 = 5.3 cm

Calculating the Standard Deviation

We will now calculate the standard deviation of a sampled population. This value, s, will allow us to determine how spread out our collected data is. An example has been done for you using the radish plant height data provided in Table 6:



∑ (x – x ) 2

Σ (x – x 2

n – 1)

27.64 9

Radish Plant Height, in cm (x)

Mean ( x ) x – x (x – x )2

5.4 7.4 –2 4

7.2 7.4 –.2 .04

4.9 7.4 –2.5 6.25

9.3 7.4 1.9 3.61

7.2 7.4 –.2 .04

8.1 7.4 .7 .49

8.5 7.4 1.1 1.21

5.4 7.4 –2 4

7.8 7.4 .4 .16

10.2 7.4 2.8 7.84

27.64

Using the formula where x = mean and n = size of the sample, we would determine the standard

deviation to be s = = 1.75

To interpret the standard deviation in the context of the radish plant data, we could conclude that the radish plants sampled are, on average, about 1.75 cm away from the mean height of 7.4 cm.

Calculating the Standard Error of the Mean in a sample data set

The standard error of the mean (SEM, or SEx) allows an experimenter to infer just how well the sample mean matches up to the true population mean. This value helps you determine the confidence in the data collected in a sample by providing the statistical probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Ideally, a random sampling of any population should produce a mean that falls within ±2 standard error (SE) 95% of the time. We would call this a 95% confidence interval.

Using the radish plant data provided in Table 1 to produce a 95% confidence interval = ± 2 SE, we would simply apply the following standard error formula:

Therefore, with a mean plant height of 7.4 cm, a standard deviation of 1.75, and a sample size of 10 plants, our SEM would be calculated as follows:

0.55 or approximately 0.6

We would interpret a 95% confidence interval (i.e., = ±2 SE from our mean of 7.4 cm) as follows:

sample mean ±2 SE = 7.4 ± 1.2 cm (95% confidence interval = 6.2 to 8.6 cm)

Analyzing Differences in Height Between Past and Present Populations of Hominids

Archaeology is the study of human history and prehistory. Scientists may study ancient tools and bones and then make inferences on the physiology and behaviors of our ancestors. As a Homo sapiens, you belong to the only extant hominid species, but several other early human or human-like species once existed. Evidence indicates that some of our ancestors and relatives utilized stone tools and behaved in other ways like modern humans.

SEx = sn

SE (7.4) = 10 1.75 =



Figure 2. Cladogram suggesting ancestral relationships between early and modern hominids

Scientists describe ancestral relationships between organisms using cladograms such as the one shown in Figure 2. Cladograms are diagrams that arrange organisms in groups sharing certain characteristics in a hierarchical format.

You have probably heard of the term “Neanderthal,” in reference to an extinct species of early human whose remains were discovered in the caves of the Neander Valley in Germany more than a century ago. The skeleton of Neanderthal shared several similarities with your own. However, there were also some significant skeletal differences between Neanderthal and Homo sapiens. It has been said that if an early human sat next to you on the bus, you may not even notice. However, many anthropologists contend that physical differences between early and modern humans would be so great that it would be impossible for individuals of species such as Homo neanderthalensis or Australopithecus afarensis to walk among modern Homo sapiens unnoticed.

In today’s Investigation, you will analyze data collected from the skeletal remains of two separate populations of early human ancestors: Homo neanderthalensis and Australopithecus afarensis. The height data for each sample population have been provided for you in the following tables. Your task is to analyze each data set to determine the mean, median, mode, and standard deviation in both sample populations. You will also measure the height of 12 randomly selected classmates to generate height data for a Homo sapiens population.

After graphing the mean height of each sample population, you will then perform a statistical test known as a t-test to compare the means of the two populations. A powerful analytical tool for scientists, a t-test’s statistical significance indicates whether or not the difference between two groups’ averages reflects a “real,” or significant, difference in the population from which the groups were sampled. In other words, your t-test results allow you to conclude with a degree of confidence whether the average height between the two groups of early human ancestors is or is not significantly different, according to the sampled data from the two populations.

Common Ancestor ?

H. Sapiens

Later Homo Sapiens

A. afarensis Modern ApesH. neanderthalensis

Mill

ions

of y

ears

H. heidelbergensis.50

.25

0

6

5

4

3

2

1

0

Simplified from graph in Nature, Vol. 371 (1994) p. 280. White, T., G. Suwa, B. Asfaw, Australopithecus ramidus, a new species of early hominid from Aramis, Ethiopia.



Hominid Comparison Data Tables

Table 7. Hominids being compared in this investigation

FeaturesHomo sapiens

Human

Homo neanderthalensis

Neanderthal

Australopithecus afarensis

Oldest known hominid

Age of Specimen Modern 30,000 YA 2.9-3.6 MYA

Location and/or Date of Fossil Discovery

Your Classmates Today Europe 1908 Africa 1975

Table 8. Homo neanderthalensis heights estimated from 12 complete skeletal fossils discovered in France

Fossilized SkeletonNumber

1 2 3 4 5 6 7 8 9 10 11 12

Height (cm) 153 168 165 158 167 166 157 154 168 159 161 167

Table 9. Australopithecus afarensis heights estimated from 12 complete skeletal fossils discovered in Africa

Fossilized SkeletonNumber

1 2 3 4 5 6 7 8 9 10 11 12

Height (cm) 105 118 151 110 107 116 134 140 108 105 111 113

Pre-laboratory Questions

1. What is the difference in age (in millions of years) between the oldest fossils collected of Australopithecus afarensis and fossils collected from Homo neanderthalensis?

2. What is a population? Provide an example.

3. Observe the height data presented in Table 8 and Table 9, would you assume that the two populations of hominids were significantly different in average height?

Materialstape measure

Procedure

There are several important parameters to consider when analyzing aspects of a population. One such parameter is population size. Scientists interested in determining the total number of individuals in a population would use the symbol N to denote the population size. However, it is not always feasible to account for every single individual in a population. For example, can you imagine trying to physically count every single fire ant (Solenopsis invicta) within a 1-km2 area in Athens, Georgia? It would make sense to take a sample of the population, denoted by the statistical symbol n, and then use statistical methods to draw inferences about the larger population (N) based on our observations of smaller samples (n).

Likewise, scientists are unable to locate and analyze every single skeleton of Homo neanderthalensis or Australopithecus afarensis. Therefore, we must make inferences about each population (N) based upon a sample (n) gathered from an archaeological dig site. You will use the height data in Table 8 and Table 9 and perform the following calculations: mean, median, mode, and standard deviation.



Calculating the Mean, Median, Mode, and Range of a Sample

1. Using data from Table 8 and Table 9, calculate the mean height for Homo neanderthalensis and for Australopithecus afarensis. Record these data in Table 11.

2. Use a tape measure to measure the height of 12 students in your class. This is your Homo sapiens sample population. Record these data in the left-hand column of the table in step 10.

3. Calculate the mean height of your sample population of Homo sapiens and record the value in Table 11.

4. Using your collected class data and the data provided in Table 8 and Table 9, calculate the median of each sampled data set. Record these data in Table 11.

5. Record the mode of each sampled hominid population in Table 11.

6. Using your collected class data and the data provided in Table 8 and Table 9, calculate the range of each set of data values and record this value in Table 11.

Instructions: Use the following charts to calculate the standard deviation for each sampled hominid population. Once calculated, record these values in Table 11.

7. Find the standard deviation of Homo neanderthalensis (Table 8) by filling in the following chart:

Height, in cm (x) Mean ( x ) x – x (x – x)2

Standard deviation: ________

∑ (x – x ) 2

Use the formula

for standard deviation

Σ (x – x 2

n – 1)



8. Find the standard deviation of Australopithecus afarensis (Table 9) by filling in the following chart:

Height, in cm (x) Mean (x) x – x (x – x)2


9. Find the standard deviation of Homo sapiens by filling in the following chart using the heights of twelve randomly selected classmates.

Height, in cm (x) Mean (x) x – x (x – x)2


∑ (x – x ) 2

∑ (x – x ) 2

Use the formula

for standard deviation

Σ (x – x 2

n – 1)



Calculating the Standard Error of the Mean for Each Sampled Human Population

Now, you will put your statistical skills to full use in interpreting the results of an experiment. Imagine that you have designed an experiment to determine whether a fertilizer treatment has a significant effect on radish plant growth. You will compare growth of two groups of radish plants. One group has the fertilizer treatment, and a control group does not. Using the statistical methods discussed up to this point in the unit, you will analyze and interpret your data in order to arrive at a scientifically sound conclusion. To begin, you need a testable hypothesis. Consider the following hypotheses:

Null Hypothesis—The addition of fertilizer treatment does not have a significant effect on average radish plant height.

Alternate Hypothesis—The addition of fertilizer does have a significant effect on average radish plant height.

Consider the hypothetical radish height data presented earlier, in Table 6. Assume that the radish data set came from the control group (radish plants grown without the addition of a fertilizer treatment) in our experiment. Remember that we were able to calculate a mean plant height of 7.4 ± 1.2 cm. Holding all other experimental variables constant, if we were to calculate a mean plant height of 10.2 ± 0.8 cm in our experimental group (radish plants grown with the addition of a fertilizer treatment), then we may be able to use our mean and standard error of the mean values to strengthen our conclusion. The data for the hypothetical experiment have been summarized in Table 10.

Control Group Radish Plant Experimental Group Radish Plant with Fertilizer Treatment

Mean (cm) 7.4 10.2

Standard Deviation 1.75 1.26

N 10 10

Standard Error (2 X SEM) ±1.2 ±0.8

We may then graph the means, including their standard errors as follows:

Notice that the ±2 SE bars (95% confidence intervals) do not overlap between the control group and fertilizer treatment. This strongly suggests that the two populations are indeed statistically significantly different from one another. However, if the error bars/confidence intervals did overlap between the groups, you could not claim a statistically significant difference. So would you accept your null hypothesis or reject it in favor of our alternate hypothesis? You would reject the null hypothesis. Our graph suggests that the fertilizer treatment does have a significant effect on radish height compared to the control.

The effect of treatment on average height in radish plants

15

10

0

Ave

rag

e P

lant

Hei

ght

(cm

)

Control Group

Fertilizer Treatment

5

Table 10. Standard error of the mean for control and experimental groups of radish plants



Statistical Analysis Using t-Tests

A t-test is a type of statistical analysis best utilized when the variable tested is analyzed using average results from multiple trials. Depending on the exact design of the experiment and the type of data collected, a student may choose to employ one of two variations of the t-test: randomized t-test or matched-pair t-test.

Randomized t-tests examine whether the difference between a control group and experimental group is statistically significant when the organisms of interest are randomly assigned into the two groups. For instance, randomly assigning radish seeds to be grown in the presence or absence of a fertilizer treatment in order to determine the growth effects of such a treatment would generate data that fit the specifications for a randomized t-test.

Conversely, a matched-pair t-test uses the same sample organisms twice, with organisms first exposed to control conditions and then exposed to experimental conditions. For instance, an experimenter may want to compare the means of AP® Biology practice exam grades in a class of students who took the exam in August and the same exam again in April. Because the experimenter is using the same student test subjects, a matched-pair t-test is appropriate.

In both cases, a t value is calculated and compared to a critical t value. If the calculated t value is greater than or equal to the critical t value, the null hypothesis is rejected and the alternate hypothesis is accepted. Using statistical software, your instructor will introduce the t-test method of statistical analysis.

Table 11. Student data table for hominid height analysis

Homo neanderthalensis Australopithecus afarensis Homo sapiens

Mean Height (cm)

Median Height (cm)

Mode (cm)

Range

Standard Deviation

Standard Error of the Mean (SEM, or SEx)


Investigation 3. Analyzing Correlation, with Height and Arm Span Data

Introduction

Correlation of Data Sets

Correlation is a statistical technique that shows how strongly pairs of data sets are related. In most cases, this is done graphically. A correlation graph shows if there is a relationship between an independent variable on the x-axis and a dependent variable on the y-axis. The graphs below indicate three types of possible relationships for two data sets.

• The first graph is a scatter plot that shows a positive correlation (and thus a positive slope) between two data sets. As the independent x variable increases, the dependent y variable increases.

• The second graph is a scatter plot that shows a negative correlation, and thus a negative slope. As the independent x variable increases, the dependent y variable decreases.

• The third graph is a scatter plot showing no order of dots to form a straight line. Clearly, there would be no correlation indicated among the data on this graph.

An upward or downward best-fit line does not automatically determine a correlation nor suggest a cause-and-effect relationship. Sometimes, two data sets just happen to form a line by chance. To indicate that one data set influences another, an experiment must do the following:

• Have one clearly defined independent variable, graphed on the x-axis.

• Have a carefully measured dependent variable on the y-axis that changes as the independent variable changes on the x-axis.

• Control all other variables that might impact the dependent variable.

A spreadsheet program can be used to graph the points and to draw a “best-fit” line. A positive or negative numerical value known as a correlation coefficient, r, can be computed. An r value of +1 represents a perfect straight line of data points with a positive correlation. A value of –1 represents a perfect straight line having a negative correlation. The closer r is to zero, the less the correlation, with zero representing no correlation.

For most spreadsheets, boxes can be checked to obtain the equation for the best-fit line (y = mx + b), where m is the slope of the line and b is the y-intercept and the square of the correlation coefficient, R2. This R2 value (the coefficient of determination) is the decimal equivalent of a percentage. For example, in the following graph, the R2 value of 0.989

0Positive Correlation Negative Correlation

x

y y y

xx

No Correlation0 0

yyy

000x x x

Positive Correlation Negative Correlation No Correlation

Student GuideNAME

DATE




Vitruvian Man Biometrics

Leonardo da Vinci was influenced by a first-century BCE Roman engineer and architect known as Vitruvius, who proposed a formula for the ideal proportions of a man. Da Vinci modified this formula to make a drawing showing a man with outstretched arms and legs fitting perfectly inside a circle and a square. As indicated in the drawing, Vitruvius believed that a man’s arm span (distance from fingertip to fingertip of outstretched arms) should equal his height (distance from head to foot). Does this correlation of height and arm span hold true for people today? You will use a correlation graph to accept or reject the following hypothesis:

The height and the arm span of a person will be of equal linear measure.

Pre-laboratory Questions 1. Data pairs comparing the length of variable A and the length of

variable B were plotted on a correlation graph. From graphical analysis, the plot yielded a correlation coefficient of 0.1035. Describe or draw what the plot would look like on a graph.

2. Variable A and variable B were plotted on a graph giving a correlation coefficient of –0.9523.

a. Sketch a graph using six points and show how a trend line would look drawn through them.

b. If A were the independent variable and increased over time, how would the dependent variable B change over time?

180

175

170

165

160

155

150

145

140

135

130

Age in Years35 40 45 50 55 60 65

Blo

od

Pre

ssur

e (m

m H

g)

Blood Pressure vs. Age

y = 1.7143x + 68.571 R2 = 0.98901

for the data points means that 98.9% of the variation in blood pressure may be explained by changes in age. Unlike the correlation coefficient, squaring r always gives a positive number for evaluating how well the data points form a straight line. An ideal correlation value would be a value of ±1.0, yielding a perfect straight line.

Leonardo da Vinci's "The Proportions of the Human Figure," also known as “Vitruvian Man”



3. Referring to the correlation coefficient in question 2, answer the following:

a. Calculate the coefficient of determination, R2.

b. On the basis of your R2 calculation, what percentage of the variation in variable B is explained by changes in variable A?

Materials

tape measure, 150 cm

Needed but not supplied:

metric ruler or meterstick

Procedure 1. Before collecting any data, sketch a graph of how height and arm span data would look if they correlated.

Treat height as the independent variable on the x-axis and arm span as the dependent variable on the y-axis.

2. Measure the height in centimeters of everyone in your lab group. Having the person stand with their back to a wall will help in getting a more accurate measurement. You can use a meterstick or a metric ruler to extend the 150-cm tape if you need to. Height is distance from the top of the head to the soles of the feet. Record height in Table 12.

3. Measure the arm span in centimeters of everyone in your lab group. Arm span is measured as the length from the tip of one middle finger to the tip of the other, when hands and arms are outstretched straight, parallel to the floor. Record measurements in Table 12.

4. Exchange data with all other lab groups in your class so that you have data for the entire class. If your class is small, your teacher may ask you to increase the size of your data sample by adding measurements of adult family members.

5. Use a spreadsheet program to enter height and arm span data. Label your first data column, “Height” and your second data column, “Arm span.”

6. Select both columns of data, click on insert, choose scatterplot, and select the equation for the trend line (y = mx + b) and R2 for evaluating correlation. The closer R2 is to 1.0, the better the data fit with a straight line indicating that 100% of the data points for height and arm span are correlated.


Student GuideNAME

DATE


Vitruvian Man Biometrics

Table 12. Student measurements

Name Height (cm) Arm Span (cm)


Student GuideNAME

DATE


Assessment Questions

1. Did you accept or reject the hypothesis that a person’s height is equal to the person’s arm span?

2. Using your correlation graph and R2 value for your graph, describe how you accepted or rejected your Vitruvian Man theory hypothesis.

3. When graphing the data collected during the Vitruvian Man Biometrics investigation, what percentage of the variability in arm span may be explained by variability in height?

4. Calculate the correlation coefficient r from the information on your Vitruvian Man Biometrics graph and determine the sign of r (i.e., either + or –).

5. Observe the data provided in tables 8 and 9 from Investigation 2, Analyzing Data Sets, with Hominid Height. Then form a testable null and alternate hypothesis about average height between the two sampled human populations.

6. Refer to your data from Investigation 2 and create a graph using the average height of the three human populations. On the axes provided, create an appropriately labeled graph to illustrate the sample means of the three populations to within 95% confidence (i.e., sample mean ± 2 SEM). Be sure to include a title and a key in addition to appropriately labeled axes.

7. Observe the graph that you created for question 6. Based on the sample means and standard errors of the means, identify the populations that are most likely to have statistically significant differences in mean height. Justify your response.

8. Perform a t-test using your height data collected from your class and the height data provided in Table 9 for the skeletal remains of A. afarensis.

a. Is there a significant difference between the average height of the two sample populations? Make sure to include the following in your response: null hypothesis, alternate hypothesis, and the resulting P value.



b. Research Australopithecus afarensis. What differences in physiology or behavior may account for any significant differences in mean height between your sampled H. sapiens population and the sampled A. afarensis population that we have studied in this investigation?

c. If a time machine brought an Australopithecus afarensis into the present, would the hominid appear out of the ordinary if he or she were to sit next to you on a bus? Explain your answer.

9. The professor of a freshman biology course at a local university claims that students who had completed an Advanced Placement® Biology class during high school scored significantly higher on their freshman college biology midterms than students who had not. Assume that you have full access to the names, midterm scores, and the high school transcripts of all students enrolled in the course.

a. Identify a statistical analysis method that could test the professor’s claim.

b. Explain why this statistical test is appropriate for the data collected and the research question itself.

10. A wildlife biologist from Oregon hypothesizes that the expected frequency of various prey items eaten by great horned owls, Bubo virginianus, is not significantly different between owl populations in Oregon and in North Carolina. The biologist collects a large sample of great horned owl pellets from each location and carefully dissects them, recording the observed frequency of voles, moles, and birds found in the pellets:

Frequency of Prey Items Found in Owl Pellets from Two Locations in the United States

Prey ItemFrequency of Prey Item in North Carolina Owl Pellets

Frequency of Prey Item in Oregon Owl Pellets

Moles 0.6 0.5

Voles 0.3 0.2

Birds 0.1 0.3

a. Identify the independent and dependent variables in this experiment.

b. Identify a statistical analysis method that could test the wildlife biologist’s hypothesis.

c. Identify the null hypothesis in this experiment.

d. Explain why the chosen statistical test is appropriate for the data collected and the research question itself.

Documents

CarolinaTM Quantitative Analysis and Statistics for AP BiologyQuantitative Analysis and Statistics Kit for AP Biology Student Guide The chi-square test is a useful method to analyze