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Insects hovering in flight expend enormous amounts of energy for their size and weight. The data shown here were taken from a much larger body of data collected by T.M. Casey and colleagues. \({ }^{15}\) They show the wing stroke frequencies (in hertz) for two different species of bees. $$\begin{array}{cc} \hline \text { Species } 1 & \text { Species } 2 \\\\\hline 235 & 180 \\\225 & 169 \\ 190 & 180 \\\188 & 185 \\\& 178 \\\& 182 \\\\\hline\end{array}$$ a. Based on the observed ranges, do you think that a difference exists between the two population variances? b. Use an appropriate test to determine whether a difference exists. c. Explain why a Student's \(t\) -test with a pooled estimator \(s^{2}\) is unsuitable for comparing the mean wing stroke frequencies for the two species of bees.

Short Answer

Expert verified
Explain why using a Student's t-test with a pooled estimator would be inappropriate in this scenario. Answer: Yes, there is evidence to suggest a difference between the population variances of the wing stroke frequencies of the two species of bees. The F-test result shows a significant difference between the population variances, with the calculated F-value (24.06) being greater than the critical F-value (6.71) at a 0.05 significance level. Using a Student's t-test with a pooled estimator would be inappropriate because it assumes equal population variances between the two groups being compared, which is not the case based on the F-test result. Instead, a statistical test that does not rely on the assumption of equal variances, such as the Welch's t-test, should be used for comparing the mean wing stroke frequencies of the two species of bees.

Step by step solution

01

a. Observed Ranges Analysis

To explore the possibility of a difference between the population variances, we can first calculate the range of the observed data for both species: Species 1 Range: (235 - 188) = 47 Hz Species 2 Range: (185 - 169) = 16 Hz The observed range of wing stroke frequencies for Species 1 is larger than that of Species 2, which indicates that a difference might exist between the population variances.
02

b. F-test for Variances

To determine if a difference exists between the population variances, we will use the F-test for comparing variances. 1. Calculate the sample variances for both species: Species 1 sample variance, \(s_1^2\):\[\frac{(235-207)^2+(225-207)^2+(190-207)^2+(188-207)^2}{4-1} = \frac{2570}{3} = 856.67\] Species 2 sample variance, \(s_2^2\):\[\frac{(180-178)^2+(169-178)^2+(180-178)^2+(185-178)^2+(178-178)^2+(182-178)^2}{6-1} = \frac{178}{5} = 35.6\] 2. Calculate the F-value: F-value = \(\frac{s_1^2}{s_2^2}\) = \(\frac{856.67}{35.6}\) = 24.06 3. Determine the degrees of freedom for both species: Species 1 degrees of freedom, \(df_1\) = 4 - 1 = 3 Species 2 degrees of freedom, \(df_2\) = 6 - 1 = 5 4. Use a significance level (\(\alpha\)) of 0.05 and compare the calculated F-value (24.06) to the critical F-value. At \(\alpha\) = 0.05 and degrees of freedom of (\(df_1\)=3, \(df_2\)=5), the critical F-value = 6.71. Since our calculated F-value (24.06) is greater than the critical F-value (6.71), we reject the null hypothesis that the population variances are equal. Therefore, there is a significant difference between the population variances of the two species' wing stroke frequencies.
03

c. Inappropriateness of Student's t-test with Pooled Estimator

A Student's t-test with a pooled estimator (such as \(s^2\)) assumes that the population variances are equal between the two groups being compared. Based on the F-test result we obtained in part (b), we have evidence to suggest that the population variances are significantly different between the two species of bees. Using a Student's t-test with a pooled estimator in this scenario would be inappropriate because the assumption of equal variances does not hold. Instead, we should use a statistical test that does not rely on the assumption of equal variances, such as the Welch's t-test, for comparing the mean wing stroke frequencies of the two species of bees.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

F-test for variances
The F-test is a statistical procedure used to compare the variances of two populations to determine if they significantly differ from each other. This test is particularly useful when comparing the variability of two different groups, like the wing stroke frequencies of two species of bees.

The F-test operates by calculating the ratio of the two sample variances. The larger the ratio, the more evidence there is that the two population variances are not equal. The calculation involves determining the sample variances for each group and then dividing the variance of the group with the higher variance by the variance of the other group to get the F-value.

Furthermore, to draw a conclusion, we compare the calculated F-value with the critical F-value from an F-distribution table at a chosen significance level usually, \(\alpha = 0.05\). If the calculated F-value is larger than the critical value, we reject the null hypothesis - in this case, that both populations have equal variances. The example of the bees' wing stroke frequencies demonstrates the application of an F-test, where a significant difference was found between the population variances of the two species.
Student's t-test
The Student's t-test is a common method of statistical hypothesis testing that is used to compare the means of two samples when the sample sizes are small, and the population variances are unknown but assumed to be equal. It is a pivotal tool for determining if there is a significant difference between the mean values from two different populations.

When conducting a t-test, there are different variations to consider, depending on the nature of your data. For equal variances, the pooled t-test is appropriate, which combines the variance estimates from two samples. However, when the variances are significantly different, as indicated by the F-test for the bee species' wing stroke frequencies, the pooled t-test would be inappropriate. In such cases, a separate variance t-test, like the Welch's t-test, should be employed.
Sample variance calculation
Sample variance is a measure of how data points in a specific sample differ from the sample mean. To calculate the variance, you first compute the mean of the sample. Then, subtract the mean from each data point to find the deviation of each point. These deviations are then squared, summed, and divided by the number of data points minus one to account for sample size bias.

The formula for calculating sample variance \(s^2\) is: \[s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}\] where \(x_i\) represents each data point, \(\bar{x}\) is the sample mean, and \(n\) is the number of data points in the sample. The bee wing stroke frequency exercise shows how variances are calculated for both species, demonstrating the process of determining whether we can use a pooled variance estimator in further analysis.
Statistical hypothesis testing
Statistical hypothesis testing is a method used to make decisions based on data analysis, performed through statistical tests. In essence, it enables researchers to determine whether there is enough evidence in a sample of data to infer that a certain condition holds for the entire population.

The process begins with formulating two hypotheses - the null hypothesis (typically denoting no effect or no difference) and the alternative hypothesis (indicating the presence of an effect or difference). An appropriate statistical test is then chosen, like the F-test for variances or the Student's t-test, depending on the data and assumptions. Finally, based on the test results and a pre-set significance level, a decision is made to either reject or not reject the null hypothesis.

In the context of our bee species example, the null hypothesis might assert that there is no difference between the variances of wing stroke frequencies, while the F-test suggests otherwise, leading to the null hypothesis being rejected, concluding that a significant variance difference does indeed exist.

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Most popular questions from this chapter

The test scores on a 100 -point test were recorded for 20 students: $$\begin{array}{lllll}71 & 93 & 91 & 86 & 75 \\\73 & 86 & 82 & 76 & 57 \\\84 & 89 & 67 & 62 & 72 \\\77 & 68 & 65 & 75 & 84\end{array}$$ a. Can you reasonably assume that these test scores have been selected from a normal population? Use a stem and leaf plot to justify your answer. b. Calculate the mean and standard deviation of the scores. c. If these students can be considered a random sample from the population of all students, find a \(95 \%\) confidence interval for the average test score in the population.

To compare the mean lengths of time required for the bodily absorption of two drugs A and \(\mathrm{B}, 10\) people were randomly selected and assigned to receive one of the drugs. The length of time (in minutes) for the drug to reach a specified level in the blood was recorded, and the data summary is given in the table: $$ \begin{array}{ll} \hline \text { Drug A } & \text { Drug B } \\ \hline \bar{x}_{1}=27.2 & \bar{x}_{2}=33.5 \\ s_{1}^{2}=16.36 & s_{2}^{2}=18.92 \\ \hline \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in mean times to absorption for the two drugs? Test using \(\alpha=.05 .\) b. Find the approximate \(p\) -value for the test. Does this value confirm your conclusions? c. Find a \(95 \%\) confidence interval for the difference in mean times to absorption. Does the interval confirm your conclusions in part a?

An experiment was conducted to study the use of \(95 \%\) ethanol versus \(20 \%\) bleach as a disinfectant in removing contamination when culturing plant tissues. The experiment was repeated 15 times with each disinfectant, using cuttings of eggplant tissue. The observation reported was the number of uncontaminated eggplant cuttings after a 4 -week storage. $$ \begin{array}{lcc} \hline \text { Disinfectant } & \text { 95\% Ethanol } & \text { 20\% Bleach } \\\ \hline \text { Mean } & 3.73 & 4.80 \\ \text { Variance } & 2.78095 & .17143 \\ n & 15 & 15 \\ & \text { Pooled variance } 1.47619 & \\ \hline \end{array} $$ a. Are you willing to assume that the underlying variances are equal? b. Using the information from part a, is there evidence of a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested?

Find the tabled value for \(a \chi^{2}\) variable based on \(n-1\) degrees of freedom with an area of a to its right. \(n=10, a=.05\)

Use the information given in Exercises \(2-7\) to find the tabled value for an \(F\) variable based on \(n_{1}-\) I numerator degrees of freedom, \(n_{2}-1\) denominator degrees of freedom with an area of \(n_{1}=25, n_{2}=26, a=.050\)

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