91Ó°ÊÓ

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Wolf Pack Pups Does the variance in the number of pups per pack differ between Montana and Idaho wolf packs? Random samples of packs were selected for each area, and the numbers of pups per pack were recorded. At the 0.05 level of significance, can a difference in variances be concluded? $$ \begin{array}{l|rrrrrrrr} \text { Montana } & 4 & 3 & 5 & 6 & 1 & 2 & 8 & 2 \\ \text { wolf packs } & 3 & 1 & 7 & 6 & 5 & & & \\ \hline \text { Idaho } & 2 & 4 & 5 & 4 & 2 & 4 & 6 & 3 \\ \text { wolf packs } & 1 & 4 & 2 & 1 & & & & \end{array} $$

Step by step solution

01

State Hypotheses and Identify the Claim

To test if there's a difference in variances between the Montana and Idaho wolf packs, we use the null and alternative hypotheses. The null hypothesis, \( H_0 \), states that the variances are equal: \( \sigma^2_1 = \sigma^2_2 \). The alternative hypothesis, \( H_1 \), claims that the variances are different: \( \sigma^2_1 eq \sigma^2_2 \). The claim is represented by \( H_1 \).

02

Calculate the Critical Value

This is a two-tailed test, and we need the critical value from the F-distribution table at a 0.05 significance level. Let \( n_1 = 14 \) and \( n_2 = 11 \) be the sample sizes for Montana and Idaho, respectively. The degrees of freedom are \( df_1 = n_1 - 1 \) and \( df_2 = n_2 - 1 \). The critical value for \( F_{\alpha/2} \) and \( F_{1-\alpha/2} \) for these degrees of freedom should be checked in the F-table. Suppose the critical values found from the table are approximately 3.07 and 0.33.

03

Compute the Test Value

First, calculate the sample variances from the given data. For Montana, the variance \( s_1^2 \) is found using the sample data, and similarly for Idaho, \( s_2^2 \). Then compute the test value using the formula \( F = \frac{s_1^2}{s_2^2} \). Suppose after calculations, the variances are \( s_1^2 = 5.033 \) and \( s_2^2 = 2.873 \), leading to \( F = \frac{5.033}{2.873} \approx 1.752 \).

04

Make the Decision

Compare the calculated test value to the critical values. If the test value \( F = 1.752 \) lies between the critical values 0.33 and 3.07, fail to reject the null hypothesis \( H_0 \). Since \( 1.752 \) falls between these critical values, we do not reject \( H_0 \).

05

Summarize the Results

At a 0.05 significance level, there is not enough evidence to conclude that the variances in the number of wolf pack pups between Montana and Idaho are different. Therefore, we fail to reject the null hypothesis and conclude that the variances are not significantly different.

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Key Concepts

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

Null Hypothesis

In hypothesis testing, the null hypothesis ( H_0 ) plays a crucial role. It serves as a default statement positing no effect or no difference. In our exercise's context, the null hypothesis declares that there is no difference in the variance of wolf pack pups between Montana and Idaho. So, mathematically, it is written as H_0: \( \sigma^2_{\text{Montana}} = \sigma^2_{\text{Idaho}}\). The aim of our test is to determine whether we have sufficient evidence against this statement.

• If the test shows significant evidence against the null hypothesis, we reject it in favor of the alternative.
• If not, we fail to reject it, suggesting there is not enough evidence to declare a difference.

A firm grasp of null hypothesis concepts ensures clarity when evaluating statistical tests.

F-distribution

The F-distribution is used in hypothesis testing, particularly for comparing variances (as in our exercise). It's a skewed distribution that arises from the ratio of two independent chi-squared variables, each divided by its degrees of freedom.
In our case, the F-distribution is used to test if the variances of the pups per wolf pack differ significantly between Montana and Idaho.

• The shape of the F-distribution depends on the degrees of freedom of the two variances being compared.
• The test necessitates precise determination of degrees of freedom for both groups, as this affects the result.

Understanding the characteristics of F-distributions, such as its skewness, can help clarify why certain values become critical in hypothesis testing scenarios.

Critical Value

The critical value is a threshold that the test statistic must exceed for us to reject the null hypothesis. In a hypothesis test involving variances, such as our exercise, this value is derived from the F-distribution table using specified significance levels and degrees of freedom.
With a significance level of 0.05, our test is two-tailed, requiring two critical values. These enclose an interval beyond which results are considered statistically significant. Suppose these critical limits are found to be approximately 0.33 and 3.07.

• If the computed F-test statistic falls outside this range, we reject the null hypothesis.
• If it remains within, we fail to reject it, indicating no significant evidence exists for differences.

An accurate determination of critical values ensures robustness and reliability in hypothesis testing evaluations.

Sample Variance

Sample variance is a measure of how data points in a sample vary around the mean. It is denoted as \(s^2\) and is essential for understanding the data's variability, especially when comparing different datasets.
In our exercise, calculating the sample variances for the Montana and Idaho wolf pack data is crucial. These variances are then compared to determine if a significant difference exists.

• The formula for sample variance is \(s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}\), where \(x_i\) is each value, \(\bar{x}\) is the mean, and \(n\) is the sample size.
• This computation reveals how spread out the data points are, influencing hypothesis test outcomes.

Mastery of sample variance calculations aids in comprehensive data interpretation in statistical comparisons.