Question

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. Carbohydrates in Candy The number of grams of carbohydrates contained in 1 -ounce servings of randomly selected chocolate and nonchocolate candy is shown. Is there sufficient evidence to conclude that there is a difference between the variation in carbohydrate content for chocolate and nonchocolate candy? Use \(\alpha=0.10 .\) $$ \begin{array}{llllllll}{\text { Chocolate }} & {29} & {25} & {17} & {36} & {41} & {25} & {32} & {29} \\ {} & {38} & {34} & {24} & {27} & {29} & {} & {} \\\ {\text { Nonchocolate }} & {41} & {41} & {37} & {29} & {30} & {38} & {39} & {10} \\ {} & {29} & {55} & {29} & {}\end{array} $$

Short answer

The test does not provide sufficient evidence to conclude that there is a difference in variation in carbohydrate content.

Step-by-step solution

State the Hypotheses

The null hypothesis is that there is no difference between the variances of carbohydrate content in chocolate and nonchocolate candies. This can be expressed as \( H_0: \sigma_1^2 = \sigma_2^2 \). The alternative hypothesis is that the variances are different, expressed as \( H_a: \sigma_1^2 eq \sigma_2^2 \). The claim is that there is a difference in variation.

Find the Critical Value

Since we're comparing variances, we'll use an F-test. Determine the degrees of freedom for each sample: \( df_1 = n_1 - 1 = 13 - 1 = 12 \) for chocolate and \( df_2 = n_2 - 1 = 11 - 1 = 10 \) for nonchocolate. Using an \( \alpha = 0.10 \), look up the critical value for \( F_{0.05,12,10} \) and \( F_{0.05,10,12} \) in the F-distribution table, noting that it's a two-tailed test.

Compute the Test Value

Calculate the sample variances. First, find the mean of each group and then use these to calculate variance. For chocolate, mean \( \mu_1 \) and variance \( s_1^2 \); for nonchocolate, mean \( \mu_2 \) and variance \( s_2^2 \). Compute the test statistic \( F = \frac{s_1^2}{s_2^2} \).

Make the Decision

Compare the computed F-value with the critical F-values obtained in Step 2. If the test value is greater than the critical value (or less than the lower critical value for the two-tailed test), reject the null hypothesis \( H_0 \). Otherwise, do not reject \( H_0 \).

Summarize the Results

Based on the decision from Step 4, summarize the results. If \( H_0 \) is rejected, state that there is sufficient evidence to conclude a difference in variation exists between the carbohydrate content of chocolate and nonchocolate candy. If \( H_0 \) is not rejected, state that there is not sufficient evidence to conclude a difference.

Key concepts

F-test

The F-test is a statistical method used to compare the variances of two populations. It's particularly useful in determining if there is a significant difference between the variability of two groups. In the context of the exercise, the F-test helps us determine if the variance in carbohydrate content is different between chocolate and nonchocolate candies.
To perform an F-test, you need two important pieces of information: the variances of each sample group and their respective degrees of freedom.

• The formula for the F-test statistic is: \( F = \frac{s_1^2}{s_2^2} \) where \( s_1^2 \) is the variance of the first group (chocolate) and \( s_2^2 \) is the variance of the second group (nonchocolate).
• The degrees of freedom are calculated as \( df_1 = n_1 - 1 \) for the first group and \( df_2 = n_2 - 1 \) for the second group.

The F-test essentially helps in testing the null hypothesis that the two sample variances are equal. It's a crucial test in situations where you're interested in understanding variability differences across groups.

Null Hypothesis

The null hypothesis is a foundational concept in hypothesis testing. It represents a statement that there is no effect or no difference in the context of the study. In our exercise, the null hypothesis is that there is no difference in the variances of carbohydrate content between chocolate and nonchocolate candies.
It can be formally stated as: \( H_0: \sigma_1^2 = \sigma_2^2 \) where \( \sigma_1^2 \) and \( \sigma_2^2 \) represent the population variances of chocolate and nonchocolate candies respectively.
By assuming the null hypothesis, we take a conservative stance: we start with the assumption that any observed differences are due to random chance. The goal of hypothesis testing is to use sample data to assess the plausibility of the null hypothesis. If our sample data provide enough evidence against the null hypothesis, we may "reject" it in favor of the alternative hypothesis, which suggests there is a difference. However, if the evidence is not strong enough, we "fail to reject" the null hypothesis.

Critical Value

A critical value is a threshold that determines whether the null hypothesis should be rejected. It's derived from a statistical distribution and corresponds to the chosen significance level, \( \alpha \), which represents the probability of making a Type I error. In the given exercise, \( \alpha \) is set at 0.10.
To find the critical value for an F-test, you'll refer to an F-distribution table using the degrees of freedom from both sample groups. For chocolate, \( df_1 = 12 \), and for nonchocolate, \( df_2 = 10 \). With these degrees of freedom and an \( \alpha = 0.10 \) for a two-tailed test, you find two critical F-values: one for \( F_{0.05,12,10} \) and another for \( F_{0.05,10,12} \).
This two-tailed approach divides the significance level into two regions on either side of the distribution. If the computed test value falls in either of these critical regions, the null hypothesis is rejected. Hence, the critical value plays an essential role in hypothesis testing by marking the boundaries of acceptance or rejection of the null hypothesis.

Two-tailed Test

A two-tailed test is a type of hypothesis test where the area of critical values is split into two equal parts on both ends of the distribution. This test is used when we're interested in detecting differences that could occur in either direction (either higher or lower than the hypothesized value).
In our exercise, since we want to see if there is any difference (not specifically greater or lesser) in variance between chocolate and nonchocolate candies, a two-tailed test is appropriate.

• It tests the null hypothesis against the alternative that there is a difference in either direction.
• The critical region, therefore, has two tails: one on the high end and one on the low end of the F-distribution.

Using a two-tailed test with \( \alpha = 0.10 \), the rejection regions lie in the top 5% and the bottom 5% of the distribution curve. This means if your calculated F-test statistic lies beyond these critical values on either side, it provides sufficient evidence to reject the null hypothesis. The two-tailed nature of the test ensures it's sensitive to differences in both directions, making it suitable for analyzing variance without a directional bias.