Question

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. 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 \(\alpha=0.10 .\) $$ \begin{array}{lllllllll} \text { Chocolate } & 29 & 25 & 17 & 36 & 41 & 25 & 32 & 29 \\ & 38 & 34 & 24 & 27 & 29 & & & \\ \text { Nonchocolate } & 41 & 41 & 37 & 29 & 30 & 38 & 39 & 10 \\ & 29 & 55 & 29 & & & & & \end{array} $$

Short answer

There is no significant difference in variance between chocolate and nonchocolate candy carb content.

Step-by-step solution

State the Hypotheses

We are testing for a difference in the variance of carbohydrate content between chocolate and nonchocolate candy. The null hypothesis (H_0) states that there is no difference in the variances: \( H_0: \sigma_1^2 = \sigma_2^2 \).The alternative hypothesis (H_1) states that there is a difference: \( H_1: \sigma_1^2 eq \sigma_2^2 \).The claim is that there is a difference in the variation of carbohydrates.

Identify the Critical Value

Since we are testing for a difference in variances and the alternative hypothesis is two-tailed, we will use the F-distribution. Given \( \alpha = 0.1 \), the critical values can be found using F-distribution tables or software.The degrees of freedom (df) for chocolate is \( n_1 - 1 = 14 - 1 = 13 \). The degrees of freedom for nonchocolate is \( n_2 - 1 = 11 - 1 = 10 \).For a two-tailed test, find the critical F-value using these degrees of freedom and \( \alpha \). The critical F-values are approximately \( F_{0.05, 13, 10} \approx 3.20 \) and \( F_{0.95, 13, 10} \approx 0.31 \).

Compute the Test Value

Calculate the sample variances for both chocolate and nonchocolate candy.The sample variance is calculated as:\[ s^2 = \frac{\sum{(x - \bar{x})^2}}{n-1} \]For chocolate:\( \bar{x}_1 = \frac{29 + 25 + 17 + 36 + 41 + 25 + 32 + 29 + 38 + 34 + 24 + 27 + 29}{13} = 30 \)\( s_1^2 = \frac{383}{13} \approx 29.46 \)For nonchocolate:\( \bar{x}_2 = \frac{41 + 41 + 37 + 29 + 30 + 38 + 39 + 10 + 29 + 55 + 29}{11} = 34.09 \)\( s_2^2 = \frac{932.82}{10} \approx 93.28 \)The test value \( F = \frac{s_1^2}{s_2^2} = \frac{29.46}{93.28} \approx 0.32 \).

Make the Decision

Compare the calculated test value with the critical values determined before:The test value \( F = 0.32 \) is greater than the lower critical value \( 0.31 \) and less than the upper critical value \( 3.20 \).Since the test value falls between the critical values, we fail to reject the null hypothesis \( H_0 \).

Summarize the Results

There is not enough evidence at the \( \alpha = 0.10 \) significance level to conclude that there is a difference in the variances of carbohydrate content between chocolate and nonchocolate candy. The claim that there is a difference in variation is not supported by the data.

Key concepts

Critical Value

In the context of hypothesis testing, the critical value is essentially a threshold that helps you decide whether to accept or reject the null hypothesis. It serves as a comparison point for your test statistic. Think of it as the dividing line between the regions where you'll consider the evidence significant or not.

For this exercise, we are comparing two datasets representing chocolate and nonchocolate candy. Given that we are assessing variance differences with a two-tailed test, our critical values are derived from the F-distribution. With the significance level, \( \alpha = 0.10 \), we check F-distribution tables or software to find the bounds.

• Degrees of freedom (df) for chocolate was 13.
• Degrees of freedom (df) for nonchocolate was 10.

Two-tailed critical values using these degrees of freedom and significance level ended up being approximately 0.31 and 3.20. Any test statistic that falls outside these values would be reason to reject the null hypothesis.

Test Value

The test value in this problem is calculated using the variances of the two groups being studied, chocolate and nonchocolate candy. In our case, the test value comes from the ratio of the sample variances. Calculating this value involves simply arranging the formula:\[ F = \frac{s_1^2}{s_2^2} \]
Where \( s_1^2 \) and \( s_2^2 \) represent the sample variances of chocolate and nonchocolate candies, respectively.

From the given data:

• Variance for chocolate candies was approximately 29.46.
• Variance for nonchocolate candies was calculated to be around 93.28.

Plug these numbers into the formula to find the F test value:
\[ F = \frac{29.46}{93.28} \approx 0.32 \] This result aids us in determining whether the variances are significantly different by comparing it against the earlier established critical values.

F-distribution

The F-distribution is critical in comparing variances because it's tailored for variance ratio tests. It’s asymmetric and stretches from 0 to infinity, with the exact shape hinging on two sets of degrees of freedom, one for each group.

In this particular test, the F-distribution helped determine if the variability (by comparing variances) in carbohydrate content between chocolate and nonchocolate candy was statistically significant.

The methodology behind using the F-distribution requires defining two parameters: \(df_1 \) for the numerator (chocolate candy) and \(df_2 \) for the denominator (nonchocolate candy).

• For chocolate candy: \( df_1 = 13 \).
• For nonchocolate candy: \( df_2 = 10 \).

Comparing variances using the F-distribution is sensitive to assumptions of normality. Deviations could influence the test's sensitivity and lead to incorrect conclusions about variance differences.

Variance

Variance is a key statistical measure that indicates the degree of spread in a dataset. Essentially, it tells you how much the data points in each group deviate from their respective means. A larger variance signifies more spread, while a smaller variance indicates that the data points cluster closely around the mean.

In this study, we calculated the variance for both the chocolate and nonchocolate candy datasets.

• The formula used was: \( s^2 = \frac{\sum{(x - \bar{x})^2}}{n-1} \).
• Chocolate candy had a variance of approximately 29.46.
• Nonchocolate candy's variance was about 93.28, indicating much more spread.

Understanding variance helps in determining whether the differences observed are due to random chance or because of underlying differences in the populations being compared. Identifying substantial differences can point to factors affecting carbohydrate content variability in these candies.