Let's go through each part of this problem step-by-step.

Part 1: State the Hypotheses

The problem asks us to test if the variance of Sample 1 ($\sigma_1^2$) is greater than or equal to the variance of Sample 2 ($\sigma_2^2$). This leads to the following hypotheses:

• Null Hypothesis ($H_0$): $\sigma_1^2 \leq \sigma_2^2$
• Alternative Hypothesis ($H_A$): $\sigma_1^2 > \sigma_2^2$

This matches Answer Choice B.

Part 2: Determine the Test Statistic (F-Test)

To determine the test statistic $F$, we use the formula:

$F = \frac{s_1^2}{s_2^2}$

where:

• $s_1^2$ is the sample variance of Sample 1
• $s_2^2$ is the sample variance of Sample 2

Step 1: Calculate Sample Variances

For each sample, we calculate the sample mean, then the sample variance.

Sample 1 Data: $11.1, 13.5, 12.2, 11.5, 14.8$

1. Mean of Sample 1: $\bar{x}_1 = \frac{11.1 + 13.5 + 12.2 + 11.5 + 14.8}{5} = 12.62$

2. Variance of Sample 1: $s_1^2 = \frac{(11.1 - 12.62)^2 + (13.5 - 12.62)^2 + (12.2 - 12.62)^2 + (11.5 - 12.62)^2 + (14.8 - 12.62)^2}{5 - 1} = 2.29$

Sample 2 Data: $10.1, 8.5, 9.2, 7.4, 11.8$

1. Mean of Sample 2: $\bar{x}_2 = \frac{10.1 + 8.5 + 9.2 + 7.4 + 11.8}{5} = 9.4$

2. Variance of Sample 2: $s_2^2 = \frac{(10.1 - 9.4)^2 + (8.5 - 9.4)^2 + (9.2 - 9.4)^2 + (7.4 - 9.4)^2 + (11.8 - 9.4)^2}{5 - 1} = 2.47$

Step 2: Calculate the $F$ Statistic

$F = \frac{s_1^2}{s_2^2} = \frac{2.29}{2.47} \approx 0.93$

Part 3: Determine the Critical Value

Since we have a one-tailed test with $\alpha = 0.05$, we need the critical $F$-value for a significance level of 0.05. With degrees of freedom $df_1 = 4$ and $df_2 = 4$, the critical value can be found from $F$-distribution tables or software. For $df_1 = 4$ and $df_2 = 4$ at the 0.05 significance level:

$F_{\text{critical}} \approx 6.39$

Part 4: Make a Decision

The decision rule for an $F$-test is:

• If $F \geq F_{\text{critical}}$, reject $H_0$.
• If $F < F_{\text{critical}}$, do not reject $H_0$.

Since $F = 0.93 < 6.39$, we do not reject the null hypothesis.

Therefore, the correct answer is D: Do not reject the null hypothesis. There is insufficient evidence to conclude that the first variance is greater than the second variance.

Summary

• Hypotheses: Answer B
• Test Statistic: $F = 0.93$
• Critical Value: $F_{\text{critical}} = 6.39$
• Decision: Do not reject the null hypothesis, Answer D

Would you like more details on any step, or do you have any questions?

Follow-up Questions

1. How would the conclusion change if the calculated $F$-statistic were greater than the critical value?
2. What is the impact of sample size on the $F$-test for equality of variances?
3. Can this test be used if the data does not follow a normal distribution?
4. How would we approach the test if we wanted a two-tailed hypothesis instead?
5. What alternative tests exist for comparing variances if the $F$-test assumptions are not met?

Tip

When using the $F$-test for variances, always check if the data meets the normality assumption, as the test is sensitive to non-normal data.