91Ó°ÊÓ

First, we need to perform an F-test for the hypothesis that the variances are equal. The null hypothesis (\(H_0\)) is that the variances are equal, and the alternative hypothesis (\(H_1\)) is that the variances are not equal. \(H_0: \sigma^{2}_{1} = \sigma^{2}_{2}\) \(H_1: \sigma^{2}_{1} \neq \sigma^{2}_{2}\) To calculate the F-test statistic, divide the larger variance by the smaller variance: \(F = \frac{\text{larger variance}}{\text{smaller variance}}\) For our data, the larger variance is 2.78095 (from the 95% ethanol disinfectant), and the smaller variance is 0.17143 (from the 20% bleach disinfectant): \(F = \frac{2.78095}{0.17143} \approx 16.229\) Next, look up the critical values of F using the F-distribution table, with degrees of freedom (df) equal to the sample sizes minus 1: df1 = \(15-1=14\) df2 = \(15-1=14\) The critical values of F for 95% confidence level are 0.387 and 2.576. Since our test statistic (16.229) falls outside this range, we reject the null hypothesis. Therefore, we are not willing to assume that the underlying variances are equal.

To determine if there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested, we will perform a t-test for independent samples with unequal variances (assuming unequal variances based on the F-test result from step 1). To perform the t-test, first state the null and alternative hypotheses: \(H_0: \mu_1 = \mu_2\) \(H_1: \mu_1 \neq \mu_2\) Next, calculate the pooled variance estimate (\(s^{2}_p\)), the test statistic (t), and the degrees of freedom (df): \(s^{2}_p = \frac{(n_1-1)s^{2}_{1} + (n_2-1)s^{2}_{2}}{n_1+n_2-2}\) \(t = \frac{(\bar{x}_{1} - \bar{x}_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\frac{s^{2}_{1}}{n_1} + \frac{s^{2}_{2}}{n_2}}}\) \(df = \frac{(\frac{s^{2}_{1}}{n_1} + \frac{s^{2}_{2}}{n_2})^{2}}{\frac{(\frac{s^{2}_{1}}{n_1})^{2}}{n_1-1} + \frac{(\frac{s^{2}_{2}}{n_2})^{2}}{n_2-1}}\) Using the given data: \(s^{2}_p = \frac{(15-1)(2.78095) + (15-1)(0.17143)}{15+15-2} \approx 1.47619\) \(t = \frac{(3.73 - 4.80) - (0)}{\sqrt{\frac{2.78095}{15} + \frac{0.17143}{15}}} \approx -4.939\) \(df = \frac{(\frac{2.78095}{15} + \frac{0.17143}{15})^{2}}{\frac{(\frac{2.78095}{15})^{2}}{14} + \frac{(\frac{0.17143}{15})^{2}}{14}} \approx 14.362\) Looking up the t-distribution table for 95% confidence level, with df = 14 (rounded down), we find the critical values to be -2.145 and 2.145. Since our test statistic (-4.939) falls outside this range, we reject the null hypothesis. Therefore, we are willing to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants tested.