For this study, we should use X^2-test Alpha = 0.5 H0: the final grades are independent from the genders H1: there is an association between genders and the final grades This is the observed frequency table: A B C Male: 14.0 29.3 41.7 Female: 21.0 43.7 62.3 The test-statistic for this data(show the answer to 3 decimal places) = ? The p-value for this sample (note it is greater then alpha. Show the answer to 4 decimal places) = ?

Step 1 ：Define the null hypothesis (H0) and the alternative hypothesis (H1). H0: the final grades are independent from the genders. H1: there is an association between genders and the final grades.

Step 2 ：Set the significance level (alpha) to 0.5.

Step 3 ：Create the observed frequency table: \[\begin{array}{ccc} & A & B & C \\ Male & 14.0 & 29.3 & 41.7 \\ Female & 21.0 & 43.7 & 62.3 \end{array}\]

Step 4 ：Perform a Chi-Square test of independence. The test statistic for a Chi-Square test of independence is calculated as: \[X^2 = \sum \frac{(O-E)^2}{E}\] where: O is the observed frequency, E is the expected frequency under the null hypothesis and is calculated as (row total * column total) / sample size.

Step 5 ：Calculate the p-value, which is the probability that you would observe such an extreme test statistic if the null hypothesis were true. It can be calculated using a Chi-Square distribution.

Step 6 ：Compare the p-value with the significance level (alpha). If the p-value is less than alpha, reject the null hypothesis.

Step 7 ：Final Answer: The final answer will be the test statistic and the p-value. The test statistic will be a single number, and the p-value will be a probability between 0 and 1. The exact values will depend on the observed frequencies given in the problem. If the p-value is less than the significance level (alpha), we reject the null hypothesis. \(\boxed{\text{Test statistic}}\), \(\boxed{\text{p-value}}\)