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The following contingency table shows the distribution of grades earned by students taking a midterm exam in an MBA class, categorized by the number of

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The following contingency table shows the distribution of grades earned by students taking a midterm exam in an MBA class, categorized by the number of hours the students spent studying for the exam. Complete parts a. and b. below. Click the icon to view the contingency table. a. Using =0.01, perform a chi-square test to determine if a student's grade on the exam and the hours spent studying for it are independent of one another. Identify the null and altemative hypotheses for a chi-square test of independence based on the information in the table. Choose the correct answer below. A. H0 : Grade and time spent studying are not independent of one another. H1 : Grade and time spent studying are independent of one another. B. H0 : The variables A,B,C, less than 3 hours, 35 hours, and more than 5 hours are independent. H1 : At least one of the variables is not independent. C. H0 : Grade and time spent studying are independent of one another. H1 : Grade and time spent studying are not independent of one another. D. H0 : The variables A, B, C, less than 3 hours, 3-5 hours, and more than 5 hours are independent. H1 : None of the variables are independent. Find the chi-square test statistic. 2= (Round to two decimal places as needed.) Contingency table The following contingency table shows the distribution of grades earned by students taking a midterm exam in an MBA class, categorized by the number of hours the students spent studying for the exam. Complete parts a. and b. below. Click the icon to view the contingency table. tecnnoiogy output, rounaing to two decimai piaces. Since the expected frequency for each cell exceeds 5, the chi-square test can proceed. Enter the chi-square test statistic from the technology output, rounding to two decimal places. x2=3.25 Enter the p-value from the technology output, rounding to three decimal places. p-value=0.517 Determine the result of the test. Compare the p-value to the significance level, =0.05, and determine if the null hypothesis should be rejected or not

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