The null hypothesis on true/false tests is that the student is guessing, and the proportion of right answers is 0.50. A student taking a five-question true/false quiz gets 4 right out of 5. She says that this shows that she knows the material, because the one-tailed p-value from the one-proportion z-test is 0.090, and she is using a significance level of 0.10. What is wrong with her approach? Choose the correct answer below. A. The student used a hypothesis test instead of a confidence interval. A confidence interval would show the range of possible values for her proportion of correct answers. If this interval does not include 0.50, there would be evidence she knows the material. B. The student is conflating statistical significance with practical significance. If her true proportion of correct answers is, say, 0.52, this does not show that she knows the material. O C. The student calculated the p-value incorrectly. The p-value in this case is actually 0.910. D. The sample size is not large enough to use the one-proportion z-test. To use that test, the tester must expect at least 10 successes and at least 10 failures. NextAccording to a reputable magazine, the dropout rate for all college students with loans is 30%. Suppose that 99 out of 300 random college students with gans drop out. Complete parts (a) and (b) below, Click the icon to view a graphical technology output for this situation. . Give the null and alternative hypotheses to test that the dropout rate is not 30%. (Type integers or decimals. Do not round.) b. Report the test statistic (z) from the output given. (Type an integer or a decimal. Round to four decimal places as needed.) X Technology Output FPropZTest - 1.1239 16 = 0 2568 Next Print