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 calculated the p-value incorrectly. The p-value in this case is actually 0.910 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. C. 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 D. 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.

Step 1 ：The null hypothesis for a true/false test is that the student is guessing, with a proportion of correct answers being \(0.50\).

Step 2 ：A student gets \(4\) out of \(5\) questions correct and claims to know the material because the one-tailed p-value from the one-proportion z-test is \(0.090\), using a significance level of \(0.10\).

Step 3 ：The one-proportion z-test requires a large enough sample size for the normal approximation to be valid, typically at least \(10\) successes and \(10\) failures.

Step 4 ：With only \(5\) questions, the sample size is not large enough to expect at least \(10\) successes and \(10\) failures, indicating an issue with using the one-proportion z-test.

Step 5 ：Calculating the correct p-value using the binomial distribution: successes = \(4\), trials = \(5\), \(p_{null} = 0.5\), resulting in a p-value of \(0.1875\).

Step 6 ：The student's reported p-value of \(0.090\) and the suggested incorrect p-value of \(0.910\) are both different from the calculated p-value of \(0.1875\).

Step 7 ：Conflating statistical significance with practical significance does not directly address the issue with the student's approach to the one-proportion z-test.

Step 8 ：Using a confidence interval instead of a hypothesis test is not inherently wrong, but it does not address the issue with the sample size for the one-proportion z-test.

Step 9 ：The correct answer is that the sample size is not large enough to use the one-proportion z-test, as indicated by option C.

Step 10 ：The final answer is \(\boxed{C}\)