Let p be the proportion of UCSD students who live on campus. You conduct a hypothesistest to test Ho: p = 0.7 against HA P0.7. The p-value for the test is 0.06. If you had considered theone-sided alternative HA :p> 0.7, the p-value would have been:a) 0.03.b) 0.03 or 0.97.c) 0.06.d) 0.97.e) None of the above.

Introduction: Denote p as the true population proportion.Explanation: The initial null and alternative hypotheses were: H0: p = 0.7; HA: p ≠ 0.7. The test was two-tailed, and it yielded a p-value of 0.06, that is, p-value (two-tailed) = 0.06. The z-test for one proportion must have been used. For a test statistic value of z*, the p-value for the two-tailed test is, p-value (two-tailed) = P (| Z | ≥ | z* |). There may be two different possibilities: either z* is positive, or z* is negative. Let us consider the cases one by one. If z* is positive, let z* = z. Then, P (| Z | ≥ | z* |) = P (Z ≥ z) + P (Z ≤ –z) = 2P (Z ≥ z). This is because, due to symmetry of the standard normal random variable Z about 0, P (Z ≥ z) = P (Z ≤ –z). If z* is negative, let z* = –z. Then, P (| Z | ≥ | z* |) = P (Z ≥ z) + P (Z ≤ –z) = 2P (Z ≤ –z). Note that, since P (Z ≥ z) = P (Z ≤ –z), 2P (Z ≥ z) = 2P (Z ≤ –z), so that the p-value for the two-tailed test is the same for both a positive and a negative test statistic value, as long as the magnitude of the test statistic is the same. Since p-value (two-tailed) = 0.06, here, we must have, either 2P (Z ≤ –z) = 0.06 (if test statistic is negative), or 2P (Z ≥ z) = 0.06 (if test statistic is positive). Now, the alternative hypothesis is changed and taken to be, HA: p > 0.7. Otherwise, the other values associated with the test remain the same. This is the alternative hypothesis for a right-tailed test. For the right-tailed test, the p-value is of the form, p-value (right-tailed) = P (Z ≥ z*). For a positive z* = z, p-value (right-tailed) = P (Z ≥ z*) = P (Z ≥ z). If, in the previous two-tailed testing setup, z* was positive, the two-tailed p-value was, p-value (two-tailed) = P (| Z | ≥ | z* |) = 2P (Z ≥ z). As a result, p-value (right-tailed) = P (Z ≥ z) = [2P (Z ≥ z)] / 2 = 0.06/2 = 0.03. For a negative z* = z, p-value (right-tailed) = P (Z ≥ z*) = P (Z ≥ –z). Now, P (Z ≥ –z) = 1 – P (Z ≤ –z). As a result, P (Z ≤ –z) = 1 – P (Z ≥ –z). If, in the previous two-tailed testing setup, z* was negative, the two-tailed p-value was, p-value (two-tailed) = P (| Z | ≥ | z* |) = 2P (Z ≤ –z) = 0.06. Then, P (Z ≤ –z) = 0.06/2 = 0.03. This implies that, P (Z ≥ –z) = 1 – P (Z ≤ –z) = 1 – 0.03 = 0.97. As a result, p-value (right-tailed) = P (Z ≥ –z) = 0.97. Since it is not known whether the test statistic was positive or negative, the p-value for the right-tailed alternative, HA: p > 0.7 can wither be 0.03, or 0.97. Hence, for the one-sided alternative, HA: p > 0.7, the p-value would have been 0.03 or 0.97. Option b is correct.