Download Hypothesis Testing and P-values in ECON 310: Solutions for Week 13 - Prof. Ignacio H. Monz and more Exams Economic statistics in PDF only on Docsity! ECON 310. TA SECTION - WEEK 13 - SOLUTIONS prepared by SANG YOON (TIM) LEE and IGNACIO MONZÓN Hypothesis Testing P-values Let {Xi}ni=1 be a sequence of i.i.d. trials with unknown mean µ. Let X̄n be the sample mean (the estimator) and let x̄ be the realized sample mean (the estimate). Our tests is given by: H0 : µ = µ0 H1 : µ > µ0 Then, the p-value is given by: p = Pr (X̄n ≥ x̄ | µ = µ0) Interpretation: “Assume that the null hypothesis is true. If so, what is the ex-ante probability of obtaining a value for the estimator that exceeds our estimate?” How to calculate p-values We turn our distribution into a standard normal (note that we are using the Central Limit Theo- rem), p = Pr (X̄n ≥ x̄ | µ = µ0) = Pr ( X̄n − µ0 s√ n ≥ x̄− µ0s√ n | µ = µ0 ) = Pr ( Z ≥ x̄− µ0s√ n ) As an example, let n = 100, s = 10 and µ0 = 1. Then, p = Pr ( Z ≥ x̄− 110√ 100 ) = Pr (Z ≥ x̄− 1) Table 1 shows p-values for different values of x̄ x̄ 1 2 3 4 p-values 50% 16% 2% 0.1% Table 1: p-values 1 Exercises 1. (VS17.13). The dean of a large business school wants to convince incoming students to major in business. He samples a random group of 82 graduating business majors. The mean starting salary of the business majors was $59, 430 with a standard deviation of $13, 450. This seems to compare favorably to the national average salary of all college graduates: $48, 920. (a) Can we say that the mean salary of business majors is greater than the national average? (b) Can we say that it is at least $10, 000 more? (c) Compute the p-values for both tests Solutions: (a) In this case, our test is as follows: H0 : µ = $48, 920 H1 : µ > $48, 920 Then, under a 5% significance level, the critical value is given by: c = µ0 + zα σ√ n = $48, 920 + 1.65 · $13, 450√ 82 = 51, 371 As a result, we believe that it is significantly greater than the national average. (b) In this case, our test is as follows: H0 : µ = $58, 920 H1 : µ > $58, 920 Then, under a 5% significance level, the critical value is given by: c = µ0 + zα σ√ n = $58, 920 + 1.65 · $13, 450√ 82 = 61, 371 As a result, we believe that it is not $10, 000 higher than the the national mean. (c) We recall that the p-value is given by, p = Pr ( Z ≥ x̄− µ0s√ n ) Consequently, in the first case, p = Pr Z ≥ $59, 430− $48, 920$13,450√ 82 = Pr (Z ≥ 7.08) ' 0 Consequently, in the second case, p = Pr Z ≥ $59, 430− $58, 920$13,450√ 82 = Pr (Z ≥ 0.34) = 0.37 2
