Chapter 7. Hypothesis Testing with One Sample

204 Chapter 7: Hypothesis Testing with One Sample38. Type I: Conclude there is sufficient evidence to support that the proportion of college graduate who smoke isless than 0.27 when in actuality p = 0.27Type II: Fail to reject that the proportion of college graduate who smoke is 0.27 when in actuality p < 0.27.39. Type I: Conclude there is sufficient evidence to support that the proportion of fatal commercial aviationcrashes is different than 0.038 when in actuality p = 0.038Type II: Fail to reject that the proportion of fatal commercial aviation crashes is 0.038 when in actuality p ≠0.038.40. Type I: Conclude there is sufficient evidence to support that the proportion of M&Ms that are blue is differentthan 0.10 when in actuality p = 0.10Type II: Fail to reject that the proportion of M&Ms that are blue is 0.10 when in actuality p ≠ 0.10.41. The alternate hypothesis is H 1 p > 0.50, which makes this a right tailed test. The sample proportion isp ˆ = 0.27, which is less than 0.50. The test statistic is then negative. The critical region is to the right of z = 0,and so the test statistic cannot fall in the critical region, and so there is no chance to reject H 0 .42. Not necessarily. Since the significance 0.05 is larger than the significance 0.01, the critical values for α = 0.05are closer to z = 0 than those for α = 0.01. It is possible that the critical value for the hypothesis test fellbetween these critical values, and so was rejected at the 5% level and not at the 1% level. It is also possiblethat the test statistic was within the critical region for α = 0.01 also, and so it may be that null hypothesis wasrejected at both significance levels.43. We follow the steps outlined in the problem:a. Step 1The significance level is 5%, this is a left tailed test, so the critical value is −z 0.05=−1.645.Step 2Solving for p ˆ , and substituting the values p=0.5, n = 1998, and q = 0.5, we findpq0.5⋅ 0.5p ˆ = p − z .05= 0.5−1.645n 1998 = 0.48Step 3We now find the area to the left of p ˆ , using the z-score based on a population proportion p = 0.45.z = p ˆ − p 0.48 − 0.45= = 2.70pq 0.45⋅ 0.55n 1998The area to the left of 2.70 is 0.9965, which is the power of the test. In this hypothesis test, it would be verylikely that we reject H 0 when the true population proportion is p = 0.45.b. β = (1 – 0.9965) = 0.003544. We will solve the problem by first solving the z score for p ˆ using the critical value for α = 0.05 and the nullhypothesis value for p = 0.04. Doing so, we see thatpq0.4 × 0.6p ˆ = p + z = 0.4 −1.645 = 0.4 − 0.80588 .n nnFor a power of 0.80, we need to find the number with 0.80 area to its left under the standard normal curve.Table A-2 gives the value z = -0.84.Now, using the alternate value for p = 0.3, we again solve for p ˆ .pq0.3× 0.7p ˆ = p + z = 0.3− 0.84 = 0.3− 0.38494 .n nnSetting the left hand sides equal to one another, we get the following equation.0.3− 0.38494 = 0.4 − 0.80589 .nn