Question

5. According to the 2010 US Census, the average number of residents per housing unit for the n=87 counties in Minnesota was 2.10, and the standard deviation was 0.38. Test whether the true mean number of residents per housing unit in Minnesota in 2010 is less than the national value of 2.34 at the level α = 0.05.

a. Show all five steps of this test.

b. What type of error could we be making in this context?

c. What is the minimum average number of residents per household needed in order to fail to reject H0? Assume the sample standard deviation is the same.

d. Suppose the true number of residents per household in Minnesota is normally distributed with a mean of 2.0 and standard deviation of 0.4. Suppose we reject null hypothesis if the sample mean number of residents is less than 2.27. What is the probability of making a type II error?

Answer 1

5. a. Let u = the true mean number of residents per housing unit in Minnesota. Null hypothesis, Ho : u = 2.34 vs. Alternative hypothesis, He: u < 2.34 vnt - 2.34) = -5.8910 Test statistic = t - score = where, n = sample size = 87, T = sample mean = 2.10, s = sample sd = 0.38 Critical value = -0.05.86 = -1.6628 Decision rule : We reject Ho at 5% level of significance if t - score < Critical value. Otherwise accept it. Decision/Conclusion : Since here, t-score < Critical value so we reject Ho at 5% level of significance and conclude that true mean number of residents per housing unit in Minnesota in 2010 is less than the national value of 2.34 at the level = 0,05.

6. Since we reject H, but in reality Ho may be true hence here Type I error is committed c. If t-score > -1.6628 then we fail to reject Ho at 5% level of significance 1877 – 2.34) = 877-2.34)/0.38 > -1.6628 or, - or, 1 > 2.3123

The minimum average number of residents per household needed in order to fail to reject H₀ =2.3123

d. X = x; + N(A, 0.42 87 = 0.0018) i=1 We reject H, if i < 2.27 P(Type II error p = 2) = P(X > 2.27 p = 2) =1-0((2.27 – 2)/V0.0018) = 1- 0(6.3640) 0.