Question

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

a)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Но: д %3 2.34 На: д < 2.34 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left tailed test is te = -1.663 The rejection region for this left-tailed test is R =t:t<-1.663 (3) Test Statistics The t-statistic is computed as follows: X t = 2.10 2.34 5.891 0.38/87 (4) Decision about the null hypothesis Since it is observed that t -5.891 < te = -1.663, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p 0, and since p 0 < 0.05, it is concluded that the null hypothesis is rejected (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean is less than 2.34, at the 0.05 significance level.

b)

since we reject null hypothesis

we could have made type I error

c)

minimum average = mu - t * s/sqrt(n)

= 2.34 - 1.663 * 0.38/sqrt(87)

= 2.2722

d)

type ii error = fail to reject the null when null is false

P(Xbar > 2.27| mu = 2 ,sigma = 0.4)

= P(Z > (2.24 - 2)/(0.4/sqrt(87))

= P(Z > 5.5964 )

= 0