Question

14. In a random sample of 121 household incomes in Winnetka, IL , the average ($1000) is 102.44 with a standard deviation of 14.3. Considering the t with 120 degrees of freedom to be equivalent to the standard normal, a 90% confidence interval for the mean population household income is:

A. (101.14, 103.74)

B. (100.11, 104.77)

C. (100.29, 104.59)

D. (101.21, 101.68)

15. Suppose it is unknown what proportion of employees at It’s-a-Living (IAL) are happy with their job. If we randomly sample 100 and 25 are happy, using the normal approximation to the binomial to compute a 95% confidence interval based on this sample for the proportion who are happy gives what result?

A. It cannot be computed due to inadequate sample size.

B. (0.25, 0.42)

C. (0.165, 0.335)

D. (0.207, 0.293)

What's the solution step??Please!!!!

Answer 1

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