Question

1) Express the confidence interval 30.2% ± 7.8% in the form of a trilinear inequality.
__% < p< __ %

2) Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 165 with 52.1% successes. Enter your answer using decimals (not percents) accurate to three decimal places. __ < p < __

3) A test was given to a group of students. The grades and gender are summarized below:

	A	B	C	Total
Male	4	19	16	39
Female	13	17	7	37
Total	17	36	23	76

Let p represent the percentage of all female students who would receive a grade of B on this test. Use a 99% confidence interval to estimate p to three decimal places.
Enter your answer using decimals (not percents). __ < p < __

Answer 1

1) Express the confidence interval 30.2% ± 7.8% in the form of a trilinear inequality.

Answer)

30.2-7.8 < p < 30.2+7.8

22.4% < P < 38%

2) Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 165 with 52.1% successes. Enter your answer using decimals (not percents) accurate to three decimal places.

Answer)

N = 165

P = 0.521

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 85.965

N*(1-p) = 79.035

Both the conditions are met so we can use standard normal z table to estimate the interval

Critical value z from z table for 99.9% confidence level from z table is 3.29

Margin of error (MOE) = Z*√{P*(1-P)}/√N = 0.12795007482

Interval is given by

P-MOE < P < P+MOE

0.393 < P < 0.649