Question

Can anyone please help me with my  Confidence Intervals for Proportions homework? Thank you so much!

1. Express the confidence interval 38.6%±9%38.6%±9% in interval form.
Express the answer in decimal format (do not enter as percents).

2. We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 26 had kids. Based on this, construct a 95% confidence interval for the proportion, p, of adult residents who are parents in this county.

Give your answers as decimals, to three places.

_____________< p <__________________

3.  Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 144 with 58.3% successes at a confidence level of 80%.

M.E. = ______________%

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places. Round final answer to one decimal place

4. Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	11	8	7	26
Female	17	4	15	36
Total	28	12	22	62

Let π represent the percentage of all male students who would receive a grade of B on this test. Use a 85% confidence interval to estimate p to three decimal places.

Enter your answer as a tri-linear inequality using decimals (not percents).

______________ < p < ________________

5. A political candidate has asked you to conduct a poll to determine what percentage of people support her.
If the candidate only wants a 3% margin of error at a 95% confidence level, what size of sample is needed?

Give your answer in whole people.

Answer 1

0 Lower Limit = 38.6% - 9% = 0.386 – 0.09 = 10.2967 upper limit = 38.6% + 9% = 0.386 +0.09 = 10.476 ) ... Confidence Intered f79.296, 0.+76) @) 98% cs for p = § 1996 CI-P h f = t = 26 = 0.26 95% CI = 0.26 + 1.96 0.26x0.74 100 = 0.26 + 0.087 = (0.174, 0.346 ) : 10.174]< p < 10:34 6]