This Question: 1 pt 5 of 15 (0 complete) This Test: 15 pts possibl What proporti

Question

This Question: 1 pt 5 of 15 (0 complete) This Test: 15 pts possibl What proportion of people get most of their news on the Internet? According to a recent poll, 40% get most of their news from the Internet. Complete parts (a) through (e) a. To conduct a follow-up study that would provide 95% confidence that the point estimate is correct to within ±0.05 of the population proportion, how large a sample size is required? A sample size ofpeople is required (Round up to the nearest integer.) b. To conduct a follow-up study that would provide 99% confidence that the point estimate is correct to within 0.05 of the population proportion, how many people need to be sampled? A sample size of people is needed (Round up to the nearest integer.) c. To conduct a follow-up study that would provide 95% confidence that the point estimate is correct to within ±0.02 of the population proportion, how large a sample size is required? A sample size ofpeople is required (Round up to the nearest integer.) d. To conduct a follow-up study that would provide 99% confidence that the point estimate is correct to within ±0.02 of the population proportion, how many people need to be sampled? A sample size of people is needed (Round up to the nearest integer.) e. Discuss the effects of changing the desired confidence level and the acceptable sampling error on sample Click to select your answer(s)

Explanation / Answer

Solution:

A)

Error E = 0.06

p = 0.42

Confidence level 1-alpha =95%

The formula for the minimum sample size n sample size required is

n > p*(1-p)*((Zalpha/2)/E)^2

From standard normal table corresponding 95% confidence level,

Zalpha/2= 1.96

Sample size n> 0.40*(1-0.40)*( 1.96/0.05)2 = 368.7

Therefore, the minimum sample size required n = 369

B)

1-alpha =99%

Zalpha/2= 2.5758

Sample size n> 0.40*(1-0.40)*( 2.5758/0.05)2 = 636.93

Therefore, the minimum sample size required n = 637

C)

E = 0.02

1-alpha =95%

Zalpha/2= 1.96

Sample size n> 0.40*(1-0.40)*( 1.96/0.02)2 = 2304.96

Therefore, the minimum sample size required n = 2305

D)

E = 0.02

1-alpha =99%

Zalpha/2= 2.5758

Sample size n> 0.40*(1-0.40)*( 2.5758/0.02)2 = 3980.8

Therefore, the minimum sample size required n = 3981

(E)

Option A and B are correct.

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