The following 5 questions are based on this information. In a poll of 500 Americ
ID: 3356869 • Letter: T
Question
The following 5 questions are based on this information. In a poll of 500 American High School students, .75% ( p ¯ =0.75 p¯=0.75 ) said that they used the Internet for social purposes at least once a day. The goal is to construct a 99% confidence interval for the percentage ( p p ) of American High School students who use the Internet for social purposes at least once per day around the period the poll was conducted. Question 4 Not yet answered Points out of 1.00 Flag question Question text The standard error (SE) of p ¯ p¯ is Select one: a. 0.019 b. 0.0004 c. 0.0002 d. 0.016 Question 5 Not yet answered Points out of 1.00 Flag question Question text The critical value (CV) needed for 99% confidence interval estimation is Select one: a. 1.96 b. 1.64 c. 1.28 d. 2.58 Question 6 Not yet answered Points out of 1.00 Flag question Question text The 99% confidence interval estimate of p p is Select one: a. 0.75 ± 0.05 b. 0.75 ± 0.15 c. 0.44 ± 0.03 d. 0.44 ± 0.002 Question 7 Not yet answered Points out of 1.00 Flag question Question text Suppose around the period the above poll was conducted, a Internet Company CEO made a personal statement saying that .85% of American High School students use the Internet daily for social reasons. In light of the sample evidence and at the 1% level of significance, Select one: a. We cannot reject the commentator's claim b. We can reject the commentator's claim Question 8 Not yet answered Points out of 1.00 Flag question Question text A principal of a high school wishes to collect new random sample with the aim of building a new confidence interval at the 99% confidence level for p p . Using the current sample proportion (from the 500 high school students poll ) as a basis, what sample size (n) would the journalist require to achieve a 10% margin of error? Select one: a. 125 b. 73 c. 250 d. 500
Explanation / Answer
4) std error =(p(1-p)/n)1/2 =0.019
option A
5) critical value (CV) needed for 99% confidence interval estimation is =2.58
option d
6) confidence interval =sample mean -/+ z*Std error =0.75 -/+0.05
option A
7)b) We can reject the commentator's claim
8)
here margin of error E =0.1
therefore required sample size n=p(1-p)*(Z/E)2 =~ 125
option A