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com X > d " Secure https://webwork.ucalgary.ca/webwork2/F2017STAT2 17L01/Stat 217 F2017 Assignment 1/11/ blem 9 Problem 10. Problem 11 Problem 12 Problem 13 (1 point) The editor of a magazine knows that 40.6% of all people who subscribe to the magazine do not renew their subscriptions. In an attempt to decrease this percentage, current subscribers are offered a renewal offer one that offers current subscribers to renew their subscription at a substantially reduced price. After a period of six-months, the magazine editor wishes to investigate if this renewal offer does decrease the percentage of subscribers who do not renew subscriptions, or p (a) Choose the correct statistical hypotheses. A. HU : p = 0.406 HA : p > 0.406 oB. Ho p-0.106 Ha:p 0.406 F. Ho : p-0.406 HA : p0.406 c, Ho : p-0.406 HA : p (b) The null hypothesis is to be tested from a random sample of n 200 magazine subscribers whose subscriptions were to be renewed in the past six months. If statistical testing is to be carried out at P(Type I Error) 0.05, what is the maximum number of subscribers out of 200 who did not renew their subscription in order to determine that the renewal offer does decrease the percentage of subscribers who do not renew. Enter your answer as an appropriate integer. 70 (c) The sample of 200 subscribers who were offered the renewal offer was taken, of which 77 didn't renew their magazine subscription in the past six months. The P-value of this result was 0.2727. What type of error could be made from this sample? A. Type I Error B. Power of a Test C. Type Il Error (d) How powerful is the statistical test outlined in (b) if the percentage of all subscribers who do not renew their subscription, p, drops by 5 percentage points? Enter your answer to three decimals. 3:54 PM Type here to search 2017-09-27Explanation / Answer
b|) for std error =(p(!-p)/n)1/2 =0.0347
for 0.05 level ; critical value z =1.645
therefore critical phat =p -z*std error =0.3489
hence maximum number of subscribers who do not renew =0.3489*200=69.78 ~69
c) type II error
d) for new p=0.406-0.05=0.356
hence std error ==(p(!-p)/n)1/2 =0.0339
therefore power =P(X<0.3489)=P(Z<(0.3489-0.356)/0.0339)=P(Z<-0.2094) =0.417
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