The following is a random sample representing the length of time between commerc
ID: 3233218 • Letter: T
Question
The following is a random sample representing the length of time between commercial breaks during a television miniseries. Find a point estimate for the average length of time between commercial breaks (sample mean) and a 90%, 95% and 99% confidence interval for the average length of time between commercial breaks. Use Analyze-Compare Means-One-Sample t Test, then use Options to change level of confidence. 5 10 14 8 7 9 5 8 8 9 5 14 11 10 11 13 12 6 8 For the following hypothesis tests: a) write out the Null and Alternate hypothesis b) write down the P-value c) state your conclusion (accept/reject) d) answer the question posed in the problem. Hawaii, where the population mean life span is 77 years. A random sample of 20 obituary notices in the Honolulu Advertiser gave the following information about life span (in Years) of Honolulu residents: sporting goods stores across the state of California. Each store was part of a large franchise of sporting goods stores. The consultant taught the manager of each store better ways to advertise and display their goods. The net sales for one month before and one month after the consultant's visit were recorded as follows for each store (in thousands of dollars): Do the data indicate that the net sales improved at the alpha = 05 level? At the alpha = .0l level? Type the information in two columns with labels Before and After. Use Analyze - Compare Meant - Paired-Sample t Test, then define the pair. Consumer Reports gave the following information about annual premiums for a $250,000 annual renewable life insurance policy for both men and women (age 45): Insurance Premiums for Men (in dollars) 430 360 439 460 521 637 272 335 453 373 393 Insurance Premiums for Women (in dollars) 282 450 500 427 487 403Explanation / Answer
point estimate=9.55
(1-alpha)*100% confidence interval for population mean=mean±t(alpha/2,n-1)*s/sqrt(n)
90% confidence interval for population mean=9.55±t(0.1/2, 19)*2.72/sqrt(20)=9.55±1.73*2.72/sqrt(20)
=9.55±1.05=(8.5,10.6)
95% confidence interval for population mean=9.55±t(0.05/2, 19)*2.72/sqrt(20)=9.55±2.09*2.72/sqrt(20)
==9.55±1.27=(8.28,10.82)
99% confidence interval for population mean=9.55±t(0.01/2, 19)*2.72/sqrt(20)=9.55±2.86*2.72/sqrt(20)
=9.55±1.74=(7.81,11.29)
following information has been generated using ms-excel
x 5 7 8 14 13 10 9 8 11 12 14 11 9 10 6 8 12 5 11 8 mean= 9.55 s= 2.72 n= 20Related Questions
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