98%(56)55 out of 56 people found this document helpful
This preview shows page 2 - 5 out of 11 pages.
We do not need to make any assumption on the distribution of of the length of inbound calls. Since the sample size is 50 which is large enough (greater than 30), Central Limit Theorem applies and the distribution of Xbar is approximately normal.
3. Use the data in BUSI1013 Bank Dataset.xlsx (from Unit 1 Exercise Question 2) to answer this question. (4 points) BUSI1013:STATISTICS FOR BUSINESS 3 a. Construct a 95% confidence interval for the mean increase in deposits. Note that the population standard deviation σis not known in this case. Instead the sample standard deviation s should be calculated from the sample and the t distribution should be used. Sample mean = 3.92 Sample standard deviation = 5.85 Sample size = 152 d.f. = 152-1 = 151 (using d.f. =100 in the t distribution table) A 95% confidence interval for the mean increase in deposits is: 3.92±1.984*5.85/sqrt(152)=3.92 ± 0.94 (2.98,4.86) b. What is the margin of error at the 95% confidence level? Margin of error = 0.94 4. According to a previous study, check out times at a supermarket are between 3 to 12 minutes (180 to 720 seconds.) A survey involving a random sample of customers of the supermarket is planned and the first order of business is to decide on an appropriate sample size. The 95% level of confidence will be used. (4 points) a. What is the planning value for the population standard deviation to be used in the sample size formula for estimating the population mean check- out time?
5. Use the data in BUSI1013 Bank Dataset.xlsx (from Unit 1 Exercise Question 2) to answer this question. (4 points) BUSI1013:STATISTICS FOR BUSINESS 4 a. Construct a 95% confidence interval to estimate the proportion of customers who would recommend the Bank to family and friends after the change. pbar = 132/152 = 0.868 A 95% confidence interval to estimate the proportion of customers who would recommend the Bank to family and friends after the change is: 0.868±1.96*sqrt(0.868*(1-0.868)/152) = 0.054 (0.814,0.922) b. What is the margin of error of the estimate at the 95% confidence level? Margin of error = 0054 6. In a survey, the planning value for the population proportion is p* = 0.28. How large should the sample size for the survey be if the desired level of confidence is 95% and the desired margin of error is 0.035? (2 points) n=1.962*0.28*(1-0.28)/0.0352=322.89≈323 always round up to 323 OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO Here is some information on topics used in this week assignment: