W6-unit 6 feedback solutions.pdf - WWW.YORKVILLEU.CA Unit 6 Exercises \u2022 \u2022 \u2022 Due no later than 11:00 p.m on Sunday of each unit Worth 25 of final

# W6-unit 6 feedback solutions.pdf - WWW.YORKVILLEU.CA Unit 6...

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BUSI 1013: S TATISTICS FOR B USINESS 1 Unit 6 Exercises Due no later than 11:00 p.m. on Sunday of each unit Worth 25% of final grade (2.5% each) Graded out of 28 points. Hello, All. Your assignments will be graded within 7 days of due date, and if necessary the feedback and comments will be provided in the comment boxes of grades. For an earlier up date and learning of your missed points, please see my solutions in the followings: 1. A simple random sample of 80 customers is selected from an account receivable portfolio and the sample mean account balance is \$1250. The population standard deviation σ is known to be \$250. ( 8 points ) a. Construct a 95% confidence interval for the mean account balance of the population. 1250±1.96*250/sqrt(80)=1250 ± 54.78 (1195.22,1304.78) b. What is the margin of error for estimating the mean account balance at the 95% confidence level. 54.78
c. Construct a 95% confidence interval for the mean account balance of the BUSI 1013: S TATISTICS FOR B USINESS 2 population if the sample mean account balance is obtained from a random sample of 150 instead of 80. 1250±1.96*250/sqrt(150)=1250 ± 40.01 (1209.99,1290.01) d. Based on the results from a. and c., what can you say about the effect of a larger sample size on the length of the confidence interval at the same confidence level? The larger the sample size, the shorter the length of the confidence interval at the same confidence level 2. A simple random sample of 50 calls is monitored at an in-bound call center and the average length of the calls is 6.5 minutes. The population standard deviation σ is unknown. Instead the sample standard deviation s is also calculated from the sample and is found to be 4.5 minutes. ( 6 points ) a. Construct a 99% confidence interval (using the t-distribution) for the average length of inbound calls. We do not know the population standard deviation Use the t distribution with 50-1=49 degrees of freedom 6.5±2.680*2.5/sqrt(50)=6.5 ± 0.95 (5.55,745) b. What is the margin of error at the 95% confidence level? 2.010*2.5/sqrt(50) = 0.71 c. Do we need to make any assumption on the distribution of the length of in- bound calls at this call center? Why or why not?