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Course- Data Analysis Homework: Sampling Distribution and Confidence interval
Total points: 20
Problem 1 [8pts] A manufacturer of small appliances employs a market research firm to estimate retail sales of its products by gathering information from a sample of retail stores. This month an SRS of 80 stores
in the Midwest sales region finds that these stores sold an average of 25 of the manufacturer's hand mixers, with standard deviation 12. a.
Give a 99% C.I. for the average number of mixers sold by a store in the region. Interpret your answer.
CI 90%=1.64 CI 95%=1.96 CI 99%=2.57.
SRS (simple Random Sample) =80
Average =25
S.D.= 12
Standard error = S.D./ sqroot(SRS)= 12/sqroot(80)= 12/8.944=1.34
Confidence interval (C.I.) for upper bound
= average +(Z-score* 1.34)= 25 +(2.57*1.34)= 28.4438
Confidence interval(C.I.) for lower bound
= average -(Z-score* 1.34)= 25 -(2.57*1.34)= 21.5562
We are 99% confident that the average number of mixers sold is 21.55 and 28.44.
b.
If you compute a 90% C.I., how does the margin of error change as the confidence level decreases? Explain.
Confidence interval (C.I.) for upper bound
= average +(Z-score* 1.34) = 25 +(1.64*1.34)= 27.1976
Confidence interval (C.I.) for lower bound
= average -(Z-score* 1.34) = 25 -(1.64*1.34)= 22.8024
As margin of errors decreases the confidence interval narrows. The interval has less room
for sample values.
c.
If you increase the sample size, and compute a new 99% C.I., do you expect the new margin of error to increase or decrease? Explain.
More the sample size, less error and better estimations.
Problem 2 [8pts] An IT professional who recently graduated from college has been interviewed for a job here in Chicago that involves development and maintenance of Web applications and the design, implement and support of electronic commerce solutions. He is offered a starting salary of 63,000 dollars per year. A recent survey reported that the average salary of a sample of 500 Internet/Intranet developers working in Chicago is $72,000 dollars with standard deviation equal
to 13,000 dollars. CI 90%=1.64 CI 95%=1.96 CI 99%=2.57.
a.
Construct a 90% confidence interval for the average salary of Web developers in Chicago. Interpret your answer in simple English
SRS= 500
Average= 72,000
S.D.= 13,000
Standard error = S.D./ sqroot(SRS)= = 13000/sqroot(500)= 13000/22.36=581.37
Confidence interval (C.I.) for upper bound
= average +(Z-score* 1.34)= 72000 +(1.64*581.37)= 72,953.4468
Confidence interval (C.I.) for lower bound
= average -(Z-score* 1.34) = 72000 - (1.64*581.37)= 71,046.5532
On average a developer in Chicago earns between $ 71,046 and $72,953.
b.
Do you conclude that the position offered to the IT professional is competitive in terms of
salary? Interpret your answer in simple English.
He is offered a starting salary of $63,000 per year and the confidence interval at 90% is from $ 71,046to $72,953. The salary of the position offered to the IT professional is definitely lower than the average salary.
c.
Explain to someone who knows no statistics what "90% confidence level" means. With 90 % of confidence interval, you have a chance of 10% of being wrong.
Problem 3 [4pts]
A bank wonders whether omitting the annual credit card fee for customers who charge at least $3,000 in a
year would increase the amount charged on their credit card. The bank makes an offer to an SRS of 500 existing credit card customers. It then compares how much these customers charge this year with the amount they charges last year. The mean increase is $565, and the standard deviation is $267.
CI 90%=1.64 CI 95%=1.96 CI 99%=2.57.
Give a 95% confidence interval for the mean amount of the increase. Interpret your answer in simple English.
S.D.= $267
Average = $565+$3000= $3565
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