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Estimation:Margin of error, Confidence interval, Sample Size

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Attached please find four Statistics problems. Please answer each question in detail, showing all intermediate steps as well as the final resolution.

Problem #4 Fortune Survey
A survey of small businesses with Web sites found that the average amount spent on a site was $11,500 per year (Fortune, March 5, 2001). Given a sample of 60 businesses and a population standard deviation of $4000, determine the margin of error. Use 95% confidence. Provide a recommendation if the study required a margin of error of $500.

Problem #5 National Retail Foundation
A National Retail Foundation survey found households intended to spend an average of $649 during the December holiday season (The Wall Street Journal, December 2, 2005). Assume that the survey included 600 households and that the sample standard deviation was $175.
a. With 95% confidence, determine the margin of error.
b. Determine the 95% confidence interval estimate of the population mean
c. The prior year, the population mean expenditure per household was $632. Discuss the change in holiday season expenditures over the one-year period.

Problem #6 Business Salaries
Annual starting salaries of college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. Determine the planning value for the population standard deviation. Determine how large a sample should be taken if the desired margin of error is:
a. $500
b. $200
c. $100
d. Would you recommend trying to obtain the $100 margin of error? Explain.

Problem #7 Towers Perrin
Towers Perrin, a New York human resources consulting firm, conducted a survey of 1100 employees at medium-sized and large companies to determine how dissatisfied employees were with their jobs (The Wall Street Journal, January 29, 2003). A total of 473 employees indicated they strongly disliked their current work experience.
a. Determine the point estimate of the proportion of the population of employees who strongly dislike their current work experience.
b. At 95% confidence, what is the margin of error?
c. Determine the 95% confidence interval for the proportion of the population of employees who strongly dislike their current work experience.
d. Towers Perrin estimates that it costs employers one-third of an hourly employee's annual salary to find a successor and as much as 1.5 times the annual salary to find a successor for a highly compensated employee. Explain the message this survey sends to employers.

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Solution Summary

Answers to 4 questions on estimation dealing with Margin of error, Confidence interval estimate, Sample size, Point Estimate

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Problem #4 Fortune Survey
A survey of small businesses with Web sites found that the average amount spent on a site was $11,500 per year (Fortune, March 5, 2001). Given a sample of 60 businesses and a population standard deviation of $4000, determine the margin of error. Use 95% confidence. Provide a recommendation if the study required a margin of error of $500.

Mean= M = $11,500
Standard deviation =s = $4,000
sample size=n= 60
s x=standard error of mean=s / square root of n= 516.4 = ( 4000 / square root of 60)
Confidence level= 95%
Therefore, Significance level= a (alpha) = 5% = 1- 0.95
No of tails= 2
This is a 2 tailed test because we are finding the margin of error

Since sample size= 60 >= 30
we will use normal distribution
Z at the 0.05 level of significance 2 tailed test = 1.96

Margin of error =z*s x= 1012.14 =1.96*516.4

If the study requires a margin of error of $500, we will have to increase the sample size

Standard deviation =s = $4,000
confidence interval= 95%
Z corresponding to 95% and two tailed test is 1.96
We have to ensure that Z* s x < $500
or s x < =500/Z
or s x < 255.102041 =500/1.96
But
s x=standard error of mean=s / square root of n
s = 4000

or n=(s ^2)/(s x^2)= 246 =4000^2/255.102041^2

Therefore, increase sample size to 246

Check:
Standard deviation =s = $4,000
sample size=n= 246
s x=standard error of mean=s / square root of n= 255.03 = ( 4000 / square root of 246)
Confidence level= 95%
Therefore, Significance level= a (alpha) = 5% = 1- 0.95
No of tails= 2
This is a 2 tailed test because we are finding the margin of error

Since sample size= 246 >= 30
we will use normal distribution
Z at the 0.05 level of significance 2 tailed test = 1.96

Margin of error =z*s x= 499.86 =1.96*255.03

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