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1) In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss. What is the width of the 95% confidence interval for the proportion of women who prefer a female boss? Please show your calculations
2)Suppose the President wants an estimate of the proportion of the population who oppose a policy toward gun control. The President wants the estimate to be within .04 of the true proportion. Assume a 95 percent level of confidence. The President’s political advisors estimated the proportion supporting the current policy to be 0.60. How large a sample is required?
3) Explain whether the width of a confidence interval would increase, decrease, or remain the same as a result of each of the following changes:
a) The sample size is doubled, from 400 to 800
b) The population size is doubled, from 25 million to 50 million
c) The level of confidence is lowered from 95% to 90%
4) A financial institution wishes to estimate the mean balances owed by its credit card customers. The population standard deviation is estimated to be $300. If a 99 percent confidence interval is used and an interval of ±$75 is desired, how many cardholders should be sampled?
5) A tire manufacturer wishes to investigate the tread life of its tires. A sample of 10 tires driven 50,000 miles revealed a sample mean of 0.32 inch of tread remaining with a standard deviation of 0.09 inch. Construct a 95 percent confidence interval for the population mean. Would it be reasonable for the manufacturer to conclude that after 50,000 miles the population mean amount of tread remaining is 0.30 inches? [compute the confidence of interval using t-distribution since σ is unknown]
Explanation & Answer
Here is the answer. Thanks
(1) Proportion of women employees prefer women boss, p = (81/810) = 0.10
Standard Error of Proportion = SQRT [{0.10*(1-0.10)}/810] = 0.010540926
The width of the 95% confidence interval for the proportion of women who prefer a female boss = ±
(1.96*0.010540926) = ± 0.0207
(2) Here,
p = 0.60
Z = 1.96
E = 0.04
Sample Size, n = p*(1-p)*(Z/E)2
= 0.60*(1-0.60)*(1.96/0.04)2
= 576.24
= 576 (Approximately)
(3) The impact of changes in scenario a, b, c on the width of a confidence interval are discussed below.
No.
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