The distribution of weekly salaries at a large company is reverse J-shaped with a mean of $1000 and a standard deviation of $370. What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75?

2 answers

The distribution of weekly salaries at a large company is reverse J-shaped with a mean of $1000 and a standard deviation of $370. What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75?
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n = [z*s/E]^2
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80 = [z*370/75]^2
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sqrt(80) = 4.93z
z = 1.81
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P(z > 1.81) = 0.035
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Ans:: P(condition described) = 2*0.035) = 7%
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Problem 5
(a) A researcher obtained estimates for the mean age, μ, of all U.S. millionaires. He randomly selected 36 U.S. millionaires and found that the mean age is 58.53 years. Assume that the standard deviation of ages of all U.S. millionaires is 13.0 years.
(i) Determine the margin of error, E, for a 98% confidence interval. (2%)
(ii) Determine the sample size required to have a margin of error of 4.5 years with a 95.44% confidence level. (2%)

(b) A U.S. researcher compared the rental rates of commercial real estates in Asia and Europe in 2018. Annual lease rates for a sample of 30 commercial properties in Hong Kong showed a mean of $1458 per square meter with a standard deviation of $269. Annual lease rates for a sample of 40 commercial properties in Paris showed a mean lease rate of $1279 per square meter with a standard deviation of $215. Obtain a 98% confidence interval for the difference between the mean annual lease rates in Hong Kong and Paris. Preliminary data analyses indicate that you can reasonably use the required procedure.