Question 1112685
A survey of 20 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $57.00 per hour. The sample standard deviation was $5.60 per hour. (Use t Distribution Table.) 
Q1) What is the best estimate of the population mean wage ?
Ans:: $57.00
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Q2) Develop a 90% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
Margin of Error = t*s/sqrt(20) = |invT(0.05,19)|*s/sqrt(20) = 1.7291*5.6/sqrt(20)
= 2.`65
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Confidence interval for the population mean wage is between 
57-2.165 and 57+2.165
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Q3)How large a sample is needed to assess the population mean with an allowable error of $1.00 at 90% confidence? (Round up your answer to the next whole
 number.) 
n = [t*s/E]^2 = [1.7291*5.6/1]^2 = 94 when rounded up
Sample size ? = 94
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Note:: I only used the t-Distribution because your post called for it.
I would normally use the z-distribution.
Cheers,
Stan H.