Question 1092192: A random sample of 22 people employed by the Florida state authority established they earned an average wage (including benefits) of $61.00 per hour. The sample standard deviation was $5.66 per hour. (Use t Distribution Table.)
a. What is the best estimate of the population mean?
Estimated population mean $___?
b.
Develop a 98% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
Confidence interval for the population mean wage is between ____ and _____.
c.
How large a sample is needed to assess the population mean with an allowable error of $1.00 at 95% confidence? (Round up your answer to the next whole number.)
Sample size ___?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Best estimate of population mean is the sample mean, or $61.00.
The SE is $5.66/sqrt(22)=1.207
CI half interval is +/- tdf=21,0.99,*SE
that is +/- 2.518*1.207=3.04
the CI is mean +/-interval or (57.96, 64.04)
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For allowable error of 1
t*SE has to be 1
use z first 1.96*5.66/sqrt(n)=1
sqrt(n)=1.96*5.66
square both sides
n=123.06 or 124. Z can be used here.
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