Question 1091936: 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! The best estimate of the population mean is $61, the sample mean.
98% CI is 61+/-t df=21,0.99*5.66/sqrt(22); the t is 2.52
the half interval is 3.04, 2.52*5.66/sqrt(22)
the 98% CI is ($57.96, $63.04)
use a z since we don't know the t for the sample size. Then check to see if this works.
half-interval is 1
z*5.66/sqrt(n)=1, and z is 1.96, so
1.96*5.66=sqrt (n)
sqrt(n)=11.09
n=123.06 or 124
check to see if this works for t. It is a little large at 1.006 half interval or (58.994, 61.006)
if one uses 125, it is +/-1.002.
n=126 gives an error of +/-0.998
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