A random sample of 22 people employed by the Florida state authority established
ID: 3151265 • Letter: A
Question
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 z 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
Explanation / Answer
A)
It is the sample mean,
X = $61.00 [ANSWER]
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b)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 61
z(alpha/2) = critical z for the confidence interval = 2.33
s = sample standard deviation = 5.66
n = sample size = 22
Thus,
Margin of Error E = 2.811648407
Lower bound = 58.18835159
Upper bound = 63.81164841
Thus, the confidence interval is
( 58.18835159 , 63.81164841 ) [ANSWER]
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c)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.96
Also,
s = sample standard deviation = 5.66
E = margin of error = 1
Thus,
n = 123.067961
Rounding up,
n = 124 [ANSWER]
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