A random sample of 27 people employed by the Florida state authority established
ID: 3380935 • Letter: A
Question
A random sample of 27 people employed by the Florida state authority established they earned an average wage (including benefits) of $63.00 per hour. The sample standard deviation was $6.11 per hour. (Use z Distribution Table.)
Develop a 95% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
How large a sample is needed to assess the population mean with an allowable error of $2.00 at 90% confidence? (Round up your answer to the next whole number.)
a. What is the best estimate of the population mean?Explanation / Answer
a) It is the sample mean, 63.00 [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 = 0.025
X = sample mean = 63
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 6.11
n = sample size = 27
Thus,
Margin of Error E = 2.304662945
Lower bound = 60.69533705
Upper bound = 65.30466295
Thus, the confidence interval is
( 60.69533705 , 65.30466295 ) [ANSWER]
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c)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.05
Using a table/technology,
z(alpha/2) = 1.644853627
Also,
s = sample standard deviation = 6.11
E = margin of error = 2
Thus,
n = 25.2509047
Rounding up,
n = 26 [ANSWER]
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