A random sample of 21 people employed by the Florida state authority established
ID: 3380988 • Letter: A
Question
A random sample of 21 people employed by the Florida state authority established they earned an average wage (including benefits) of $59.00 per hour. The sample standard deviation was $5.63 per hour. (Use z Distribution Table.)
A. What is the best estimate of the population mean?
B. Develop a 95% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
C. 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.)
C. 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.)
Explanation / Answer
a)
It is the sample mean, 59.00 [ANSWER]
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 = 59
z(alpha/2) = critical z for the confidence interval = 1.96
s = sample standard deviation = 5.63
n = sample size = 21
Thus,
Margin of Error E = 2.407990775
Lower bound = 56.59200922
Upper bound = 61.40799078
Thus, the confidence interval is
( 56.59200922 , 61.40799078 ) [ANSWER]
**************************
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 = 5.63
E = margin of error = 0.9
Thus,
n = 105.8732596
Rounding up,
n = 106 [ANSWER]
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.