A sample of 100 postal employees found that the average time these employees had

Question

A sample of 100 postal employees found that the average time these employees had for the postal service was 8 years (sample mean). Assume that we know that the standard deviation of the population of times postal service employees have spent with the postal service is 5 years (population standard deviation). A 95% confidence interval for the mean time the population of postal service employees have spent with the postal service is used. Which of the following would produce a confidence interval with a larger margin of error?        A. using a sample of 1,000 postal employees        B. using a confidence level of 90%        C. using a confidence level of 99%        D. using a different sample of 100 employees, ignoring the results of the previous sample.

Explanation / Answer

Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=8
Standard deviation( sd )=5
Sample Size(n)=34
Margin of Error = Z a/2 * 5/ Sqrt ( 34)
= 2.58 * (0.857)
= 2.212

When using a confidence 99% it produces the larger margin of error

[ANSWER]
C. using a confidence level of 99%

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