Asked by Rald
A human resources manager wants to determine a confidence interval estimate for the mean test score
for the next office-skills test to be given to a group of job applicants. In the past, the test scores have been
normally distributed with a mean of 74.2 and a standard deviation of 30.9. Determine a 95% confidence
interval estimate if there are 30 applicants in the group.
A. 63.14 to 85.26
B. 68.72 to 79.68
C. 13.64 to 134.76
D. 64.92 to 83.48
for the next office-skills test to be given to a group of job applicants. In the past, the test scores have been
normally distributed with a mean of 74.2 and a standard deviation of 30.9. Determine a 95% confidence
interval estimate if there are 30 applicants in the group.
A. 63.14 to 85.26
B. 68.72 to 79.68
C. 13.64 to 134.76
D. 64.92 to 83.48
Answers
Answered by
Stace
You need to find the error in the problem. You should have a formula that determines error. You also need to transform the 95% confidence interval into a z score in order to solve. Mean plus your error, mean minus your error should give you the correct confidence interval for the problem. Does this help?
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