Asked by Monica
A company is considering offering child care for their employees. They wish to
estimate the mean weekly child-care cost of their employees. A sample of 10 employees
reveals the following amounts spent last week in dollars.
107 92 97 95 105 101 91 99 95 104
Develop a 90% confidence interval for the population mean. Interpret the result.
=??, S=??, =??, Range of = (??) xt2/.
.
estimate the mean weekly child-care cost of their employees. A sample of 10 employees
reveals the following amounts spent last week in dollars.
107 92 97 95 105 101 91 99 95 104
Develop a 90% confidence interval for the population mean. Interpret the result.
=??, S=??, =??, Range of = (??) xt2/.
.
Answers
Answered by
PsyDAG
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation (SD) = square root of variance
Range = highest score - lowest
90% interval = mean ± 1.645SD
I'll let you do the calculations.
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation (SD) = square root of variance
Range = highest score - lowest
90% interval = mean ± 1.645SD
I'll let you do the calculations.
Answered by
ree
5s1a
Answered by
Dan
company is considering offering child care for their employees. They wish to
estimate the mean weekly child-care cost of their employees. A sample of 10 employees
reveals the following amounts spent last week in dollars.
estimate the mean weekly child-care cost of their employees. A sample of 10 employees
reveals the following amounts spent last week in dollars.
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