Hello,
I am having difficulty getting the right answer for the following problem. It reads as follows:
Merrill Lynch Securities and Health Care Retirement, Inc., are two large employers in downtown Toledo, Ohio. They are considering jointly offering child care for their employees. As a part of the feasibility study, they wish to estimate the mean weekly child-care cost of their employees. A sample of 10 employees who use child care reveals the following amounts spent last week.
$107, $92, $97, $95, $105, $101, $91, $99, $95, $104
Develop a 90 percent confidence interval for the population mean. Interpret the result.”
Any help would be greatly appreciated.
Thanks again!
J
1 Answer
ANSWER: 90% Confidence Interval = [94.6, 102.6]
Why???
SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
x-bar = Sample mean [98.6]
s = Sample standard deviation [5.541]
n = Number of samples [10]
df = degrees of freedom [10 – 1 = 9]
For confidence level of 90%, two-sided interval (“look-up” from Table) “t critical value” [2.262]
Resulting Confidence Interval for “true mean”: x-bar +/- (t critical value) * s/SQRT(n) = 98.6 +/- 2.262 * 5.541/SQRT(10) = [94.6, 102.6]