# Help with a mean confidence interval problem please?

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

• M
1 month ago

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]