A company that provides care for the elderly is contemplating offering childcare services for its employees on an aggregate basis. As part of the project's feasibility study, they wish to calculate the average weekly cost of childcare. A sample of 10 employees using this service reveals the following amounts spent in the past week. 150, 187, 190, 197, 192, 195, 181, 191, 199, 174. Construct the 99% confidence interval of the population mean. Interpret the result
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To construct the 99% confidence interval for the population mean, we can use the t-distribution since the sample size is less than 30.
First, we need to calculate the sample mean and standard deviation:
Sample mean (x̄) = (150 + 187 + 190 + 197 + 192 + 195 + 181 + 191 + 199 + 174) / 10 = 187.6
Sample standard deviation (s) = √[((150-187.6)^2 + (187-187.6)^2 + (190-187.6)^2 + (197-187.6)^2 + (192-187.6)^2 + (195-187.6)^2 + (181-187.6)^2 + (191-187.6)^2 + (199-187.6)^2 + (174-187.6)^2) / 9] ≈ 8.12
Next, we need to determine the critical value from the t-distribution table for a 99% confidence level with 9 degrees of freedom. The critical value is approximately 3.250.
Now we can calculate the margin of error:
Margin of error = Critical value * (s/√n) = 3.250 * (8.12/√10) ≈ 8.78
Finally, we can construct the confidence interval:
Confidence interval = Sample mean ± Margin of error = 187.6 ± 8.78
Thus, the 99% confidence interval for the population mean is approximately 178.82 to 196.38.
Interpretation: We are 99% confident that the true average weekly cost of childcare for all employees using this service falls between $178.82 and $196.38.
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