Construct a 98% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A study of 14 car owners showed that their average repair bill was $192 with a standard deviationof $8. Round to the nearest cent.
A) ($186.33, $197.67)
B) ($222.33, $256.10)
C) ($328.33, $386.99)
D) ($115.40, $158.80)

Question

Quizplus · Accepted Answer

To construct a 98% confidence interval for the population mean with a sample size of 14, an average of $192, and a standard deviation of $8, we use the formula for the confidence interval: mean ± (critical value * standard error). The standard error is the standard deviation divided by the square root of the sample size. For a 98% confidence level and a sample size smaller than 30, we use the t-distribution. The degrees of freedom (df) is 13 (n-1), and the critical t-value for a 98% confidence level with 13 df is approximately 2.650 (value might slightly vary depending on the t-distribution table used). The standard error (SE) is $8/√14 ≈ 2.14. Thus, the margin of error (ME) is 2.650 * 2.14 ≈ 5.67. The confidence interval is $192 ± $5.67, which gives us ($186.33, $197.67) when rounded to the nearest cent.

Construct a 98% Confidence Interval for the Population Mean, μ