Provide an appropriate response.
-Construct a $98 \%$ confidence interval for the population mean, $\mu$. Assume the population has a normal distribution. A study of 14 car owners showed that their average repair bill was $\$ 192$ with a standard deviation of $\$ 8 .$
A) $( \$ 186.30 , \$ 197.70 )$
B) $( \$ 222.30 , \$ 256.10 )$
C) $( \$ 328.30 , \$ 386.90 )$
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: $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level, $\sigma$ is the standard deviation, and $n$ is the sample size. For a 98% confidence level and a sample size less than 30, we use the t-distribution. The degrees of freedom (df) is $n-1 = 13$. Looking up the t-table for 98% confidence and 13 df, the t-value is approximately 2.650. Plugging in the values: $192 \pm 2.650 \times \frac{8}{\sqrt{14}}$, we get the interval $(\$186.30, \$197.70)$.

Provide an Appropriate Response 98%98 \%98% Confidence Interval for the Population Mean

Provide an Appropriate Response $98 \%$ Confidence Interval for the Population Mean