Provide an appropriate response.
-Construct a $99 \%$ confidence interval for the population mean, $\mu$. Assume the population has a normal distribution. A group of 19 randomly selected employees has a mean age of $22.4$ years with a standard deviation of $3.8$ years.
A) $( 19.9,24.9 )$
B) $( 16.3,26.9 )$
C) $( 17.2,23.6 )$
D) $( 18.7,24.1 )$

Question

Quizplus · Accepted Answer

The confidence interval for the population mean $\mu$ can be calculated using the formula: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the t-score for the desired confidence level (99% in this case), $s$ is the sample standard deviation, and $n$ is the sample size. Given $\bar{x} = 22.4$, $s = 3.8$, and $n = 19$, and knowing the t-score for 99% confidence level with 18 degrees of freedom (approximated since the exact value depends on tables or software), the calculation results in a confidence interval of approximately $(19.9, 24.9)$.

Provide an Appropriate Response 99%99 \%99% Confidence Interval for the Population Mean

Provide an Appropriate Response $99 \%$ Confidence Interval for the Population Mean