Provide an appropriate response.
-Construct a $95 \%$ confidence interval for the population mean, $\mu$. Assume the population has a normal distribution. A sample of 20 part-time workers had mean annual earnings of $\$ 3120$ with a standard deviation of $\$ 677$.
A) $( \$ 2803 , \$ 3437 )$
B) $( \$ 1324 , \$ 1567 )$
C) $( \$ 2135 , \$ 2567 )$
D) $( \$ 2657 , \$ 2891 )$

Question

Quizplus · Accepted Answer

To construct a 95% confidence interval for the population mean with a normal distribution, the formula is $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level (for 95%, $z = 1.96$), $\sigma$ is the sample standard deviation, and $n$ is the sample size. Plugging in the given values: $3120 \pm 1.96\frac{677}{\sqrt{20}}$ results in the interval $(2803, 3437)$.

Provide an Appropriate Response 95%95 \%95% Confidence Interval for the Population Mean

Provide an Appropriate Response $95 \%$ Confidence Interval for the Population Mean