Question

A simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is x Overbar, while the standard deviation is s. What is the 99% confidence interval for the population mean? Use the table below to help you answer the question. Confidence Level 90% 95% 99% z*-score 1. 645 1. 96 2. 58.

Answer 1

Using the z-distribution, as we have the standard deviation for the population, it is found that the 99% confidence interval for the population mean is given by:

$\overline{x} \pm 2.58\frac{s}{\sqrt{n}}$

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• s is the standard deviation for the population.

99% confidence level, hence$\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so $z = 2.58$.

The other parameters are symbolic, hence the interval is given by:

$\overline{x} \pm 2.58\frac{s}{\sqrt{n}}$

More can be learned about the z-distribution at https://brainly.com/question/25890103