Question

The 95 % confidence interval (use z*) is:
____ to ____
(to 3 decimals)

Answer 1

Using the z-distribution, it is found that the 95% confidence interval is given by:

$\overline{x} - 1.96\frac{\sigma}{\sqrt{n}}$ to $\overline{x} + 1.96\frac{\sigma}{\sqrt{n}}$

What is a z-distribution confidence interval?

The confidence interval is:

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

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

In this problem, we have a 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

Hence, the interval is:

$\overline{x} - 1.96\frac{\sigma}{\sqrt{n}}$ to $\overline{x} + 1.96\frac{\sigma}{\sqrt{n}}$

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

#SPJ1