Question

The standard deviation for actuary salaries has a standard deviation of $36,730 . You collect a simple random sample of n=36 salaries and find that x bar = $111,000 . Use this information for both parts. Part 1: What is the lower limit of this 95% confidence interval? (Round to the nearest whole number)

Answer 1

Using the z-distribution, it is found that the lower limit of the 95% confidence interval is of $99,002.

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.

The other parameters are given as follows:

$\overline{x} = 111000, \sigma = 36730, n = 36$

Hence, the lower bound of the interval is:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 111000 - 1.96\frac{36730}{\sqrt{36}} = 99002$

The lower limit of the 95% confidence interval is of $99,002.

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

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