Question

A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence interval for the population mean. Round your answers to one decimal place. ( , ) b. Assume that the same sample mean was obtained from a sample of items. Provide a confidence interval for the population mean. Round your answers to two decimal places. ( , ) c. What is the effect of a larger sample size on the interval estimate? Larger sample provides a - Select your answer - margin of error.

Answer 1

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

$\sigma_x = 1.825$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.825$

$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 3.577$

$Lower = 46.423$

$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

So, the 95% confidence interval is (46.4, 53.6)

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

So, the 95% confidence interval is (47.9, 52.1)

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$     $E = 3.577$

In (b), we have:

$n = 90$    $E = 2.06584$

Notice that the margin of error decreases when the sample size increases.