8.2.63 one-mean-z-interval procedure

A sample​ mean, sample​ size, population standard​ deviation, and confidence level are provided. Use this information to complete parts​ (a) through​ (c) below.

\(\bar{x}=24, n=37, \sigma=3, confidence level = 95%\)

(a). Use the​ one-mean z-interval procedure to find a confidence interval for the mean of the population from which the sample was drawn.

First, we need to get the data from the question

x = 24
n = 37
sigma = 3
alpha = .05

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We do not use the empirical rule like in question 8.1.11

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To find confidence interval, we use formulas \(\bar{x} \pm z_{\frac{\alpha}{2}}.\frac{\sigma}{\sqrt{n}}\)

First we find \(z_{\frac{\alpha}{2}}\) by using qnorm()

zalpha2 = abs(qnorm(alpha/2))
zalpha2

## [1] 1.959964

To find 95% confidence interval, we run

x - zalpha2 * sigma/sqrt(n)

## [1] 23.03335

x + zalpha2 * sigma/sqrt(n)

## [1] 24.96665

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Round to one decimal places as needed

round(x - zalpha2 * sigma/sqrt(n), 1)

## [1] 23

round(x + zalpha2 * sigma/sqrt(n), 1)

## [1] 25

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(b).Obtain the margin of error by taking half the length of the confidence interval.What is the length of the confidence​ interval?

Length of confidence interval

round(x + zalpha2 * sigma/sqrt(n), 1)-round(x - zalpha2 * sigma/sqrt(n), 1)

## [1] 2

Margin of error = 1/2 distance of two endpoints of confidence interval

(round(x + zalpha2 * sigma/sqrt(n), 1)-round(x - zalpha2 * sigma/sqrt(n), 1)) /2

## [1] 1

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(c). Obtain the margin of error by using the formula \(E=z_{\frac{\alpha}{2}}.\frac{\sigma}{\sqrt{n}}\)

Identify the critical value.

zalpha2

## [1] 1.959964

round(zalpha2,2)

## [1] 1.96

We can find margin of error

E = zalpha2 * sigma/sqrt(n)
E

## [1] 0.9666483

round(E, 1)

## [1] 1

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Hope that helps!