Chapter 11: Q. 11.72 (page 462)

In this Exercise, we have given the number of successes and the sample size for a simple random sample from a population. In each case,
a. use the one-proportion plus-four $z$ -interval procedure to find the required confidence interval.
b. compare your result with the corresponding confidence interval found in Exercises $11.25 - 11.30$ , if finding such a confidence interval was appropriate.
$x = 40, n = 50, 95 % level$

Short Answer

Expert verified

(a) The one-proportion $z$ -interval technique is adequate because $x$ and $n - x$ are both $5$ or more.

(b) The confidence interval can be estimated to be between $0.689$ and $0.911$ with a $95 %$ confidence level.

Step by step solution

Part(a) Step 1: Given Information

The size of a simple random sample from a population, as well as the number of successes.

$x = 40 and n = 50, 95 % level$

role="math" localid="1651400180077" $∴ n - x = 50 - 40 = 10$ , here $x$ and $n - x$ are both $5$ or greater.

Part(a) Step 2: Explanation

The sample proportion $p^{'} = \frac{x}{n}$ is calculated from the data.

$\frac{40}{50} = 0.8$

Part(b) Step 1: Given Information

The size of a simple random sample from a population, as well as the number of successes.

$x = 40 and n = 50, 95 % level$

$∴ n - x = 50 - 40 = 10$ , here $x$ and $n - x$ are both $5$ or greater.

Part(b) Step 2: Explanation

The confidence interval is $95 %$ , which means $α = 0.05$ .

It is discovered that $z_{a / 2} = z_{0.05 / 2} = 1.96$

The $p$ confidence interval is of the form

$p^{'} - z_{α / 2} \sqrt{\frac{p^{'} (1 - p^{'})}{n}} to p^{'} + z_{α / 2} \sqrt{\frac{p^{'} (1 - p^{'})}{n}}$

i.e. $0.8 - 1.960 \sqrt{\frac{0.8 (1 - 0.8)}{50}} to 0.8 - 1.960 \sqrt{\frac{0.8 (1 - 0.8)}{50}}$

i.e. $(0.8 - 0.111) to (0.8 + 0.111)$

i.e. $0.689 to 0.911$

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Short Answer

Step by step solution

Part(a) Step 1: Given Information

Part(a) Step 2: Explanation

Part(b) Step 1: Given Information

Part(b) Step 2: Explanation

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