Chapter 11: Q. 11.74 (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.
$x = 3, n = 100, 99 % level$

Short Answer

Expert verified

(a) The one-proportion $z$ -interval technique is appropriate since role="math" localid="1651401865385" $x$ is not larger than $5$ and $n - x > 5$ is not greater than $5$ or greater.

(b) It is possible to be $99 %$ certain that the confidence interval is between $0.137$ and $0.363$ .

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.

$x = 3 and n = 100, 99 % level$

$∴ n - x = 100 - 3 = 97$ , 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.

$\frac{3}{100} = 0.03$

Part(b) Step 1: Given Information

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

$x = 3 and n = 100, 99 % level$

$∴ n - x = 100 - 3 = 97$ , here $x$ and $n - x$ are both $5$ or greater.

Part(b) Step 2: Explanation

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

It is discovered that $z_{a / 2} = z_{0.1 / 2} = 1.645$

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. $0.25 - 1.645 \sqrt{\frac{0.25 (1 - 0.25)}{40}} to 0.25 + 1.645 \sqrt{\frac{0.25 (1 - 0.25)}{40}}$

i.e. $(0.25 - 0.113) to (0.2 + 0.124)$

i.e. $0.137 to 0.363$

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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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