Chapter 11: Q. 11.71 (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 Exercise $11.25 - 11.30$ , if finding such a confidence interval was appropriate.
role="math" localid="1651327118166" $x = 35, n = 50, 99 % level$

Expert verified

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

(b) It is possible to be $90 %$ certain that the confidence interval is between $0.533$ and $0.867$ .

Step by step solution

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

$x = 35, n = 50, 99 % level$

$∴ n - x = 50 - 35 = 15$ , here $x$ and $n - x$ are both $5$ or greater.

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

$\frac{35}{50} = 0.7$

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

$x = 35, n = 50, 99 % level$

$∴ n - x = 50 - 35 = 15$ , here $x$ and $n - x$ are both $5$ or greater.

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

It is discovered that $z_{a / 2} = z_{0.01 / 2} = 2.576$

The $p$ confidence interval is of the form

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

i.e. role="math" localid="1651328225031" $0.7 - 2.576 \sqrt{\frac{0.7 (1 - 0.7)}{50}} to 0.7 + 2.576 \sqrt{\frac{0.7 (1 - 0.7)}{50}}$

i.e. $(0.7 - 0.167) to (0.7 + 0.167)$

i.e. $0.533 to 0.867$

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