Question

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.

$x=16,n=20,90\%\text{level}$

Short answer

(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 $0.137$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.

$x=16\text{and}n=20,90\%\text{level}$

$$\begin{gathered} \therefore n-x=20-16 \\ =4 \end{gathered}$$, here $x$and $n-x$ are both $5$ or greater.

Part(a) Step 2: Explanation

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

$\frac{16}{20}=0.8$

Part(b) Step 1: Given Information

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

$x=16\text{and}n=20,90\%\text{level}$

$$\begin{gathered} \therefore n-x=20-16 \\ =4 \end{gathered}$$, here $x$and $n-x$ are both $5$ or greater.

Part(b) Step 2: Explanation

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

It is discovered that role="math" localid="1651401333333" $$\begin{gathered} z_{a/2}=z_{0.1/2} \\ =1.645 \end{gathered}$$

The $p$confidence interval is of the form

$p^{\text{'}}-z_{\alpha /2}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}\text{to}{p}^{\text{'}}+z_{\alpha /2}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$

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

i.e. $(0.25-0.113)\text{to}(0.2+0.124)$

i.e.$0.137\text{to}0.363$