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 Exercise $11.25-11.30$, if finding such a confidence interval was appropriate.

role="math" localid="1651327118166" $x=35,n=50,99\%\text{level}$

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 $0.533$and $0.867$.

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.

$x=35,n=50,99\%\text{level}$

$$\begin{gathered} \therefore n-x=50-35 \\ =15 \end{gathered}$$, 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.

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

Part(b) Step 1: Given Information

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

$x=35,n=50,99\%\text{level}$

$$\begin{gathered} \therefore n-x=50-35 \\ =15 \end{gathered}$$, 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$$\begin{gathered} z_{a/2}=z_{0.01/2} \\ =2.576 \end{gathered}$$

The $p$confidence interval is of the form

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

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

i.e. $(0.7-0.167)\text{to}(0.7+0.167)$

i.e.$0.533\text{to}0.867$