$x=8,n=40,95\%level$

We have given the number of successes and the sample size for a simple random sample from a population. In each case, do the following tasks.

a. Determine the sample proportion.

b. Decide whether using the one-proportion $z$-interval procedure is appropriate.

c. If appropriate, use the one-proportion $z$-interval procedure to find the confidence interval at the specified confidence level.

d. If appropriate, find the margin of error for the estimate of $p$and express the confidence interval in terms of the sample proportion and the margin of error

Given in the question that,

$x=8andn=40,95\%level$. we have to determine the sample proportion

The number of success is$x=8$,the sample size of a sample random sample from a population is$20and90\%level$

The formula of sample proportion $\hat{p}=\frac{x}{n}$

Substitute $x=16\&n=20$

$\hat{p}=\frac{8}{40}$

We have to decide whether using the one-proportion $z$-interval procedure is appropriate.

There are 2 basic assumptions:

1- A basic random sample should be used.

2-Both the number of successes $\left(x=8\right)$and failures$\left(n-x\right)$should be at least $5$.

Here the number of success ,$x=8$is larger than $5$.

The number of failure is,

$$\begin{gathered} n-x=40-8 \\ =32 \end{gathered}$$

The number of failure $n-x$is larger than $5$.

The number of failure is larger than $5$as the result, the one-proportion $z$interval procedure is appropriate.

We have to find out that if appropriate, use the one-proportion $z$-interval procedure to find the confidence interval at the specified confidence level.

From part(a) $\hat{p}=0.2$

The value of $z_{0.25}=1.96$

$\hat{p}\pm z_{\frac{a}{2}}·\sqrt{\hat{p}(1-\hat{p})/n}=0.2\pm 1.960·\sqrt{0.2(0.8)/40}$

$$\begin{gathered} =0.2\pm 1.960·\sqrt{0.16/40} \\ =0.2\pm 1.960·\sqrt{0.004} \\ =0.2\pm 1.960·(0.0632) \\ =0.2\pm 0.1239 \\ =(0.2-0.1239,0.2+0.1239) \\ \approx (0.076,0.324) \end{gathered}$$

Thus the confidence interval is$\left(0.076,0.324\right)$

We have to find out that If appropriate, find the margin of error for the estimate of $p$and express the confidence interval in terms of the sample proportion and the margin of error.

The formula of margin of error:$E=z_{\frac{\alpha }{2}}·\sqrt{\frac{\hat{p}(1-\dot{\hat{p}})}{n}}$

Here, $\alpha =0.05and\hat{p}=0.2$and $n=40$

$$\begin{gathered} E=z_{\frac{\text{0.05}}{2}}·\sqrt{\frac{0.2(1-0.2)}{40}} \\ =z_{0.025}·\sqrt{\frac{0.2(0.8)}{40}} \\ =2_{0.025}·\sqrt{\frac{0.16}{40}} \\ =z_{0.025}·\sqrt{0.004} \end{gathered}$$

The value of $z_{0.025}=1.96$

$$\begin{gathered} E=1.960\left(0.0632\right) \\ =0.1239 \end{gathered}$$

As the result, the margin of error is$P\pm Ethatis\left(0.2\pm 0.1239\right)$