Question

In each of Exercises 11.122-11.127, we have given the numbers of successes and the sample sizes for simple random samples for independent random samples from two populations. In each case,

a. use the two-proportions plus-four $z$-interval procedure to find the required confidence interval for the difference between the no population proportions.

b. compare your result with the corresponding confidence interval found in parts (d) of Exercises 11.100-I1.105, if finding such a confidence interval was appropriate.

$x_1=18,n_1=40,x_2=30,n_2=40;80\mathrm{\%}$confidence interval

Short answer

1. For the difference between the two-population proportion, the needed confidence interval is $-0.421$to $-0.159-0.421$to $-0.159$
   
2. The results are in line with the stated exercise outcomes, with an$80\%$ confidence level.

Step-by-step solution

Part (a) Step 1: Given Information

Given in the question that, $x_1=18,n_1=40,x_2=30,n_2=40;80\mathrm{\%}$confidence interval.

We have to determine the needed confidence interval for the difference in the proportions of the two populations.

Part (a) Step 2: Explanation

Let's compute the value of ${\tilde{p}}_1$ as follow:

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1+1}{n_1+2} \\ =\frac{18+1}{40+2} \\ =0.45 \end{gathered}$$

Consider the formula for ${\tilde{p}}_2$as follow:

${\tilde{p}}_2=\frac{x_2+1}{n_2+2}$

So, the value of ${\tilde{p}}_2$is:

localid="1651426030441" $$\begin{gathered} {\tilde{p}}_2=\frac{x_2+1}{n_2+2} \\ =\frac{30+1}{40+2} \\ =0.74 \end{gathered}$$

Part (a) Step 3: Calculate the value of required confidence interval

Let's find the value of $\alpha$first,

$$\begin{gathered} 80=100(1-\alpha ) \\ \alpha =0.2 \end{gathered}$$

We observed that the values of $z$at $a/2$from the table of $z$score is $1.282$

For the difference between the two-population proportion, the needed confidence interval is determined as,

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\alpha /2}\cdot \sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.45-0.74)\pm 1.282\times \sqrt{\frac{0.45(1-0.45)}{40+2}+\frac{0.74(1-0.74)}{40+2}}$

$$\begin{gathered} =-0.29\pm 0.131 \\ =-0.421\text{to}-0.159 \end{gathered}$$

Part (b) Step 1: Given Information

Given in the question that, $x_1=18,n_1=40,x_2=30,n_2=40;80\mathrm{\%}\text{confidence interval}$

We have to determine the needed confidence interval for the difference in the proportions of the two populations.

Part (b) Step 2: Explanation

Using the two-proportions plus-four $z$-interval technique, the needed confidence interval for the difference between the two-population proportion is $-0.421$to $-0.159$.