we have given the members of successes and the sample sizes for simple nunulom samples for independent random samples from two populations. In each case,

a. use the noo-proportions plus-four z-internal procedure to find the wquimal confidence interval for the difference between the nov population proportions.

b. compare your resalt with the corresponding confidence interval found in pards (d) of Exencien 11. 100-11.105, if finding such $d$confidence intenal was appropriate.

$x_1=10,n_1=20,x_2=18,n_2=30;80\%$confidence interval

The given values are,

$$\begin{gathered} x_1=10, \\ {n}_1=20, \\ {x}_2=18, \\ {n}_2=30\text{,} \\ \text{and}80\%confidenceinterval. \end{gathered}$$

The formula for ${\tilde{p}}_1$is given by,

${\tilde{p}}_1=\frac{x_1+1}{n_1+2}$

The formula for ${\tilde{p}}_2$is given by,

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

The value of ${\tilde{p}}_1$is calculated as,

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1+1}{n_1+2} \\ =\frac{10+1}{20+2} \\ =0.5 \end{gathered}$$

The value of ${\tilde{p}}_2$is calculated as,

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

$$\begin{gathered} =\frac{18+1}{30+2} \\ =0.59 \end{gathered}$$

Calculate the value of $\alpha ,$

$80=100(1-\backslash alpha)$

$\backslash alpha=0.2$

The value of $z$at$\alpha /2$from the $z$-score table is $1.282.$

The required confidence interval for the difference between the two-population proportion is calculated as,

$=-0.09\pm 0.176$

The given values are,

$$\begin{gathered} x\_\left\{1\right\}=10, \\ n\_\left\{1\right\}=20, \\ x\_\left\{2\right\}=18, \\ n\_\left\{2\right\}=30 \\ 80\backslash \% \end{gathered}$$

The formula for p_1is given by,

$=\frac{x_{1+1}}{n_1+2}$

The formula for p_2is given by,

$=\frac{x_{2+1}}{n2+2}$

The required confidence interval for the difference between the two-population proportion is -0.266 to 0.086 by using the two-proportions plus-four z-interval procedure. The results are somewhere are appropriate with the given exercise results which is 80% confidence.