91Ó°ÊÓ

Given in the question that, Since$1973,$Gallup has asked Americans how much confidence they have in a variety of U.S. institutions. One question asked of those polled is whether they have a great deal of confidence in banks. In $2007$, of a random sample of$1008$adult Americans, $413$said yes; and, in $2013$, of a random sample of $1529$adult Americans, $398$said yes. we need to find and interpret a $95\%$confidence interval for the difference between the percentages of adult Americans who had a great deal of confidence in banks For the two years.

we need to use the two-proportions plus four z-interval procedure to find the required confidence interval.

The given values are, $x_1=413,n_1=1008,x_2=398,n_2=1529$, and $95\%$confidence interval.

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

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

Substitute $x_1=413,n_1=1008$we get,

$=\frac{413+1}{1008+2}$

$=0.4099$

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

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

Substitute $x_2=398,n_2=1529$we get,

$=\frac{398+1}{1529+2}$

$=0.2606$

Calculate the value of $\mathrm{Î\pm }$,

$95=100(1-\mathrm{Î\pm })$

The value of $z$ at $\mathrm{Î\pm }/2$ from the $z$-score table is$1.96.$

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

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)Â\pm z_{\mathrm{Î\pm }/2}·\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.4099-0.2606)Â\pm 1.96$

$.\sqrt{\frac{0.4099(1-0.4099)}{1008+2}+\frac{0.2606(1-2606)}{1529+2}}$

$=0.1493Â\pm 0.0375$

$=0.1118$to $0.1868$

As a result, the difference in adult-American percentages is $0.1118$ to $0.1868$.