Find and interpret 95 % confidence interval for the proportion of all US adults who never clothes-shop online.

A total of 1000 adults in the U S were polled regarding their online vs. in-store clothing purchase habits.

32% of respondents never shop for garments online.

The one-proportion z-interval technique is appropriate because x and n are both 5 or more.

The confidence interval is 95 percent, which equals
$\alpha$=0.05.

It discovers that $z_{\alpha /2}=z_{0.05/2}=1.96$

For p, the confidence interval is of the type

$\begin{array}{rl}& p^{\mathrm{\prime }}-z_{a/2}\sqrt{\frac{p^{\mathrm{\prime }}(1-p^{\mathrm{\prime }})}{n}}\text{to}{p}^{\mathrm{\prime }}+z_{a/2}\sqrt{\frac{p^{\mathrm{\prime }}(1-p^{\mathrm{\prime }})}{n}}\\ & \text{i.e.}0.32-1.960\sqrt{\frac{0.32(1-0.32)}{1000}}\text{to}0.32-1.960\sqrt{\frac{0.32(1-0.32)}{1000}}\\ & \text{i.e.}0.32\pm 1.960\left(0.0148\right)\\ & 0.32\pm 0.0290\\ & (0.291,0.349)\end{array}$

As a result, the 95% confidence interval for the fraction of all US adults who never shop for garments online is (0.291,0.349).

interpretation:

The proportion of all US adults who never buy for garments online has a 95 percent confidence interval between 0.291 and 0.349.