Question

Construct a $95\%$ confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph and calculate the error bound.

Short answer

The confidence interval: $CI:(0.532,0.668)$

Margin of error $=0.0679$

Step-by-step solution

Given Information

A $95\%$confidence interval for the population proportion of households.

Explanation

Here we need to find the margin of errors,

That are,

$\text{Margin Of Error}=\left[Z_{\frac{\alpha }{2}}\times \sqrt{\frac{p^{\mathrm{\prime }}\left(1-p^{\mathrm{\prime }}\right)}{n}}\right]$

$=\left[Z_{\frac{0.05}{2}}\times \sqrt{\frac{\frac{120}{200}\left(1-\frac{120}{200}\right)}{200}}\right]$

$=[1.96\times 0.03464]$

$=0.0679$

Now we need to find the confidence interval

$\text{Confidence Interval}=\left[p^{\mathrm{\prime }}\pm Z_{\frac{a}{2}}\times \sqrt{\frac{p(1-p)}{n}}\right]$

$=\left[\frac{120}{200}\pm Z\frac{0.05}{2}\times \sqrt{\frac{\frac{120}{200}\left(1-\frac{120}{200}\right)}{200}}\right]$

$=\left[\frac{120}{200}\pm 1.96\times 0.03464\right]$

$=[0.532,0.668]$

The confidence interval is $CI:(0.532,0.668)$.

The graph: