# In a survey, 598 out of 1,399 US adults said they drink at least 4 cups of coffee a day. Find a...

## Question:

In a survey, 598 out of 1,399 US adults said they drink at least 4 cups of coffee a day.

Find a point estimate (P) for the population proportion of US adults who drink at least 4 cups of coffee a day, then construct a 95% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day.

## Confidence Interval for a Proportion:

Confidence interval gives upper and lower bounds true population proportion is most likely to lie below or above the beat point estimate (the P-hat). The length of the confidence interval is determined by sample size, sample variability and level of confidence chosen.

Given that;

{eq}x=598\\n=1399 {/eq}

The point estimate of population proportion is the P-hat:

{eq}\displaystyle \hat p=\frac{x}{n}=\frac{598}{1399}=0.43 {/eq}

Use equation below to construct 95% confidence interval for a proportion:

{eq}\displaystyle \left(\hat p\pm z_{\frac{\alpha}{2}}\times \sqrt{\frac{p(1-p)}{n}}\right) {/eq}

Find critical value z that corresponds to 95% level of confidence:

{eq}\displaystyle \frac{\alpha}{2}=\frac{1-0.95}{2}=0.025\\Z_{0.025}=\pm 1.96 {/eq}

Plug in values into the equation and solve for the upper and lower bounds:

{eq}\displaystyle \left(0.43\pm 1.96\times \sqrt{\frac{0.43(1-0.43)}{1399}}\right)\\(0.43\pm 0.026)\\(0.404, 0.456) {/eq}

The 95% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day is contained between 0.404 and 0.456.