Question

Out of 80 customers surveyed, 25 ordered cheese pizza. What is the 99% confidence interval for the true proportion of customers who ordered a cheese pizza?

Answer 1

Out of 80 customers surveyed, 25 ordered cheese pizza. What is the 99% confidence interval for the true proportion of customers who ordered a cheese pizza?

Solution

A sample mean is

$$\bar{x} = \frac{25}{80} = 0.3125.$$

A sample variance is

$$Var(x) = \sqrt{\frac{\left[\left(\frac{25}{80}\right) * \left(\frac{80 - 25}{80}\right)\right]}{80}} = 0.05182.$$

The confidence interval = sample mean ± zₐ/₂ * (sample variance).

Level of confidence = 1 - α.

In our case:

Level of confidence = 1 - α = 99% → α = 1%.

A standard error:

$$z_{\alpha/2} = z_{.005} = 2.576.$$

The 99% confidence interval = sample mean ± 2.576 * (sample variance).

99% C.I. = 0.3125 ± 2.576 * 0.05182 = 0.3125 ± 0.1335.

99% C.I. = (0.179; 0.446).

Answer: (0.179; 0.446).