Question

An Internet service provider sampled 540 customers and found that 55 of them experienced an interruption in their service during the previous month. Construct a 95% confidence interval for the proportion of all customers who have experienced a service interruption. (0.080, 0.123) (0.873, 0.924) (0.076, 0.127) (0.102, 0.898)

Answer 1

Answer: a 95% confidence interval for the proportion of all customers who have experienced a service interruption: (0.076, 0.127).

Explanation:

Let p be the population proportion of all customers who have experienced a service interruption.

Confidence interval for p:
`\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`, where
`\hat{p}` = sample proportion, n= sample size, z* = critical z-value.

Given: n = 540

`\hat{p}=(55)/(540)=0.102`

z* = 1.96

The required confidence interval:

`0.102\pm (1.96)\sqrt{(0.102(1-0.102))/(540)}\\\\=0.102\pm (1.96)(0.013024)\\\\=0.102\pm$$0.02552704\\\\=(0.102-$$0.02552704, 0.102+0.02552704)\\\\\approx(0.076, 0.127)`

Hence, a 95% confidence interval for the proportion of all customers who have experienced a service interruption: (0.076, 0.127).