Question

In a random sample of 150 customers of a high-speed internet provider, 63 said that their service had been interrupted one or more times in the past month. Find a 95% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month.

Answer 1

Answer:

The correct answer is "0.3410, 0.4990".

Step-by-step explanation:

Given values are:

$n=150$

$p=\frac{63}{150}$

  $=0.42$

At 95% confidence interval,

C = 95%

z = 1.96

As we know,

⇒ $E=z\sqrt{\frac{p(1-p)}{n} }$

By substituting the values, we get

       $=1.96\sqrt{\frac{0.42\times 0.58}{150} }$

       $=1.96\sqrt{\frac{0.2436}{150} }$

       $=0.0790$

hence,

The confidence interval will be:

= $p \pm E$

= $0.42 \pm 0.079$

= $(0.3410,0.4990)$