In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 100, p = 0.42$

92% confidence level

So $\alpha = 0.08$ , z is the value of Z that has a pvalue of $1 - \frac{0.08}{2} = 0.96$ , so $Z = 1.75$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.3336$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.5064$

The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).

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3 years ago