Question

Suppose a marketing company wants to determine the current proportion of customers who click on ads on their smartphones. It was estimated that the current proportion of customers who click on ads on their smartphones is 0.42 based on a random sample of 100 customers.
Required:
Compute a 92% confidence interval for the true proportion of customers who click on ads on their smartphones and fill in the blanks appropriately.

Answer 1

Answer:

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

Step-by-step explanation:

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).