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 0.64. How many customers should the company survey in order to be 95% confident that the margin of error is 0.21 for the confidence interval of true proportion of customers who click on ads on their smartphones?

Please explain in detail how you get your values. I am having a very hard time understanding where Z is coming from.

Answer 1

Solution :

Given that,

$\hat p$ = 0.64

1 - $\hat p$ = 0.36

margin of error = E = 0.21

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96 Using z table,

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.21)² * 0.64 * 0.36

= 20.07

sample size = 21