Question

Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones

Answer 1

Answer:

The value is  $n =271$  

Step-by-step explanation:

From the question we are told that

  The margin of error is  $E = 5\% = 0.05$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 90 ) \%$

=>   $\alpha = 0.10$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } =  1.645$

Generally we will assume that the sample proportion is $\^ p = 0.5$ since we have no information on the proportion of  of customers who click on ads on their smartphones

Generally the sample size is mathematically represented as

      $n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=>    $n = [\frac{1.645 }}{0.05} ]^2 * 0.5 (1 - 0.5 )$

=>    $n =271$    

Answer 2

Using the z-distribution, as we are working with a proportion, it is found that 271 customers should be sampled.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

The margin of error is given by:

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

In this problem:

• 90% confidence level, hence the critical value is z = 1.645.

• No prior estimate, hence p = 0.5.

• Within five percentage points, hence M = 0.05.

Then:

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

$0.05 = 1.645\sqrt{\frac{0.5(0.5)}{n}}$

$0.05\sqrt{n} = 0.5 \times 1.645$

$\sqrt{n} = [10(1.645)]$

$(\sqrt{n})^2 = [10(1.645)]^2$

$n = 270.6$

Rounding up, 271 customers should be sampled.

More can be learned about the z-distribution at https://brainly.com/question/25890103