Question

Suppose an automotive repair company wants to determine the current percentage of customers who keep up with regular vehicle maintenance. How many customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within 2 percentage points of the true population proportion of customers who keep up with regular vehicle maintenance?

Answer 1

Answer: 1692

Step-by-step explanation:

Formula to find the sample size :

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

Given : Confidence level : $(1-\alpha)=0.90$

⇒ significance level =$\alpha= 0.10$

z-value for 90% confidence interval (using z-table)=$z_{\alpha/2}=1.645$

Prior estimate of the population proportion (p) of customers who keep up with regular vehicle maintenance is unknown.

Let we take p= 0.5

Margin of error : E= 2%=0.02

Now, the required sample size will be :

$n=0.5(1-0.5)(\dfrac{1.645}{0.02})^2$

Simplify , we get

$n=(0.25)(6765.0625)=1691.265625\approx1692$

Hence, the required sample size = 1692

Answer 2

Answer: 2401

Step-by-step explanation: