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 95% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion of customers who keep up with regular vehicle maintenance?
Answer: 385
Given the information in the question, EBP=0.05 since 5%=0.05 and zα2=z0.025=1.96 because the confidence level is 95%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore,
n=z2p′q′EBP2=1.962(0.5)(0.5)0.052=384.2
Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 385 customers.

Sample proportion must be within 5 percentage points of the true population proportion. The difference between sample proportion and true population proportion is refereed to as Margin of Error. So this means,

Margin of Error = 5% = 0.05

We have to find the minimum sample size required to keep the margin of error upto 5 percentage points. The formula to calculate the margin of error for a sample proportion p is:

[tex]M.E=z_{\frac{\alpha }{2}} \times \sqrt{\frac{p(1-p)}{n} }[/tex]

Here,

[tex]z_{\frac{\alpha }{2}}[/tex] is the critical value for the given confidence level. The critical value, as seen from the z-table, for 95% confidence level is 1.96

p = sample proportion

In case the value of sample proportion is not given in the question, we always assume it to be equal to 0.50

So, p = 0.5

n = sample size , which we need to find out

Using the values we have in the above formula:

[tex]0.05=1.96 \times \sqrt{\frac{0.5(1-0,5)}{n} } \\\\\frac{0.05}{1.96} =\sqrt{\frac{0.5(1-0,5)}{n} } \\\\(\frac{0.05}{1.96} )^{2}=\frac{0.5(0,5)}{n}\\\\ n \times (\frac{0.05}{1.96} )^{2} = 0.25\\\\ n=0.25 \times (\frac{1.96}{0.05} )^2\\\\ n=384.16[/tex]

In all such cases, we round up the value of sample size to the nearest whole number. Therefore, the minimum number of sample size required to keep the margin of error to 5 percentage points at 95% confidence level is 385.