Question

Suppose an advertising company wants to determine the current percentage of customers who read print magazines. How many customers should the company survey in order to be 98% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion of customers who read print magazines?

Answer 1

Answer: 543

Step-by-step explanation:

Given : Level of confidence = 0.98

Significance level : $\alpha=1-0.98=0.02$

By using the normal distribution table,

Critical value : $z_{\alpha/2}=2.33$

Margin of error : $E=5\%=0.05$

The formula to find the population proportion if prior proportion of population is unknown  :-

$n=0.25(\dfrac{z_{\alpha/2}}{E})^2$

$\Rightarrow n=0.25(\dfrac{2.33}{0.05})^2=542.89\approx543$

Hence, the company survey minimum sample having size  =543

Answer 2

Answer:

542

Step-by-step explanation:

Given the information in the question,  ME=0.05  since  5%=0.05  and  

z  α/2 = z0.01 = 2.326  

because the confidence level is  98% . 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   ′  =0.5 , then  q ′ =1−0.5=0.5 . Therefore,

n   = z^2  p ′ q/  M E^2

  = 2.326^2  (0.5)(0.5)/ 0.05^2  

=541.0276 (Round Up) = 542