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

542

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

Answer 2

Answer: 543

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((z_(\alpha/2))/(E))^2`

`\Rightarrow n=0.25((2.33)/(0.05))^2=542.89\approx543`

Hence, the company survey minimum sample having size =543