A recent survey conducted by Wall Street Journal indicates that the average reading time of 100 its readers/participants was 45 minutes and the standard deviation was 20 minutes. The journal is now planning another survey study. If the acceptable error is 3 minutes and a confidence level of 99% is desired, what would be the minimum sample size needed for the upcoming survey?

Given A recent survey conducted by Wall Street Journal indicates that the average reading time of 100 its readers/participants was 45 minutes and the standard deviation was 20 minutes

sample size n= 100

mean =45 min and standard deviation = 20 minutes

If the acceptable error is 3 minutes and a confidence level of 99% is desired

Given margin of error is '3' min

Step(ii):-

Margin error

We know that margin of error = [tex]z_{0.99} \frac{S.D}{\sqrt{n } }[/tex]

Now we determine the sample size

[tex]n = (\frac{z_{0.99}S.D }{M.E} )^2[/tex]

The 99% of level of significance 'Z' value = 2.57

[tex]n = (\frac{2.57 (20) }{3} )^2[/tex]

n = (17)^2

n = 289

Conclusion:-

The minimum sample size needed for the upcoming survey

n = 289

Verification:-

Margin error = [tex]Margin error=z_{0.99} \frac{S.D}{\sqrt{n } }=\frac{2.57 (20)}{\sqrt{289} }= 3.02[/tex]

Given data of margin error also =3

so both are equal

There fore the minimum sample size needed for the upcoming survey

n = 289