Answer to Question #263681 in Statistics and Probability for Tim

Question

A statistics professor wants to compare today’s students with those 25 years ago. All his current students’ marks are stored on a computer so that he can easily determine the population mean. However, the marks 25 years ago reside only in his rusty files. He does not want to retrieve all the marks and will be satisfied with a 95% confidence interval estimate of the mean mark 25 years ago. If he assumes that the population standard deviation is 12, how large a sample should he take to estimate the mean to within 2 marks ?

5(b) n = ? for B = 1 mark:

 

A) 553

B) 355

C) 335

D) 333

Answer 1

5(a):

Standard deviation (s) = 12

Margin of error (B) = 2

Z value at 95% confidence interval = 1.96

Following formula can be used to calculate sample size:

$n=\frac{(Z\times s)^{2}}{m^{2}}$

$n=\frac{(1.96\times 12)^{2}}{2^{2}}=138.3$

Largest sample size should take to estimate the mean to within 2 marks is 138.

5(b):

If margin of error (B) is 1 then sample size will be:

$n=\frac{(1.96\times 12)^{2}}{1^{2}}=553.2$

In this case same size should be 553.