Question

An advertising agency for a cereal company wants to know the age of people watching "Fists of Fury Television" on Saturday mornings. A random sample of 50 viewers showed an average age of 12.7 years. From previous experience it is known that r= 2.9 years. Find a 90% confidence interval for the mean age all viewers. How large should our sample be in the previous question, if we want to be sure that the sample mean is written 0.5 years of the true population mean? (use C and r from the problem.)

Answer 1

We have standard deviation = 2.9 years, sample mean = 12.7 years and sample size n = 50

z critical for 90% level = 1.645 (using z table)

Using standard deviation (sigma) = 2.9, Margin of error (E) = 0.5 and z critical = 1.645 (for 90% level)

sample size n = [(z*)/E]²

=[(1.645*2.9)/0.5]²

= [4.7705/0.5]²

= 91.031

= 92 (rounded up to nearest integer)