Find confidence interval

Question

A certain union wants to estimate the mean hourly wage μ of its members. A random sample of 15 members yields a mean equal to 6.5 dollars an a sample standard deviation equal to 0.9 dollars. Construct a 95% confidence interval for μ.

A) 6.5 +- 0.455

B) 6.5 +- 0.382

C) 6.5 +- 0.498

D) 0.9 +- 0.598

I used the confidence interval for μ when n is large or the population is normal, as the next question asked what assumption must be made to validate the results of the previous question. But when I use the formula, I end up finding the answer to be A, but the answer key say C. So I don't know if the answer key is wrong or I am missing something.

Answer 1

The equation that you used is valid when you know the population standard deviation.  In this problem, the only value that you are given is the sample standard deviation, not the population standard deviation.  So in this problem, you should use the critical value from a Student's t-distribution with n-1 degrees of freedom.  When n=15, this critical value is 2.145, which corresponds to an interval of +-0.498.

 

Some references allow the use of the Normal distribution critical values when n is large enough (e.g., greater than 30) even if the population standard deviation is unknown.  In this situation, the Student's t-distribution is still the correct distribution although there is a much smaller difference between the two critical values, so it may be considered acceptable to use the Normal distribution critical values.