Question

Example Problem: [Ch 8, Q14] A simple random sample with n=54 provided a sample mean of 22.5 and a sample standard deviation of 4.4. Develop a 90% confidence interval for the population mean using the relevant fact: i) for a tdistribution with 53 degrees of freedom, ?(? > 1.674) = 0.05; ii) for a tdistribution with 54 degrees of freedom, ?(? > 1.297) = 0.10.

Answer 1

We have the following sample information

• n=54 is the sample size
• is the sample mean
• is the sample standard deviation

Since we do not know the population standard deviation, we will estimate it using the sample SD

• 
• is the standard error of mean
• The value of significance level for 90% confidence interval is
• The t value is given by .
  • The degrees of freedom for t is n-1 = 54-1=53
  • We have been given i) for a tdistribution with 53 degrees of freedom, ?(? > 1.674) = 0.05
  • Hence

90% confidence interval for the population mean is

90% confidence interval for the population mean is [21.4977,23.5023]