Question 638:

Asked on February 23, 2009

Tags: 1-sample t-test

1

Answer:

No answer provided yet.

The Null Hypothesis (Ho) is that the mean waist size for males under 40 is not different than 34.

 

The alternative Hypothesis (Ha) is that the mean waist size for males under 40 is different than 34.

 

Since we are comparing a sample mean against a test value, we can use the 1-sample t-test. First we find the mean and standard deviation for this sample. We get a mean of 33.83 and standard deviation of 2.526.

 

The test statistics t* is found by dividing the difference between the observed mean and test mean 33.83-34 = -.16667 and and dividing by the standard error of the mean (SEM). The standard error of the  mean is the standard deviation divided by the square root of the sample size = 2.526/SQRT(18) = .595462. Our test statistics is then -.16667/.595462 = -.27989.

 

Next we need to find the probability associated with this t-value by using a t-table or the  Percentiles from the t-distribution calculator. http://www.usablestats.com/calcs/tdist We enter the t-value -.27989 and 17 degrees of freedom on 2 sided area. We get the p-value .7829.  This p-value is well above the common cut-off value of .05 so we would NOT reject the null hypothesis. Given this sample of data we cannot conclude the average waist size is different than 34.

 

If we needed to calculate the critical value of t given an alpha of .05 and 17 degrees of freedom we can also use a t-table or the inverse t distribution calculator http://www.usablestats.com/calcs/tinv and enter .05 and 17 degrees of freedom. We get the critical value of 2.1098. We'd need to get a test statistic with an absolute value greater than 2.1098 to reject the Null Hypothesis.

 

Note: The note: "The standard deviation has increased from its previous value of 15" doesn't seem to apply here. First, the standard deviation for the sample is only 2.526, so it well less than 15. Check the question and be sure information wasn't left out or improperly entered.