Question 680:

Asked on April 17, 2009

Tags: test statistic

1

Answer:

No answer provided yet.

We would conduct a 1-sample t-test here. There are 18 values with a mean of 33.83 and standard deviation of 2.526. The t-statistic is equal to the difference between the sample mean and the test-value 34, divide by the standard error of the mean (SE). So the numerator is -.16667. The SE equals the standard deviation divided by the square root of the sample size = 2.526/SQRT(18) = .5954. This makes the t-value of the test statistic -.2798.

 

While the question doesn't ask us, given this data we would NOT reject the NULL and we cannot conclude there is a difference from 34.