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Question 680:



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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.

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