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The mid-distance running coach, Zdravko Popovich, for the Olympic team of an eastern European country claims that his six-month training program significantly reduces the average time to complete a 1500-meter run. Four mid-distance runners were randomly selected before they were trained with coach Popovich’s six-month training program and their completion time of 1500-meter run was recorded (in minutes). After six months of training under coach Popovich, the same four runners’ 1500 meter run time was recorded again.

The results are (for the 4 runners):

Completion time before training 6.0, 7.5, 6.2, 6.8

Completion time after training 5.5, 7.1, 6.2, 6.4

Assume that the times before training and after training are normally distributed.

(A) Construct 95% confidence interval for the difference between the actual mean of completion time before training and the actual mean of completion time after training.

(B) Based on the confidence interval in part (A) can we conclude that there is no significant difference between the actual mean of completion time before training and the actual mean of completion time after training. Hint: Check and see if your confidence interval in part (A) includes 0.

My work:

(A) H0: μ1-μ2 = 0 and Ha: μ1-μ2 > 0

First off I found the mean before training: x̄=6.6250, and then the mean after training: x̄=6.3. Then I found the standard deviation before training = 0.6751, and the standard deviation after training = 0.6583.

Following this, I solved for the test statistic: (6.6250-6.3)/{sqrt[41.79x(1/4+1/4)]} = 0.3250/4.5711 = 0.0711, as well as the critical value to be t(6,0.05) = 1.943

Since 0.0711 > 1.943 is false, do not reject H0.

(B) Therefore there is no significant difference between the actual mean of completion time before and after training.

I'm getting a bad feeling about my answer of 0.0711 but I'm not sure where I'd be going wrong in finding the test statistic's value. All help is appreciated, thank you!

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  • $\begingroup$ I have not checked your numbers, but one can hardly ever reject the null hypothesis on the basis of an experiment with $4$ people. $\endgroup$ Commented Dec 1, 2015 at 1:41

1 Answer 1

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The means and standard deviations are correct, you just need to remember to include both of them in solving the test statistic because you are trying to estimate the difference between two means.

test statistic: (mean before - mean after) ± za/2 √(sd before^2 / n)+(sd after^2 / n) = (6.625-6.3) ± 1.96 √(0.6751^2 / 4) + (0.6583^2 / 4) = 0.325 ± 1.96 √(0.1139+0.10834) = -0.598989< μ1-μ2 < 1.248989

Therefore this does include 0 because it is contained within the interval

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  • $\begingroup$ Thank you very much for the help! Would 0.598989 not be negative though because of 0.325 - (1.96*0.4714) = 0.325 - 0.9239 = -0.5989 ? $\endgroup$ Commented Dec 1, 2015 at 20:00
  • $\begingroup$ Yeah you're right, thanks. So it does contain 0 which makes more sense in context. $\endgroup$ Commented Dec 1, 2015 at 20:12
  • $\begingroup$ Perfect. Since the interval contains 0 would this mean that there is not a significant difference? Thank you! $\endgroup$ Commented Dec 1, 2015 at 20:20

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