Question

During the off-season (January – October), animal scientists at the North Pole train the reindeer to...

During the off-season (January – October), animal scientists at the North Pole train the reindeer to improve their take-off speed. During the most recent off-season, a new training device was used to improve the average time required for reindeer to sprint 200 yards. Each of the current reindeer was timed in the 200-yard dash before and after using the device. The results (in sec) are given in the table below.

Reindeer Time Before Training Time After Training
1 8.19 7.83
2 8.62 8.42
3 7.98 7.65
4 8.41 7.89
5 9.15 9.01
6 7.45 6.69
7 8.55 7.98
8 7.11 6.88
9 9.02 8.56

Using this data, the animal scientists wish to test the following hypotheses using a level of significance of 0.05.

Note that mu1 denotes the average time required for an untrained reindeer to sprint 200 yards and mu2 denotes the average time required for a trained reindeer to sprint 200 yards.

Compute the test statistic. (

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Answer #1

We frame the Null Hypothesis

он : There is no Significant difference between the two Sample i.e before and after.

To test the он we use the Paired t - test

t = \frac{\left | \bar{d} \right |}{S.E}

X Y d = X - Y (d - \bar{d}) (d - \bar{d})^{2}
8.19 7.83 0.36 -0.0367 0.0013
8.62 8.42 0.2 -0.1967 0.0387
7.98 7.65 0.33 -0.0667 0.0044
8.41 7.89 0.52 0.1233 0.0152
9.15 9.01 0.14 -0.2567 0.0659
7.45 6.69 0.76 0.3633 0.1320
8.55 7.98 0.57 0.1733 0.0300
7.11 6.88 0.23 -0.1667 0.0278
9.02 8.56 0.46 0.0633 0.0040
TOTAL 0.3194
MEAN 0.3967

\Sigma (d - \bar{d})^{2} = 0.3194

S^{2}= \frac{1}{n-1}*\Sigma (d - \bar{d})^{2}

S^{2}= \frac{1}{9-1}*0.3194

\Rightarrow S^{2}= \frac{0.3194}{8}

\Rightarrow S^{2}= 0.0399

Therefore

S.E = \sqrt{\frac{S^{2}}{n}}

\Rightarrow S.E = \sqrt{\frac{0.0399}{9}}

\Rightarrow S.E = \sqrt{0.0044}

\therefore \, \, \, S.E = 0.0666

Therefore

t = \frac{\left | \bar{d} \right |}{S.E}

\Rightarrow t = \frac{\left | 0.3967 \right |}{0.0666}

t_{cal} = 5.9565

Given level of significance = 0.05.= 5%

t_{cri} = t_{(n-1),\alpha } = t_{(9-1),0.05} = t_{8,0.05} = 2.306

Since t_{cal} > t_{cri}; So we Reject он at 5% Level of Significant.

NOTE: The Critical value of t has been extracted from the tabulated value of t; which is posted below.

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