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Math Help - Hypothesis test

  1. #1
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    Hypothesis test

    After running about 17 miles, marathon runners encounter a form of physiological stress which they call "hitting the wall". To better pinpoint where in the race to expect this phenomenon, a sports physiologist has 50 marathon runners race until each feels the stress. The mean number of miles the 50 runners run until the stress occurs is 16.69.

    a) With population standard deviation=0.850 miles, perform a one-sample z-test to determine if the sports physiologist's claim that the mean distance needed to run is more than the 17 miles found in the population. Use a 0.05 level of significance.

    Ho: mu = 17
    Ha: mu > 17

    z=16.69 - 17 / 0.850/sqr rt of 50 = -0.31/0.12 = -2.58

    So, since -2.58 < 1.65 (0.05), do not reject Ho.... "the data does not provide enough evidence in favor of Ha but does provide enough evidence in favor of Ho:mu=17 miles.

    How does that look? Am I supposed to be drawing some sort of a bell curve?

    b) Based on the information given above, find a 95% confidence interval for the mean number of miles run until the runners feel the stress.

    16.69 - 1.96(0.850 / sqr rt of 50) < mu < 16.69 + 1.96(0.850 / sqr rt of 50)

    = 16.69 - .238 < mu < 16.69 + .238 = 16.452 miles, 16.928 miles.
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  3. #3
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    Quote Originally Posted by Jimmy23
    After running about 17 miles, marathon runners encounter a form of physiological stress which they call "hitting the wall". To better pinpoint where in the race to expect this phenomenon, a sports physiologist has 50 marathon runners race until each feels the stress. The mean number of miles the 50 runners run until the stress occurs is 16.69.

    a) With population standard deviation=0.850 miles, perform a one-sample z-test to determine if the sports physiologist's claim that the mean distance needed to run is more than the 17 miles found in the population. Use a 0.05 level of significance.

    Ho: mu = 17
    Ha: mu > 17

    z=16.69 - 17 / 0.850/sqr rt of 50 = -0.31/0.12 = -2.58

    So, since -2.58 < 1.65 (0.05), do not reject Ho.... "the data does not provide enough evidence in favor of Ha but does provide enough evidence in favor of Ho:mu=17 miles.

    How does that look? Am I supposed to be drawing some sort of a bell curve?

    b) Based on the information given above, find a 95% confidence interval for the mean number of miles run until the runners feel the stress.

    16.69 - 1.96(0.850 / sqr rt of 50) < mu < 16.69 + 1.96(0.850 / sqr rt of 50)

    = 16.69 - .238 < mu < 16.69 + .238 = 16.452 miles, 16.928 miles.
    I only checked some of the numbers, but this looks correct to me, one-tailed test and everything. Good job!
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