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A nutritional scientist was trying to get a new herbal supplement approved by the Food and Drug Administration. Her belief is that the herbal supplement can improve endurance in humans. The scientist took 30 mice and split them into 15 groups of two mice. The two mice in each group are of the same species, have same diet patterns, and same size.
One of the mice in each group was selected randomly and given the herbal supplement while the other mice was not given the supplement. The scientist recorded the total running time of the two mice. The scientist wanted to prove that there is a difference between the mean endurance time of a mouse with the supplement and the mean endurance time of a mouse without the supplement. The parameter for this test is m.
For this test mn represents the mean endurance time for mice not given herbal supplement and ms = represents the mean endurance time for mice given herbal supplement. The test is looking for the difference between the two parameters. I will denote this as md. Therefore md is ms subtracted from mn. The null hypothesis for this test is the difference between the two parameters equals zero (Ho: md = 0). The alternative hypothesis for this test is the difference between the two parameters does not equal zero (H 1: md? 0).
This hypothesis test was a matched-pairs sample with two-population mean. There are several reasons that this test was chosen. The size, species, and dieting habits for all 30 mice could not possibly be the same, and therefore would factor in the outcome of the test. There is no guarantee that the mice that were chosen to get the supplement were not all heavy eaters, large mice, or white mice.
The only way that there could be a fair test is if two mice of the same characteristics are paired up and one given the supplement and the other was not given the supplement. Also one mice form each group was chosen at random to get the supplement. The last reason this type of hypothesis test was chosen is because the population in normally distributed, which has to be true for this to work. In this test it is necessary to produce a test statistic. The test statistic for this test is (d-do) / (sd/ (n^ (1 / 2) ) ). Where d is?
di / n and sd is ( ( (? di^ 2) - ( (? di) ^ 2 /n) ) / (n- 1). On the histogram that I have printed up for this test the difference between the mean is on the horizontal axis while the percentage is on the vertical axis.
By the graph you can tell that approximately 40 % of the mice without supplements had a slightly greater endurance than the mice with the supplements. About 55 % of the mice with the supplement had a slightly greater endurance than the mice without the supplement. Also about 10 % of the mice with the supplement had much higher endurance than the mice without the supplement. For this the decision is to fail to reject the null hypothesis because the p-value is greater than alpha (. 05). The conclusion is that at alpha equal to. 05 (a = . 05) there is insufficient evidence to conclude that the mean difference between the two types of mice (ones with the supplement and ones without the supplement) is equal to zero.
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Research essay sample on Null Hypothesis N 1