One of the simpler ways to test a hypothesis is to use the sign test. It decides whether a sample that gives different information to the average is different enough to be more than a random fluctuation. This example illustrates how it is used:

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The average weight of a rugby reserve squad is 85 kg. The actual on-field team has the following weights: 89, 72, 95, 93, 102, 75, 86, 83, 85, 89, 72, 89, 86 kg. Are the on-field team members heavier than the reserve squad members? |
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Solution |
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Using the average, ‘+’s and ‘–’s are assigned to the on-field weights according to whether they are more or less than the average weight of the reserve squad.
Where values are the same as the average, they are ignored/excluded, and the number ‘n’ representing the total number of players is reduced. For this question, ‘n’ would be reduced from 13 to 12 (since we are ignoring one of the values that equals the average). Of these 12, only 4 are ‘–’s. You can then use a binomial distribution table to calculate how unlikely this is – these tables are in the back of most textbooks. Look up the table for ‘n = 12’ and the two probabilities both equalling 0.5, and add up the following values: Probability of 4 or less ‘–’s = probability of 0 ‘–’s + probability of 1 ‘–’ + probability of 2 ‘–’s + probability of 3 ‘–’s + probability of 4 ‘–’s. = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729 = 7.29%. For us to accept the hypothesis that the on-field team is heavier than the reserve squad, this percentage would have to be less than 5%. Since it is over, we cannot support the hypothesis. If we accepted it, there would be a 7.29% chance of it being wrong. The 5% significance level tells us we can only accept the hypothesis if there is a 5% or less chance of it being wrong. In other words, using the table, we calculated that there is a 7.29 % chance that the difference between the on-field team’s average weight and the reserve squad’s average weight could have occurred through natural random variation. |