Topic 15 Distribution Free Tests For Comparing Two Population Means 15 1 Topic 15 DISTRIBUTION FREE TESTS FOR COMPARING TWO POPULATION MEANS OR TREATMENT MEANS A Two Population Means Using Independent Samples In Topic 13 we had several tests all of which required that the distributions of the sample statistics x1 and x 2 or p1 and p 2 be at least approximately normal For proportions this really isn t a very stringent assumption since usually either the sample sizes are sufficiently large or the population proportions are not too far from 0 5 For sample means this can sometimes be a difficult assumption to meet often the populations from which the samples are taken are very skewed and or the sample sizes are quite small As a result t tests and C I s based on T critical values are WRONG Alternative tests do exist I ll show one such test which does not require any particular shape for the Topic 15 Distribution Free Tests For Comparing Two Population Means 15 2 sampling distributions of x1 and x 2 that s why they are called distribution free tests The assumptions of these tests for comparing two population treatment means are 1 the two samples are independently and randomly taken and 2 the shape of the two sampling distributions must be the same it doesn t matter what shape they are as long as they are the same shape Correct Incorrect Topic 15 Distribution Free Tests For Comparing Two Population Means 15 3 Wilcoxon Rank Sum Test Of The Difference Of Two Population Means To use this test we need to convert first all of the sample data to ranks EXAMPLE Nitrogen loss experiment We originally had 15 observations for the UN treatment and 13 for the U treatment Let s suppose we only had 7 from the UN sample and 5 from the U sample Fertilizer UN U Percentage N loss 10 8 10 5 14 0 13 5 8 0 9 5 12 8 6 3 7 3 14 1 9 8 7 1 Hypotheses Ho 1 2 0 HA 1 2 0 Significance level 0 05 If the null hypothesis is true the 12 observations are actually a single sample from a single population with identical distributions same mean and shape and standard deviation Topic 15 Distribution Free Tests For Comparing Two Population Means 15 4 For this test we first sort the values of both samples combined and rank them from low to high ties are given the average rank within the ties X Sample Ranked X 6 3 U 1 7 1 U 2 7 3 U 3 8 UN 4 9 5 UN 5 9 8 U 6 10 5 UN 7 10 8 UN 8 12 8 UN 9 13 5 UN 10 14 UN 11 14 1 U 12 Now if Ho were true then the ranks should be scattered randomly among the two samples When the alternative hypothesis is true the ranks will be on average higher in one of the samples than in the other The test statistic is the sum of the ranks in the smaller sample which is the U sample Then the ranks are 1 2 3 6 12 and the sum of those ranks is Topic 15 Distribution Free Tests For Comparing Two Population Means Test Statistic 15 5 Rank Sum 1 2 3 6 12 24 Under the null hypothesis the mean value of the n n n2 1 Rank Sum is Rank Sum 1 1 and its 2 n1n2 2 variance is Rank n1 n2 1 Sum 12 Intuitively if the sample Rank Sum is not close to the mean value then the two samples are not from the same distribution If it is then they are likely to be from the same distribution Hence we need to compare the test statistic Rank Sum to cutoff values that are small or high to see if we should reject the null hypothesis If the test statistic Rank Sum falls between these two cutoff values then it is reasonably close to its mean and we conclude that the null hypothesis is true If it falls below the lower cutoff then we conclude that population 1 has a mean smaller than the mean of population 2 If it falls above the upper cutoff then we conclude that population 1 has a mean larger than population 2s mean To find if we should reject the null hypothesis we need to know the 2 sample sizes the value of the Rank Sum whether we are doing a 1 or 2 tailed test and the type I error rate Topic 15 Distribution Free Tests For Comparing Two Population Means 15 6 Now n1 5 n2 7 and the sample 1 Rank Sum 24 So look in the row in the table with these two sample sizes in this order and under the 2 columns for the two tailed test The column for n1 5 intersected with the row for n2 7 gives us TU 45 and TL 20 for a 1 tailed 0 025 or a 2 tailed 0 05 Conclusion For a 2 tailed test we fail to reject the null hypothesis since TL 20 Rank Sum 24 TU 45 Based on these two samples there is insufficient evidence that the percentage nitrogen loss differs for the two nitrogen treatments on agricultural fields The more formal statement of the testing procedure is Topic 15 Distribution Free Tests For Comparing Two Population Means 15 7 Wilcoxon Rank Sum Test Of The Difference Of Two Population Means Based On Independent Samples Null hypothesis H0 1 2 0 Alternative Hypothesis is one of three a HA 1 2 0 upper tailed b HA 1 2 0 lower tailed c HA 1 2 0 two tailed Test Statistic Rank Sum sum of the ranks in the smaller sample The two samples are combined and then the values are ranked If some observations have the same value their rank is the average of the ranks of the tied values Decision Rule a Reject H0 if the Rank Sum TU b Reject H0 if the Rank Sum TL c Reject H0 if the Rank Sum TU or TL Topic 15 Distribution Free Tests For Comparing Two Population Means 15 8 Assumptions the two samples are independent and randomly taken and the shapes of the two distributions being sampled are the same EXAMPLE A blood lead level of 70 mg L or less is accepted as safe Unfortunately there is some evidence of neurophysiological symptoms of lead poisoning at levels below 70 mg L Nerve conduction velocities were measured in two groups of individuals those exposed to small amounts of lead and a control group not exposed The data are Exposure Nerve Conduction Velocities Yes 2 46 46 43 41 38 36 31 Control 1 54 50 5 46 45 44 42 41 Test the hypothesis that exposure to lead reduces the mean velocity compared to non exposure Use 0 05 Hypotheses Ho 1 2 0 HA 1 2 0 Significance level Test Statistic 0 05 Topic 15 Distribution Free Tests For Comparing Two Population Means …
View Full Document
Unlocking...