The problem is to test, based on the level of mercury, whether the two batches, X and Y, originated from the same catch. The statistical method used to perform this kind of test is the Mann-Whitney U-test.
Mann-Whitney U-test is used to compare whether two independent samples have been drawn from a population with the same characteristics. And it is used only when there are two samples to compare. In case there are two independent samples, Kruskal-Wallis test is used to investigate whether the samples were drawn from the population with similar characteristics.
The following steps would be followed when doing a Mann-Whitney u-test:
- Levels of mercury content in both batches are ranked in the order of lowest to the highest while indicating the batch that they originate from as shown in the table below:
N1= number of items in batch X, N1=32
N2=number of items in batch Y, N2=26
R1=sum of ranks of all the items in batch X, R1=2+4+5+6+7+8+9+12+13+14+15+16+18+36+20+22+24+26+27.5+27.5+29+32+34+40+41+44+45+48.5+51+54+56=786.5
R2=sum of ranks of all the items in batch Y, R2=1+3+10+11+21+23+25+30+31+33+35+36+37+38+39+42+43+46.5+46.5+48.5+50+52+53+55+57+58=924.5
Now get the Computed U-statistic given by the formula;
U = N1N2 +N1 (N1+1)/2 –R1
U = (32*26) +32(32+1)/2 -786.5
=573.5
Then if the two samples came from a population with the same characteristics, the mean of the U-statistic is given by;
Mean of U-statistics = N1N2/2 = (26*32)/2=416
Now getting the standard error of the U-statistics, we have;
Standard error =N1N2(N1+N2+1)/√12
=32*26(59)/12
=67.84
The test follows a normal distribution when both samples are greater than 10 .that is N1 =32, N2=26
- Stating the hypothesis:
Null hypothesis; H0; µ1 =µ2 (µ1-mean for batch X and µ2-mean for Y) {Batches from the same catch}
Alternative hypothesis; HA; µ1≠µ2{Batches from different catch}
- Testing for significance:
Mann-whitney U-test is always a Two-tailed kind of test.
At 5% level of significance, that is α =0.05, the normal distribution value is 1.96 (z =1.96) obtained from the normal tables.
The confidence interval is given as;
Upper limit = mean of U-stat +1.96*standard error of U-stat
=416 +1.96(67.84)
=548.97
Lower limit = 416 – 1.96(67.84)
=283.03
The computed U-statistic is 573.5 which falls in the rejection region. Therefore reject the null hypothesis
Conclusion
The computed U-statistics which is 573.5 falls outside the confidence interval (283.03 to 548.97). The value of the computed U-statistics, 573.5, falls in the rejection region. Therefore, the null hypothesis, which stated that the two batches originated from the same catch, is rejected and the alternative hypothesis, which stated that the two batches did not originate from the same catch, is accepted. This means that based on the level of mercury content, the two batches did not originate from the same catch at 5% level of significance (α=5%). This means that the researcher is 95% confident that the two batches did not originate from the same catch. Normally, in a normal distribution curve, if the computed value of Mann-Whitney U-statistics falls outside the acceptance region, the null hypothesis is rejected. Mann-Whitney U-statistics test was preferred because the responses were ordinal. This means that it is impossible to quantify the results.
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