1. In the unit notes, I state that non-parametric tests are less powerful- that
ID: 3296813 • Letter: 1
Question
1. In the unit notes, I state that non-parametric tests are less powerful- that is, they require more evidence than a parametric test to reject the null hypothesis. In your own words, why do you think this is? Justify your answer.
2. The file "barley.sav" contains data on the yield of a barley from two different years. Different regions of the field are sampled at year one and the total weight of the barley yield is recorded. The same regions are sampled again a year later. Is there a significant difference in barley yield between the two years? Select the appropriate test and run an analysis. Explain what test you chose and why, state the null and alternative hypothesis, report the results, and the meaning of the results.
Year1 Year2 81.0 80.7 105.4 82.3 119.7 80.4 109.7 87.2 98.3 84.2 146.6 100.4 142.0 115.5 150.7 112.2 191.5 147.7 145.7 108.1 82.3 103.1 77.3 105.1 78.4 116.5 131.3 139.9 89.6 129.6 119.8 98.9 121.4 61.9 124.0 96.2 140.8 125.5 124.8 75.7 98.9 66.4 89.0 49.9 69.1 96.7 89.3 61.9 104.1 80.3 86.9 67.7 77.1 66.7 78.9 67.4 101.8 91.8 96.0 94.1Explanation / Answer
1]
The nonparametric tests are sometimes called distribution-free tests because they are based on fewer assumptions. The nonparametric tests are generally less powerful than their parametric test. That is when the alternative is true, they may be less likely to reject H0. Parametric tests involve specific probability distributions.
And General the Null and Alternative hypothesis is fixed in nonparametric test such as
The null hypothesis for each test is H0: Data follow a normal distribution versus
Alternative H1: Data do not follow a normal distribution.
Hence non-parametric tests are less powerful than parametric test.
2]
Here we use two sample t test for testing. Because there is population statndard deviations are not known and sample size are 30.
Null hypothesis H0 : µ1-µ2 = 0. that is there is not significant difference in barley yield between the two years.
Alternative hypothesis HA: µ1-µ2 not equal to 0. that is there a significant difference in barley yield between the two years.
USING MINITAB
Choose Stat > Basic Statistics > 2-Sample t.
Choose Samples in different columns
In First, enter the column containing the first sample.
In Second, enter the column containing the other sample.
If you like, use any dialog box options, and click OK.
Check Assume equal variances. Click OK.
Session window output
MTB > TwoSample 'Year1' 'Year2';
SUBC> Pooled.
Two-Sample T-Test and CI: Year1, Year2
Two-sample T for Year1 vs Year2
N Mean StDev SE Mean
Year1 30 109.0 28.7 5.2
Year2 30 93.1 24.3 4.4
Difference = mu (Year1) - mu (Year2)
Estimate for difference: 15.91
95% CI for difference: (2.18, 29.64)
T-Test of difference = 0 (vs not =): T-Value = 2.32 P-Value = 0.024 DF = 58
Both use Pooled StDev = 26.5661
Decision rule: 1) If p-value < level of significance (alpha) then we reject null hypothesis
2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.
Here p value = 0.024 < 0.05 so we used first rule.
That is we reject null hypothesis and accept alternative hypothesis.
Conclusion: At 5% level of significance there is significant difference in barley yield between the two years.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.