Nine universities participated in testing the yield capabilities of two new vari
ID: 2948380 • Letter: N
Question
Nine universities participated in testing the yield capabilities of two new varieties of wheat. Each variety was planted on plots of equal area at each university and the yields (kg per plot) are shown in the table below.
(a) Some argued that variety 2 was likely to be the better performer. Test this alternative hypothesis assuming the differences of yields to be approximately normally distributed.
(b) Under the hypothesis of “no difference” between the two varieties, the distribution of “+’s” formed from V2-V1 should behave as what distribution? Completely specify with distribution name and parameter(s).
(c) If the test were carried out in terms of the distribution in (b), what exact p-value would result?
Variety
1
2
3
4
5
6
7
8
9
V1
38
23
35
41
44
29
37
31
38
V2
45
25
31
38
50
33
36
40
43
Variety
1
2
3
4
5
6
7
8
9
V1
38
23
35
41
44
29
37
31
38
V2
45
25
31
38
50
33
36
40
43
Explanation / Answer
Nine universities participated in testing the yield capabilities of two new varieties of wheat. Each variety was planted on plots of equal area at each university and the yields (kg per plot) are shown in the table below.
(a) Some argued that variety 2 was likely to be the better performer. Test this alternative hypothesis assuming the differences of yields to be approximately normally distributed.
Here we have to test the hypothesis that,
H0 " mu1 = mu2 Vs H1 : mu1 < mu2
where mu1 and mu2 are two population means of V1 and V2 respectively.
(b) Under the hypothesis of “no difference” between the two varieties, the distribution of “+’s” formed from V2-V1 should behave as what distribution? Completely specify with distribution name and parameter(s).
Here test statistic follows t-distribution with n1+n2-2 degrees of freedoms.
We can do this test in MINITAB.
steps :
ENTER data into MINITAB sheet --> STAT --> Basic statistics --> 2-Sample t for the mean --> Each sample inits own column --> Select both the samples --> Options --> Confidence level : 95.0 --> Hypothesized difference : 0.0 --> Alternative hypothesis : < -->Assume equal variances --> ok --> ok
Two-Sample T-Test and CI: V1, V2
Two-sample T for V1 vs V2
N Mean StDev SE Mean
V1 9 35.11 6.47 2.2
V2 9 37.89 7.66 2.6
Difference = ? (V1) - ? (V2)
Estimate for difference: -2.78
95% upper bound for difference: 3.06
T-Test of difference = 0 (vs <): T-Value = -0.83 P-Value = 0.209 DF = 16
Both use Pooled StDev = 7.0877
Test statistic = -0.83
P-value = 0.209
P-value > alpha
Accept H0 at 5% level of significance.
Conclusion : There is not sufficient evidence to say that variety 2 was likely to be the better performer.
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