Question

We want to compare the soil water content (% water by volume) of two fields growing bell peppers.
The claim is that the two fields have different soil water content.
Use the data below to test the hypothesis that the fields have different soil water content.
Field 1 is in list1 and Field 2 is in list2.
We do not know whether the water content values are normally distributed or not, but their variances are equal.
Provide your answers below.
Use alpha = 0.05

F1 F2
15.1 12.1
11.2 10.2
10.3 13.6
10.8 8.1
16.6 13.5
8.3 7.8
9.1 11.8
12.3 7.7
9.1 8.1
14.3 9.2
10.7 14.1
16.1 8.9
10.2 13.9
15.2 7.5
8.9 12.6
9.5 7.3
9.6 14.9
11.3 12.2
14 7.6
11.3 8.9
15.6 13.9
11.2 8.4
13.8 13.4
9 7.1
8.4 12.4
8.2 7.6
12 9.9
13.9 26
11.6 7.3
16 7.4
9.6 14.3
11.4 8.4
8.4 13.2
8 7.3
14.1 11.3
10.9 7.5
13.2 9.7
13.8 12.3
14.6 6.9
10.2 7.6
11.5 13.8
13.1 7.5
14.7 13.3
12.5 8
10.2 11.3
11.8 6.8
11 7.4
12.7 11.7
10.3 11.8
10.8 7.7
11 12.6
12.6 7.7
10.8 13.2
9.6 13.9
11.5 10.4
10.6 12.8
11.7 7.6
10.1 10.7
9.7 10.7
9.7 10.9
11.2 12.5
9.8 11.3
10.3 10.7
11.9 13.2
9.7 8.9
11.3 12.9
10.4 7.7
12 9.7
11 9.7
10.7 11.4
8.8 11.9
11.1 13.4

One Tail or Two? - [ Choose ]TwoCU2.046907e-02One1472.0FTR HoReject Ho2080.01.7938214546.5571369e-027.261351e-021.80869884016

Test Statistic = [ Choose ]TwoCU2.046907e-02One1472.0FTR HoReject Ho2080.01.7938214546.5571369e-027.261351e-021.80869884016

p = [ Choose ]TwoCU2.046907e-02One1472.0FTR HoReject Ho2080.01.7938214546.5571369e-027.261351e-021.80869884016

Decision = [ Choose ]TwoCU2.046907e-02One1472.0FTR HoReject Ho2080.01.7938214546.5571369e-027.261351e-021.80869884016

Enter CU - [ Choose ]TwoCU2.046907e-02One1472.0FTR HoReject Ho2080.01.7938214546.5571369e-027.261351e-021.80869884016

Answer 1

using minitab>stat>basic stat>2 sample t

we have

Two-Sample T-Test and CI: F1, F2

Two-sample T for F1 vs F2

N Mean StDev SE Mean
F1 72 11.42 2.08 0.25
F2 72 10.63 3.07 0.36

Difference = μ (F1) - μ (F2)
Estimate for difference: 0.790
95% CI for difference: (-0.073, 1.654)
T-Test of difference = 0 (vs ≠): T-Value = 1.81 P-Value = 0.073 DF = 142
Both use Pooled StDev = 2.6216

Ho:μ (F1) = μ (F2)

Ha:μ (F1) μ (F2)

this is two tailed test

Test Statistic = 1.81

p = 0.073

Decision = since p value is greater than 0.05 so we do not reject H0