Answer to Question #259094 in Statistics and Probability for ItzHammadK

Question

An experiment was conducted to investigate the effect of two treatment T1, T2 on wheat growth. The experiment was carried out in two fields. 40 randomly selected plants were treated(20 with T1 and 20 withT2) in each field. The height of the mature plants were recorded as below (in cms). The interset is to answer the following questions.

1) Do the treatments appear to differ in their effect on the growth?

2) Is growth different in the fields?

3) Does the difference between the treatment differ between two fields?

Field 1 (Treatment 1)

90, 123, 108, 93, 100

108, 47, 105, 94, 123

90, 87, 124, 90, 77

110, 65, 93, 54, 60

Field 2 (Treatment 1)

56, 70, 66, 78, 74

72, 108, 87, 99, 61

98, 50, 53, 78, 76

74, 87, 81, 67, 61

Field 1 ( Treatment 2)

55, 54, 38, 72, 73

40, 66, 88, 54, 70

73, 53, 50, 24, 66

51, 46, 42, 40, 21

Field 2 (Treatment 2)

40, 40, 83, 30, 46

53, 50, 50, 24, 25

24, 22, 43, 54, 30

34, 35, 36, 38, 30

Answer 1

1)

hull hypothesis:

$H_0:\mu_1=\mu_2$ , treatments do not appear to differ in their effect on the growth

alternative hypothesis:

$H_a:\mu_1\neq\mu_2$ , treatments appear to differ in their effect on the growth

$\mu_1,s_1$ - for field with treatment 1

$\mu_2,s_2$ - for field with treatment 2

$\mu_1=83.4,s_1=20.9$

$\mu_2=46.6,s_2=17.2$

$t=\frac{\mu_1-\mu_2}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{83.4-46.6}{\sqrt{20.9^2/40+17.2^2/40}}=8.600$

for test with unequal variances:

$df=\frac{(s_1^2/n_1+s^2_2/n_2)^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}}=\frac{(20.9^2/40+17.2^2/40)^2}{\frac{(20.9^2/40)^2}{39}+\frac{(17.2^2/40)^2}{39}}=\frac{335.5}{3.06+1.40}=75$

for $\alpha=0.05$ :

$t_{crit}=1.99$

Since $t>t_{crit}$ , we reject the hull hypothesis. So, treatments appear to differ in their effect on the growth.

2)

hull hypothesis:

$H_0:\mu_1=\mu_2$ ​, growth is not different in the fields

alternative hypothesis:

$H_a:\mu_1\neq\mu_2$ ​, growth is different in the fields

$\mu_1,s_1$ ​- for field 1

$\mu_2,s_2$ - for field 2

$\mu_1=72.9,s_1=27.6$

$\mu_2=57.1,s_2=23.2$

$t=\frac{\mu_1-\mu_2}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{72.9-57.1}{\sqrt{27.6^2/40+23.2^2/40}}=2.7715$

for test with unequal variances:

$df=\frac{(s_1^2/n_1+s^2_2/n_2)^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}}=\frac{(27.6^2/40+23.2^2/40)^2}{\frac{(27.6^2/40)^2}{39}+\frac{(23.2^2/40)^2}{39}}=\frac{1056.25}{9.30+4.64}=76$

for $\alpha=0.05$ :

$t_{crit}=1.99$

Since $t>t_{crit}$ , we reject the hull hypothesis. So, growth is different in the fields.

3)

hull hypothesis:

$H_0:\Delta_1=\Delta_2$ ​​, difference between the treatment does not differ between two fields

alternative hypothesis:

$H_a:\Delta_1\neq\Delta_2$ ​​, difference between the treatment differ between two fields

$\Delta_1=92.05-53.08=38.97$ ​​- for field 1

$\Delta_2=74.80-39.35-=45.45$ - for field 2

$t=\frac{\Delta_2-\Delta_1}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{45.45-38.97}{\sqrt{27.6^2/40+23.2^2/40}}=1.137$

for test with unequal variances:

$df=\frac{(s_1^2/n_1+s^2_2/n_2)^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}}=\frac{(27.6^2/40+23.2^2/40)^2}{\frac{(27.6^2/40)^2}{39}+\frac{(23.2^2/40)^2}{39}}=\frac{1056.25}{9.30+4.64}=76$

for $\alpha=0.05$ :

$t_{crit}=1.99$

Since $t<t_{crit}$ , we accept the hull hypothesis. So, difference between the treatment does not differ between two fields.