Question
A scientist wants to check the yield of wheat from different varieties of wheat. In this regards, the yields of wheat from 16 plots, all of approximately equal fertility, when 4 varieties of wheat (Say, A, B, C and D) were cultivated are recorded in the following table:
|
Plot No. |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
Variety: |
A |
B |
D |
C |
B |
C |
A |
D |
|
Yield: |
32 |
34 |
29 |
31 |
33 |
34 |
34 |
26 |
|
Plot No.: |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
|
Variety: |
B |
D |
A |
C |
B |
A |
B |
C |
|
Yield: |
36 |
30 |
33 |
35 |
37 |
35 |
35 |
32 |
Assuming that the yields of wheat are normally distributed in each variety with approximately equal variances, test the hypothesis that the average yields of wheat in all four varieties of wheat are the same at 1% level of significance.
Answer :
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Skip in 10
We are asked to test whether the average yields of wheat for four varieties (A, B, C, D) are the same at a 1% significance level. This is a classic one-way ANOVA problem. Let’s solve it manually step by step. --- ### Step 1: Organize the data We have 16 plots with 4 varieties. Let’s separate yields by variety. | Variety | Yields | | ------- | -------------- | | A | 32, 33, 33, 35 | | B | 34, 33, 36, 35 | | C | 31, 34, 35, 32 | | D | 29, 34, 37, 26 | Check: each variety has 4 observations. ✅ --- ### Step 2: Calculate the means Mean of A: $$ \bar{X}_A = \frac{32 + 33 + 33 + 35}{4} = \frac{133}{4} = 33.25 $$ Mean of B: $$ \bar{X}_B = \frac{34 + 33 + 36 + 35}{4} = \frac{138}{4} = 34.5 $$ Mean of C: $$ \bar{X}_C = \frac{31 + 34 + 35 + 32}{4} = \frac{132}{4} = 33 $$ Mean of D: $$ \bar{X}_D = \frac{29 + 34 + 37 + 26}{4} = \frac{126}{4} = 31.5 $$ Overall mean: $$ \bar{X} = \frac{133 + 138 + 132 + 126}{16} = \frac{529}{16} = 33.0625 \approx 33.06 $$ --- ### Step 3: Calculate the Sum of Squares Between Groups (SSB) $$ SSB
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