In: Statistics and Probability
A dairy man thinks that the average weight gain of his cows depends on two factors : the type of grain which they are fed and the type of grass which they are fed. The dairyman has four different types of grain from which to choose and three different types of grass from which to choose. He would like to determine if there is a particular combination of grain and grass which would lead to the greatest weight gain on average for his cows. He randomly selects three one- year-old cows and assigns them to each of the possible combinations of grain and grass. After one year he records the weight gain for each cow with the following results :
Grass A | Grass B | Grass C | |
Grain A | 175,160,185 | 225,215,230 | 250,240,260 |
Grain B | 190,185,195 | 245,240,255 | 275,260,285 |
Grain C | 210,220,200 | 255,245,265 | 300,310,295 |
Grain D | 225,235,220 | 275,270,280 | 350,360,345 |
Perform the following hypothesis tests at 5 % level of significance. Show all four steps and show complete work. Use P- Value Method. Provide the complete ANOVA TABLE :
Is there a significant interaction between the independent variables (factors) : grass and grain?
b) Is there is a significant difference in average weight gain for the cows among the four different types of grain ?
c) Is there a significant difference in average weight gain for the cows among the three different types of grass?
The given data are elaborately expressed as
Grain-Grass | Grass A | Grass B | Grass C |
Grain A | 175 | 225 | 250 |
Grain A | 160 | 215 | 240 |
Grain A | 185 | 230 | 260 |
Grain B | 190 | 245 | 275 |
Grain B | 185 | 240 | 260 |
Grain B | 195 | 255 | 285 |
Grain C | 210 | 255 | 300 |
Grain C | 220 | 245 | 310 |
Grain C | 200 | 265 | 295 |
Grain D | 225 | 275 | 350 |
Grain D | 235 | 270 | 360 |
Grain D | 220 | 280 | 345 |
Above is the ANOVA-2 way with 3 observations per cell and its analysis is as follws:
Anova: Two-Factor With Replication | ||||||
SUMMARY | Grass A | Grass B | Grass C | Total | ||
Grain A | ||||||
Count | 3 | 3 | 3 | 9 | ||
Sum | 520 | 670 | 750 | 1940 | ||
Average | 173.3333 | 223.3333 | 250 | 215.5556 | ||
Variance | 158.3333 | 58.33333 | 100 | 1215.278 | ||
Grain B | ||||||
Count | 3 | 3 | 3 | 9 | ||
Sum | 570 | 740 | 820 | 2130 | ||
Average | 190 | 246.6667 | 273.3333 | 236.6667 | ||
Variance | 25 | 58.33333 | 158.3333 | 1418.75 | ||
Grain C | ||||||
Count | 3 | 3 | 3 | 9 | ||
Sum | 630 | 765 | 905 | 2300 | ||
Average | 210 | 255 | 301.6667 | 255.5556 | ||
Variance | 100 | 100 | 58.33333 | 1640.278 | ||
Grain D | ||||||
Count | 3 | 3 | 3 | 9 | ||
Sum | 680 | 825 | 1055 | 2560 | ||
Average | 226.6667 | 275 | 351.6667 | 284.4444 | ||
Variance | 58.33333 | 25 | 58.33333 | 3015.278 | ||
Total | ||||||
Count | 12 | 12 | 12 | |||
Sum | 2400 | 3000 | 3530 | |||
Average | 200 | 250 | 294.1667 | |||
Variance | 504.5455 | 418.1818 | 1635.606 | |||
ANOVA-TABLE | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Grains | 23097.22 | 3 | 7699.074 | 96.4058 | 1.59E-13 | 3.008787 |
Between Grasses | 53272.22 | 2 | 26636.11 | 333.5304 | 3.08E-18 | 3.402826 |
Interaction | 3127.778 | 6 | 521.2963 | 6.527536 | 0.000348 | 2.508189 |
Within | 1916.667 | 24 | 79.86111 | |||
Total | 81413.89 | 35 |
(a) step 1
H0(Interaction): There is no significant interaction between Grass and Grain
Ha(Interaction): There is a significant interaction between Grass and Grain
step 2
Calculated F = 6.5275 (See the above last ANOVA-TABLE)
Step 3
p-value of F for 6.5275 with(6, 24) degree of freedom is 0.000348 (See the above last ANOVA-TABLE)
Step 4
Since p value(0.000348) is much less than the 5%(=0.05) significance level, then we reject H0(Interaction) and accept Ha(Interaction) and conclude that there is a significant interaction between Grass and Grain.
(b) step 1
H0(grains): There is no significant difference in average weight gain for the cows among the four different grains
Ha(grains): There is a significant difference in average weight gain for the cows among the four different grains
step 2
Calculated F = 96.4058(See the above last ANOVA-TABLE)
Step 3
p-value of F for 96.4058 with(3, 24) degree of freedom is 0(See the above last ANOVA-TABLE)
Step 4
Since p value(0) is much less than the 5%(=0.05) significance level, then we reject H0(grains) and accept Ha(grains) and conclude that there is a significant difference in average weight gain for the cows among the four different grains.
(c) step 1
H0(grasses): There is no significant difference in average weight gain for the cows among the three different grasses
Ha(grains): There is a significant difference in average weight gain for the cows among the three different grasses
step 2
Calculated F = 333.5304(See the above last ANOVA-TABLE)
Step 3
p-value of F for 333.5304 with(2, 24) degree of freedom is 0(See the above last ANOVA-TABLE)
Step 4
Since p value(0) is much less than the 5%(=0.05) significance level, then we reject H0(grasses) and accept Ha(grasses) and conclude that there is a significant difference in average weight gain for the cows among the three different grasses.