In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST = 320, SSTR = 130,
nT = 19.
Set up the ANOVA table. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)
Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square |
F | p-value |
---|---|---|---|---|---|
Treatments | |||||
Error | |||||
Total |
Test for any significant difference between the mean output levels of the three treatments. Use
α = 0.05.
State the null and alternative hypotheses.
H0: Not all the population means are
equal.
Ha: μ1 =
μ2 =
μ3H0: At least two of the
population means are equal.
Ha: At least two of the population means are
different. H0:
μ1 = μ2 =
μ3
Ha: Not all the population means are
equal.H0: μ1 ≠
μ2 ≠ μ3
Ha: μ1 =
μ2 =
μ3H0:
μ1 = μ2 =
μ3
Ha: μ1 ≠
μ2 ≠ μ3
Find the value of the test statistic. (Round your answer to two decimal places.)
Find the p-value. (Round your answer to four decimal places.)
p-value =
State your conclusion.
Do not reject H0. There is sufficient evidence to conclude that the means of the three treatments are not equal.Do not reject H0. There is not sufficient evidence to conclude that the means of the three treatments are not equal. Reject H0. There is sufficient evidence to conclude that the means of the three treatments are not equal.Reject H0. There is not sufficient evidence to conclude that the means of the three treatments are not equal.
H0: μ1 =
μ2 = μ3
Ha: Not all the population means are equal.
Level of Significance : = 0.05
Anova table :
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value |
Treatments | 130 | K - 1 = 3-1 = 2 | 65 | 5.4737 | 0.015446 |
Error | 190 | - K = 16 | 11.875 | ||
Total | 320 | - 1 = 18 |
Here, Mean Square = Sum of Squares / Degrees of freedom
F = Mean Square of Treatment / Mean Square of
error
And p-value is obtained from the F tables.
Since p-value < 0.05, we reject H0 and conclude that there is sufficient evidence to conclude that the means of the three treatments are not equal.
Hope this answers your query!
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