Answer:
P-value > 0.100
Step-by-step explanation:
Degree of freedom Group = df1 = I - 1 = 3 - 1 = 2
Degree of freedom Error = df2 = I * (J - 1) = 3 * (9 - 1) = 24
MSTR = SSTR / df1 = 594.8 / 2 = 297.4
MSE = SSE / df2 = 4777.8 / 24 = 199.075
F = MSTR / MSE = 297.4 / 199.075 = 1.49
Using R command,
> 1 - pf(1.49, 2, 24)
[1] 0.2454887
P-value = 0.2455
check the attatchment for better understanding
The hypotheses of interest is stated as; Null Hypothesis; H0: μ₁ = μ₂ = μ₃
Alternative Hypothesis; Ha: at least 2 μi's are unequal
1) Since we have 3 differenet brands of lightbulbs having same wattage, the we can state the hypotheses as;
μi = true average lumen output for brand i bulbs
Null Hypothesis; H0: μ₁ = μ₂ = μ₃
Alternative Hypothesis; Ha: at least 2 μi's are unequal
B) We are given that there are 3 treatments which are the different brands of lightbulbs having the same wattage. Thus;
Degree of freedom for the treatments; df₁ = 3 - 1 = 2
Now, we are told that 9 bulbs of each brand were tested. Thus;
Degree of freedom for the possible Error; df₂ = 3 * (9 - 1) = 24
Mean of squares for treatments;
MS_Tr = SSTr/df₁ = 594.8/2 = 297.4
Mean of squares for errors;
MS_E = SSE/df₂ = 4777.8/24 = 199.075
Ratio of the 2 mean squares is;
F = MS_Tr/MS_E
F = 297.4 / 199.075
F = 1.49
3) From online anova table, using the values above, we have;
p-value > 0.1
4) We fail to reject the null hypothesis and conclude that there is staticstically no differences in the lumen output
Read more about hypotheses at; https://brainly.com/question/15980493
