Question

The lumen output was determined for each of

I = 3

different brands of lightbulbs having the same wattage, with J = 7 bulbs of each brand tested. The sums of squares were computed as SSE = 4772.5 and SSTr = 595.3.

Use the F test of ANOVA

(α = 0.05)

to decide whether there are any differences in true average lumen outputs among the three brands for this type of bulb by obtaining as much information as possible about the P-value. (Round your answer to two decimal places.)

Answer 1

Solution:

Number of treatments = I = 3

dftreatments = I - 1 = 3 -1 = 2

Number of observations of each treatment = j = 7

Thus total observations = N = I * J = 3 * 7 = 21

dftotal = N - 1 = 21 - 1 = 20

thus

dferror = dftotal - dftreatments

dferror = 20 - 2

dferror = 18

SSE = 4772.5 and SSTr = 595.3.

Thus

SST = SSTr+ SSE

SST =  595.3 + 4772.5

SST = 5367.8

Thus

MSTr = SSTr / dftreatments

MSTr = 595.3 / 2

MSTr = 297.65

MSE = SSE / dferror

MSE = 4772.5 / 18

MSE = 265.14

Thus

F = MSTr / MSE

F = 297.65 / 265.14

F = 1.12

P-value =........?

Use following Excel command:

=F.DIST.RT( F , dftreatments , dferror )

=F.DIST.RT(1.12,2,18)

=0.3480

Thus P-value = 0.3480

Source	df	SS	MS	F	P-value
Treatment	2	595.30	297.65	1.12	0.3480
Error	18	4772.50	265.14
Total	20	5367.80

Since P-value = 0.3480 > α = 0.05, we fail to reject null hypothesis that: there is no differences in true average lumen outputs among the three brands for this type of bulb.

Thus we conclude that: true average lumen outputs among the three brands for this type of bulb is equal.