Suppose as part of your job you are responsible for installing emergency lightin

Question

Suppose as part of your job you are responsible for installing emergency lighting in a series of state office buildings. Bids have been received from four manufacturers of battery-operated emergency lights. The costs are about equal, so the decision will be based on the length of time the lights last before failing. A sample of four lights from each manufacturer has been tested with the following values (time in hours) recorded for each manufacturer. Assume that the data are approximately normal.

2 3650 1 8150 e-4250 p1211 T-11 2626 e6412 p2343 T1-11 0088 e1441 p0120 T1-11

Explanation / Answer

Step 1

null, Ho: µ1 =µ2 =µ3 =µ4

alternative: H1 : atleast one of th emean is diffrent

Step 2

Degrees of freedom between = k - 1 = 4 - 1 = 3

Degrees of freedom Within = n - k = 16 - 4 = 12

Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.49

Step 3

Grand Mean = G / N = 1104+1336.5+1156 +1012.75 / 4 = 1152.313

SST = ( Xi - GrandMean)^2 = (1010-1152.313)^2 + (1140-1152.313)^2 + (1248-1152.313)^2 + ……..& so on = 202231.364

SS Within = (Xi - Mean of Xi ) ^2 =,(1010-1104)^2 + (1140-1104)^2 + (1248-1104)^2 + ……..& so on = 57141

SS Between = SST - SS Within = 202231.364 - 57141 = 145090.364

Step 4

Mean Square Between = SS Between / df Between = 145090.364/3 = 48363.455

Mean Square Within = SS Within / df Within = 57141/12 = 4761.75

Step 5

F Cal = MS Between / Ms Within = 48363.455/4761.75 = 10.157

We got |F cal| = 10.157 & |F Crit| =3.49

MAKE DECISION

Hence Value of |F cal| > |F Crit|and Here We Reject Ho

|F cal| > |F Crit|and Here We Reject Ho

| F cal | > 3.49 we reject Ho other wise do not reject Ho

Treatments ONE WAY ANOVA Mean = X /n TYPE A 1010 1140 1248 1018 1104 TYPE B 1262 1346 1412 1326 1336.5 TYPE C 1148 1221 1155 1100 1156 TYPE D 913 986 1050 1102 1012.75

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