Question

Service Type	Time
1	65
3	147
4	190
4	200
2	90
2	97
2	95
1	61
3	125
1	49
4	143
1	47
1	38
2	78
4	179
1	33
1	30
3	93
3	95
4	111
2	80
1	30
1	24
1	22
2	66
2	90
4	240
2	99
4	270
3	120
3	150
3	166
3	130
4	156
3	136
4	178
4	146
4	245
4	243
3	124
2	92
2	88
2	75
1	11

Using a level of significance of α=0.05, test whether the means of the four types of services are equal. Assume the samples for the four types were independently drawn and that the four populations are normally distributed.

Answer 1

Here we need to use one way ANOVA. Following table shows the data according to service types:

Service type 1	Service type 2	Service type 3	Service type 4
65	90	147	190
61	97	125	200
49	95	93	143
47	78	95	179
38	80	120	111
33	66	150	240
30	90	166	270
30	99	130	156
24	92	136	178
22	88	124	146
11	75		245
			243

Hypotheses are:

H0: The means of the four types of services are equal.

Ha: At least one population mean out of four types of services is different.

Goto excel- Data analysis ->select single factor ANOVA -> select data and click ok

Following is the output generated by excel:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Service type 1	11	410	37.27272727	278.8182
Service type 2	11	950	86.36363636	106.2545
Service type 3	10	1286	128.6	528.4889
Service type 4	12	2301	191.75	2418.568
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	147723.4	3	49241.13939	55.93776	2.26E-14	2.838745
Within Groups	35211.38	40	880.2844318
Total	182934.8	43

The F test statistics is

F = 55.94

The p-value is: 0.0000

Since p-value is less than level of significance of α=0.05 so we reject the null hypothesis. That is we cannot conclude that the means of the four types of services are equal.