Question

Use =0.05 to test if these 4 Bus lines has a same average (mean) # of passengers on it at different random times of day given the sample data from one day.

6-Use =0.05 to test if these 4 Bus lines has a same average (mean) # of passengers on it at different random times of day given the sample data from one day. passengers Line U 25 22 Line M 15 12 11 22 2 Line N 35 38 41 15 Line L 9 5 2 35 10 29 11 35 45

Answer 1

The best way to test the equality of more than 2 group means is ANOVA(Analysis of Variance) test.

Step 1: Stating the null and alternative hypothesis

$H_{0} : \mu _{U} = \mu _{M} = \mu_{N} = \mu_{L}$

, The means are not equal

H1 Not Ho

Step 2 : Level of significance

g= 0,05

Step 3 : Test Statistic

F statistic for ANOVA

where , F = MSB/MSE

Step 4 : Computing the test statistic

First , we need to compute the group means

	Line U	Line M	Line N	Line L
n	5	5	5	5
Group sum	122	62	174	61
Group mean	24.4	12.4	34.8	12.2

The overall mean = (122+62+174+61)/20 = 419/20 = 20.95

Now we need to compute SSB and SSE

where , SSB = Sum of Squares between the groups , and

SSE = Sum of Squares within the groups

$SSB = \sum_{j=1}^{4}n_{j}\left ( \overline{x_{j}}- \overline{x} \right )$

= 5*(24.4 - 20.95)² + 5*(12.4 - 20.95)² + 5*(34.8 - 20.95)² + 5*(12.2 - 20.95)²

= 59.5125 + 365.5125 + 959.1125 + 382.8125

= 1766.95

Now , computing SSE for each group

SSE for group Line U ,where  $\overline{x_{1}} = 24.4$

Line U	$(x-\overline{x_{1}})$	$(x-\overline{x_{1}})^{2}$
25	0.6	0.36
22	-2.4	5.76
29	4.6	21.16
11	-13.4	179.56
35	10.6	112.36
Total		319.2

Similarly , for all other groups

SSE for group Line M , where  $\overline{x_{2}} = 12.4$

Line M	$(x-\overline{x_{2}})$	$(x-\overline{x_{2}})^{2}$
15	2.6	6.76
12	-0.4	0.16
11	-1.4	1.96
22	9.6	92.16
2	-10.4	108.16
Total		209.2

SSE for group Line N , $\overline{x_{3}} = 34.8$

Line N	$(x-\overline{x_{3}})$	$(x-\overline{x_{3}})^{2}$
35	0.2	0.04
38	3.2	10.24
41	6.2	38.44
15	-19.8	392.04
45	10.2	104.04
Total		544.8

SSE for group L , where $\overline{x_{4}} = 12.2$

Line L	$(x-\overline{x_{4}})$	$(x-\overline{x_{4}})^{2}$
9	-3.2	10.24
5	-7.2	51.84
2	-10.2	104.04
35	22.8	519.84
10	-2.2	4.84
Total		690.8

SSE = Sum of all the SSE for each groups = 319.2 + 209.2 + 544.8 + 690.8 = 1764

constructing the ANOVA table

	SS	DF	MS	F
SSB	1766.95	4-1=3	1766.95/3 = 588.983	588.983/110.25 = 5.342
SSE	1764	20-4=16	1764/16 = 110.25
Total	3530.95	20-1 = 19 1

Step 5 : Decision

Critical value of F at (16,3) for $\alpha$ = 0.05  = 3.24

Calculated value of F = 5.342

Since Calculated value (5.342) > Critical Value (3.24)

We reject Null hypothesis(H₀)

i.e we don't have significant evidence to show that 4 bus lines have same mean of number of passengers on it at different random times of day given the sample data from one day .

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