Question

Five cigarette manufacturers claim that their product has low tar content. Independent random samples of cigarettes are taken from each manufacturer and the following tar levels (in milligrams) are recorded.

Brand Tar Level (mg)

A 4.2, 4.8, 4.6, 4.0, 4.4

B 4.9, 4.8, 4.7, 5.0, 4.9, 5.2

C 5.4, 5.3, 5.4, 5.2, 5.5

D 5.8, 5.6, 5.5, 5.4, 5.6, 5.8

E 5.9, 6.2, 6.2, 6.8, 6.4, 6.3

Can the differences among the sample means be attributed to chance?

Answer 1

We will use ANOVA for testing weather

differences among the sample means be attributed to chance

In the below table data has been shown of tar level of five manufacturer

A	B	C	D	E
4.2	4.9	5.4	5.8	5.9
4.8	4.8	5.3	5.6	6.2
4.6	4.7	5.4	5.5	6.2
4	5	5.2	5.4	6.8
4.4	4.9	5.5	5.6	6.4
	5.2		5.8	6.3

	A	B	C	D	E	Total (T')
	4.2	4.9	5.4	5.8	5.9
	4.8	4.8	5.3	5.6	6.2
	4.6	4.7	5.4	5.5	6.2
	4	5	5.2	5.4	6.8
	4.4	4.9	5.5	5.6	6.4
		5.2		5.8	6.3
Total	22	24.6	21.4	27.9	31.9	Total=127.8
Mean	4.4	4.92	5.35	5.58	6.38	Overall mean=5.326

We have to test whether there is significant difference among the sample means attributed to chance

Let us set up null and alternative hypothesis given as below

Ho:there is no evidence to claim that mean number of tar levels for five cigratee manufacturers is different

Ha:there is enough  evidence to claim that mean number of tar levels for five cigratee manufacturers is different

From the given data

n1=5,n2=6,n3=5,n4=6,n5=6

Total number of observations are

N=5+6+5+6+6=26

from the above data correction term for the mean will be

Let us find the total sum of squares

=814.08-583.31

=230.77

=588.18-583.31

=4.87

SSE=SST-SS(Tr)=230.77-4.87=225.9

Degree of freedom for treatment =k-1=5-1=4

Degree of freedom for error=N-K=28-5=23

Total=N-1=28-1=27

Sum of Squares of treatment=4.87

Sum of squares of error=225.9

Total sum of squares=230.77

Mean square=SS(Tr)/Df=4.87/4=1.21

Mean square error=225.9/23=9.82

Value of F will be =1.21/9.82=0.123

Using F table critical value of F for (4,23) degree of freedom will be 2.20 at level of significance 0.10

F stat<F critical

So we will fail to reject the null hypothesis

We have no evidence to conclude that mean number of tar level of five manufacturer are different

Hence mean number of tar levels are same