Question

The length of odontoblasts (cells responsible for tooth growth) is recorded for 10 guinea pigs treated with orange juice and 10 guinea pigs treated with ascorbic acid. The results are shown below:

Orange Juice : 15.2, 21.5, 17.6, 9.7, 14.5, 10.0, 8.2, 9.4, 16.5, 9.7    ( n o = 10 , x ¯ o = 13.23 , s o = 4.46 )

Ascorbic Acid : 4.2, 11.5, 7.3, 5.8, 6.4, 10.0, 11.2, 11.2, 5.2, 7.0   

a. Assume the samples are approximately normal (do not waste time checking this assumption).

Find the test statistic for this variance test.

b. Find the p-value for the variance test.

c. When conducted at a significance level of 0.05, what is the conclusion of this variance test and the reason it was made?

d. Write the inference for the variance test.

Answer 1

a) The null and alternative hypothesis

where represent the population variances for length of odontoblast for treatment with orange juice and ascorbic acid is given below.

Test statistic

where

Thus,

F= 7.544 / 19.889 = 2.63

b) degrees of freedom = (10-1, 10-1) = (9,9)

P value = 0.0830

Note : P value using excel "=F.DIST.RT(2.63,9,9)"

c) Since P value > 0.05

We fail to reject H0

d) At 0.05 level of significance ,there is not sufficient evidence to conclude that variances of lengths of odontoblasts for two treatments are not equal.

Calcualtion

	x1	(x1-)2		x2	(x2-)2
	15.2	3.8809		4.2	14.2884
	21.5	68.3929		11.5	12.3904
	17.6	19.0969		7.3	0.4624
	9.7	12.4609		5.8	4.7524
	14.5	1.6129		6.4	2.4964
	10	10.4329		10	4.0804
	8.2	25.3009		11.2	10.3684
	9.4	14.6689		11.2	10.3684
	16.5	10.6929		5.2	7.7284
	9.7	12.4609		7	0.9604
sum/ss	132.3	179.001	sum/ss	79.8	67.896
=sum/10	13.23		=sum/10	7.98
=ss/9		19.889	=ss/9		7.544