Question

Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the α = 0.05 level of significance? Assume normality. (Use Diet B - Diet A.)

Diet A	9	8	7	14	10	7	8	11	5	14
Diet B	8	21	20	19	17	8	10	19	11	12

(a) Find t. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

Answer 1

Solution:

Here, we have to use paired t test.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: The mean weight gained on diet B was same as the mean weight gained on diet A.

Alternative hypothesis: Ha: The mean weight gained on diet B was greater than the mean weight gained on diet A.

We use Diet B - Diet A.

H0: µd = 0 versus Ha: µd > 0

This is a right tailed test.

(a) Find t.

Test statistic for paired t test is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

From given data, we have

Dbar = 5.2

Sd = 5.2873

n = 10

df = n – 1 = 9

α = 0.05

t = (Dbar - µd)/[Sd/sqrt(n)]

t = (5.2 – 0)/[ 5.2873/sqrt(10)]

t = 3.11

(ii) Find the p-value.

The p-value by using t-table is given as below:

P-value = 0.0063

P-value < α

So, we reject the null hypothesis

There is sufficient evidence to conclude that the mean weight gained on diet B was greater than the mean weight gained on diet A.