Question

Suppose you have two groups, with Group 1's sample size being 76 and Group 2's sample size being 68. The average of the first group is 31.85 and a sample standard deviation of 10.44. The average of the second group is 29.16 and a sample standard deviation of 11.1. Can you reject the null that there is no difference between the two groups at the 99% significance level?

H0: μ_1,0 - μ_2,0 = 0

HA: μ_1,0 - μ_2,0 ≠ 0

What is the Standard Error for this situation? SE =  

What is the t-score for this situation? t =

Answer 1

Solution:

Let 1 and 1 be the mean and standard deviation of first group respectively.

Let 2 and 2 be the mean and standard deviation of second group respectively.

Let n1 and n2 be sample size of first and second group respectively.

Given:n1 =76, n2 =68,  1=31.85, 2 =29.16, 1 =10.44 , 2=11.1, Level of significance,=99%

Formula to find Standard error, S.E=( 12/n1 )+( 22/n2 )

=(10.442/76)+(11.12/68)

=(1.4341+1.8112 )

=1.8

Since n1 and n2 are greater than 30, we calculate z-score.

z-score, z=(1-2)/S.E=(31.85-29.16)/1.8=1.5

Critical region for both sided alternatives at 1% level of significance is |z|2.58.

Since Z-score value=1.5 is less than Z critical value=2.58, we accept null hypothesis and conclude that there is no difference between two means, i.e,μ1 - μ2 = 0