Question

In: Statistics and Probability

Suppose you have two groups, with Group 1's sample size being 76 and Group 2's sample...

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 =

Expert Solution

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

orchestra answered 1 year ago

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