In: Statistics and Probability
Consider the hypothesis statement shown below using
alphaαequals=0.050.05 and the data to the right from two independent samples.Upper H 0 : mu 1 minus mu 2 equals 0H0: μ1−μ2=0 Upper H 1 : mu 1 minus mu 2 not equals 0H1: μ1−μ2≠0 a) Calculate the appropriate test statistic and interpret the result. b) Calculate the p-value and interpret the result. |
x overbarx1 |
equals= |
231231 |
x overbarx2 |
equals= |
209209 |
sigmaσ1 |
equals= |
6565 |
sigmaσ2 |
equals= |
5353 |
n1 |
equals= |
4444 |
n2 |
equals= |
3636 |
Given information:
=231 , =209
=65 , =53
n1 =44 , n2 = 36
Hypothesis :
Two tailed test.
Here population standar deviations are known so we have to use two sample z test.
Test statistic -
z = 1.6676
P-value -
P- value for this two tailed test is,
P-value = 2*P( z >1.6676)
P( z >1.6676) = 1- P( z < 1.6676)
Using excel function,. =NORMSDIST( z)
P( z<1.6676) = NORMSDIST( 1.6676) =0.9523
So P(z > 1.6676)= 1 - 0.9523 = 0.0477
P- value = 2*0.0477 = 0.0954
P-value = 0.0954
Decision about null hypothesis-
Rule - Reject null hypothesis if p-value less than significance level
Here = 0.05
It is observed that p-value (0.0954) greater than significance level
So fail to reject null hypothesis.
Conclusion-
There is sufficient evidence to conclude that there is difference between population means.