In: Statistics and Probability
Using the following output and α = 0.05, test that the mean for group 1 is greater than the mean for group 2.
group n mean sample std dev
1 10 4.5 0.9
2 10 3.5 0.9
n1 = 10
= 4.5
s1 = 0.9
n2 = 10
= 3.5
s2 = 0.9
Claim: The mean for group 1 is greater than the mean for group 2.
The null and alternative hypothesis is
For doing this test first we have to check the two groups have population variances are equal or not.
The null and alternative hypothesis is
Test statistic is
F = largest sample variance / Smallest sample variances
F = 0.9^2 / 0.9^2 = 0.81 / 0.81 = 1
Degrees of freedom => n1 - 1 , n2 - 1 => 10- 1 , 10 - 1 => 9 , 9
Critical value = 3.179 ( Using f table)
Critical value > test statistic so we fail to reject null hypothesis.
Conclusion: The population variances are equal.
So we have to use here pooled variance.
Test statistic is
Degrees of freedom = n1 + n2 - 2 = 10 + 10 - 2 = 18
Critical value = 1.734 ( Using t table)
| t | > critical value we reject null hypothesis.
Conclusion:
The mean for group 1 is greater than the mean for group 2.