# Using the following output and

Using the following output and α = 0.05, testthat the mean for group 1 is greater than the mean for group2.

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 group2.

The null and alternative hypothesis is

For doing this test first we have to check the two groups havepopulation 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 nullhypothesis.

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.

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