Arandom sample of n1 =12 students majoring in accounting in a college of business has a mean grade-point average of 2.70 (where A = 4.0) with a sample standard deviation of 0.40. For the students majoring in computer information systems, a random sample of n2 = 10 students has a mean grade-point average of 2.90 with a standard deviation of 0.30. The grade-point values are assumed to be normally distributed. Test the null hypothesis that the mean-grade point average for the two categories of students is not different, using the 5 percent level of signiﬁcance.

In which [tex]\overline{Y_{1}}, \overline{Y_{2}}[/tex] are the sample means, [tex]N_{1}, N_{2}[/tex] are the sample sizes and [tex]s_{1}, s_{2}[/tex] are the standard deviations of the sample.

In this problem, we have that:

[tex]\overline{Y_{1}} = 2.7, s_{1} = 0.4, N_{1} = 12, \overline{Y_{2}} = 2.9, s_{2} = 0.3, N_{2} = 10[/tex]

So

[tex]T = \frac{2.7 - 2.9}{\sqrt{\frac{0.4}^{2}{12} + \frac{0.3}^{2}{10}}} = -1.3383[/tex]

What to do with the null hypothesis?

We will reject the null hypothesis, that is, that the means are equal, with a significante level of [tex]\alpha[/tex] if

[tex]|T| > t_{1-\frac{\alpha}{2},v}[/tex]

In which v is the number of degrees of freedom, given by

[tex]v = \frac{(\frac{s_{1}^{2}}{N_{1}} + \frac{s_{2}^{2}}{N_{2}})^{2}}{\frac{\frac{s_{1}^{2}}{N_{1}}}{N_{1}-1} + \frac{\frac{s_{2}^{2}}{N_{2}}}{N_{2} - 1}}[/tex]

Applying the formula in this problem, we have that:

[tex]v = 20[/tex]

So, applying t at the t-table at a level of 0.975, with 20 degrees of freedom, we find that

[tex]t = 2.086[/tex]

We have that

[tex]|T| = 1.3383[/tex]

Which is lesser than t.

So we accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.