Answer:
We accept the null hypothesis that the mean gpa's are equal, with the 5 percent level of significance.
Step-by-step explanation:
We have these following hypothesis:
Null
Equal means
So
![\mu_{1} = \mu_{2}](https://tex.z-dn.net/?f=%5Cmu_%7B1%7D%20%3D%20%5Cmu_%7B2%7D)
Alternative
Different means
So
![\mu_{1} \neq \mu_{2}](https://tex.z-dn.net/?f=%5Cmu_%7B1%7D%20%5Cneq%20%5Cmu_%7B2%7D)
Our test statistic is:
![\frac{\overline{Y_{1}} - \overline{Y_{2}}}{\sqrt{\frac{s_{1}^{2}}{N_{1}} + \frac{s_{2}^{2}}{N_{2}}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Coverline%7BY_%7B1%7D%7D%20-%20%5Coverline%7BY_%7B2%7D%7D%7D%7B%5Csqrt%7B%5Cfrac%7Bs_%7B1%7D%5E%7B2%7D%7D%7BN_%7B1%7D%7D%20%2B%20%5Cfrac%7Bs_%7B2%7D%5E%7B2%7D%7D%7BN_%7B2%7D%7D%7D%7D)
In which
are the sample means,
are the sample sizes and
are the standard deviations of the sample.
In this problem, we have that:
![\overline{Y_{1}} = 2.7, s_{1} = 0.4, N_{1} = 12, \overline{Y_{2}} = 2.9, s_{2} = 0.3, N_{2} = 10](https://tex.z-dn.net/?f=%5Coverline%7BY_%7B1%7D%7D%20%3D%202.7%2C%20s_%7B1%7D%20%3D%200.4%2C%20N_%7B1%7D%20%3D%2012%2C%20%5Coverline%7BY_%7B2%7D%7D%20%3D%202.9%2C%20s_%7B2%7D%20%3D%200.3%2C%20N_%7B2%7D%20%3D%2010)
So
![T = \frac{2.7 - 2.9}{\sqrt{\frac{0.4}^{2}{12} + \frac{0.3}^{2}{10}}} = -1.3383](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2.7%20-%202.9%7D%7B%5Csqrt%7B%5Cfrac%7B0.4%7D%5E%7B2%7D%7B12%7D%20%2B%20%5Cfrac%7B0.3%7D%5E%7B2%7D%7B10%7D%7D%7D%20%3D%20-1.3383)
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
if
![|T| > t_{1-\frac{\alpha}{2},v}](https://tex.z-dn.net/?f=%7CT%7C%20%3E%20t_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%2Cv%7D)
In which v is the number of degrees of freedom, given by
![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}}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B%28%5Cfrac%7Bs_%7B1%7D%5E%7B2%7D%7D%7BN_%7B1%7D%7D%20%2B%20%5Cfrac%7Bs_%7B2%7D%5E%7B2%7D%7D%7BN_%7B2%7D%7D%29%5E%7B2%7D%7D%7B%5Cfrac%7B%5Cfrac%7Bs_%7B1%7D%5E%7B2%7D%7D%7BN_%7B1%7D%7D%7D%7BN_%7B1%7D-1%7D%20%2B%20%5Cfrac%7B%5Cfrac%7Bs_%7B2%7D%5E%7B2%7D%7D%7BN_%7B2%7D%7D%7D%7BN_%7B2%7D%20-%201%7D%7D)
Applying the formula in this problem, we have that:
![v = 20](https://tex.z-dn.net/?f=v%20%3D%2020)
So, applying t at the t-table at a level of 0.975, with 20 degrees of freedom, we find that
![t = 2.086](https://tex.z-dn.net/?f=t%20%3D%202.086)
We have that
![|T| = 1.3383](https://tex.z-dn.net/?f=%7CT%7C%20%3D%201.3383)
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.