Question

In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were reported for a sample of students who worked and for a sample of students who did not work. The samples were selected at random from working and nonworking students at a university. (Use a statistical computer package to calculate the P-value. Use μemployed − μnot employed. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)

STUDENTS WHO ARE EMPLOYED

Sample Size 176

Mean GPA 3.22

Standard Deviation 0.485

STUDENTS WHO ARE NOT EMPLOYED

Sample Size 116

Mean GPA 3.33

Standard Deviation 0.514

t =

df =

P =

Does this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed? Use a significance level of 0.05. Yes or No

Answer #1

group 1 | group 2 | |||

sample mean x = | 3.220 | 3.330 | ||

std deviation s= | 0.485 | 0.514 | ||

sample size n= | 176 | 116 | ||

std error se=s/√n= | 0.037 | 0.048 | ||

degree
freedom=(se_{1}^{2}+se_{2}^{2})^{2}/(se_{1}^{2}/(n_{1}-1)+se_{2}^{2}/(n_{2}-1))= |
236 |

Point estimate =x1-x2= | -0.110 | ||

standard
error of difference
Se=√(S^{2}_{1}/n_{1}+S^{2}_{2}/n_{2})= |
0.0601 | ||

test
statistic t =(x_{1}-x_{2}-Δ_{o})/Se= |
-1.8298 | ||

p value : = | 0.0343 |
from excel: tdist(1.8298,236,1) |

from above:

**t =-1.83**

**df =236**

**p value =0.034**

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