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 who did not work.( university of central florida undergraduate research journal, spring 2005).
sample s. gpa standard d
students employed, 184, 3.12, .485
not employed 114, 3.23, .524
the samples are selected at random from working and nonworking students at the University of Central Florida. 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 ( significance level =.05)?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z score.
To determine if the information supports the hypothesis that students who are not employed have a higher mean GPA than those who are employed, we can perform a hypothesis test.
Step 1: Define the null and alternative hypotheses:
Null Hypothesis (H0): The mean GPA of employed students is equal to or higher than the mean GPA of non-employed students.
Alternative Hypothesis (H1): The mean GPA of employed students is lower than the mean GPA of non-employed students.
Step 2: Set the significance level (α):
The significance level, denoted by α, is the probability of rejecting the null hypothesis when it is true. In this case, the significance level is given as α = 0.05.
Step 3: Calculate the test statistic:
We can use a t-test to compare the means of two independent samples. The test statistic is calculated as:
t = (mean1 - mean2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where mean1 and mean2 are the means of the two samples, s1 and s2 are the corresponding standard deviations, and n1 and n2 are the sample sizes.
In this case, for the employed students:
Mean1 = 3.12
Standard Deviation1 = 0.485
Sample Size1 = 184
For the non-employed students:
Mean2 = 3.23
Standard Deviation2 = 0.524
Sample Size2 = 114
Substituting these values into the formula, we get:
t = (3.12 - 3.23) / sqrt((0.485^2 / 184) + (0.524^2 / 114))
Step 4: Determine the critical value:
Since the alternative hypothesis is that the mean GPA of employed students is lower than the mean GPA of non-employed students, we will use a one-tailed test. With a significance level of α = 0.05, we need to find the critical value from the t-distribution table using the degrees of freedom (df) calculated as (n1 - 1) + (n2 - 1).
df = (184 - 1) + (114 - 1)
Look up the critical value for the given degrees of freedom and significance level in the t-distribution table.
Step 5: Compare the test statistic with the critical value:
If the absolute value of the calculated test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is evidence to support the alternative hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Make a conclusion:
Based on the comparison between the test statistic and the critical value, we can make a conclusion about the hypothesis. If the calculated test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is evidence to support the alternative hypothesis. If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis.
By following these steps and calculating the test statistic, you should be able to determine if the information provided supports the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed.