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.)
Sample Size Mean GPA Standard Deviation
Students Who
Are Employed 172 3.22 0.475

Students Who
Are Not Employed 116 3.33 0.524

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.

Answer 1

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μemployed(μ1) be the sample mean of students who are employed and μnot employed(μ2) be the sample mean of students who are not employed

The random variable is μ1 - μ2 = difference in the mean of the employed and unemployed students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 3.22

μ2 = 3.33

s1 = 0.475

s2 = 0.524

n1 = 172

n2 = 116

t = (3.22 - 3.33)/√(0.475²/172 + 0.524²/116)

t = - 1.81

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.475²/172 + 0.524²/116]²/[(1/172 - 1)(0.475²/172)² + (1/116 - 1)(0.524²/116)²] = 0.00001353363/0.00000005878

df = 230

We would determine the probability value from the t test calculator. It becomes

p value = 0.036

Since alpha, 0.05 > than the p value, 0.036, then we would reject the null hypothesis.

Therefore, at a 5% significant level, 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