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
and would end up having a word problem that says somthing like "Jimmy has 60 apples and eats 12 a day how many days go by before Jonny has no more apples ?"