Question

You are testing the claim that the mean GPA of night students is greater than the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.36 with a standard deviation of 0.96 You sample 60 day students, and the sample mean GPA is 2.19 with a standard deviation of 0.66 Calculate the test statistic, rounded to 2 decimal places

Answer 1

Z = 0.87

Step-by-step explanation:

Given the following data;

Sample 1:

n1 = 30

Mean, X = 2.36

Standard deviation, Ox = 0.96

Sample 2:

n2 = 60

Mean, Y = 2.19

Standard deviation, Oy = 0.66

The formula for test statistics for two population is;

`Z = \frac{X-Y}{\sqrt{(\frac{Ox^2} {n_1} } +(Oy^2)/(n_2) )}}`

Substituting the values, we have;

`Z = \frac{2.36-2.19}{\sqrt{(\frac{0.96^2} {30} +(0.66^2)/(60) )}}`

`Z = \frac{0.17}{\sqrt{(\frac{0.9216} {30} +(0.4356)/(60) )}}`

`Z = (0.17)/(√((0.03072 +0.00726)))`

`Z = (0.17)/(√(0.03798))`

`Z = (0.17)/(0.19488)`

Z = 0.8723

The test statistics to 2 d.p is 0.87

Therefore, Z = 0.87