GPA students mean

Question

You are testing the claim that the mean GPA of night students is less than the mean GPA of day students.

You sample 40 night students, and the sample mean GPA is 2.14 with a standard deviation of 0.89

You sample 45 day students, and the sample mean GPA is 2 with a standard deviation of 0.66

Calculate the test statistic, rounded to 2 decimal places

Answer 1

In this case we want to perform a two-sample t-test. Since we don't have the actual data, we cannot perform a test of equal variance, so we'll have to assume equal variance. The two-sample t-test is performed with the following formula:

(x₁ - x₂) / (s_p × √((1 / n₁) + (1 / n₂)))

where:

x₁ = mean of the first group

x₂ = mean of the second group

s₁ = standard deviation of the first group

s₂ = standard deviation of the second group

n₁ = size of the first group

n₂ = size of the second group

s_p = pooled standard deviation

the pooled variance is calculated with the following formula:

(((n₁ - 1) × (s₁²)) + ((n₂ - 1) × (s₂²))) / (n₁ + n₂ - 2)

Plugging in the following values:

x₁ = 2.14, s₁ = 0.89, n₁ = 40

x₂ = 2, s₂ = 0.66, n₂ = 45

We first calculate our pooled standard deviation:

s_p = (((40 - 1) × (0.89²)) + ((45 - 1) × (0.66²))) / (40 + 45 - 2) ≈ 0.7776

Then we calculate the two-sample t-statistic:

t = (2.14 - 2) / (0.7776 × √((1 / 40) + (1 / 45))) ≈ 0.8296

To go the extra step of checking whether this t-statistic is significant at the 95% confidence interval we calculate our degrees of freedom:

degrees of freedom = (n₁ + n₂ - 2) = (45 + 40 - 2) = 83

Then we use our degrees of freedom [83] with our t-statistic [0.8296] to look up our p-value in a t-table and find that our p-value is 0.40914. So we do not reject the null hypothesis that the two population means are equal.