Answer:
Step-by-step explanation:
We would set up the hypothesis test.
For the null hypothesis,
µ ≥ 3.5
For the alternative hypothesis,
µ < 3.5
It is a left tailed test.
Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = population standard deviation
n = number of samples
From the information given,
µ = 3.5
x = 3.6
σ = 0.4
n = 36
z = (3.6 - 3.5)/(0.4/√36) = 1.5
Looking at the normal distribution table, the probability corresponding to the z score is 0.933
The p value gotten is to the right of the normal curve. Since it is a left tailed test, we need the p value to the left of the curve. Therefore,
p = 1 - 0.933 = 0.067
Since alpha, 0.05 < than the p value, 0.067, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, there is no significant evidence that mean GPA of these graduates does not exceed 3.50
