Question

A group of n - 9 students was selected for a comparative study that involved their Exam 1 scores [variable - X] and their overall course grades [variable Y] Educators intend to draw inferences about the differences D = Y-X) assuming that differences are normally distributed with unknown parameters. The sample summaries are presented below. Summaries Values Average X X-bar]-74.98 | AverageYSample SD [D] [Y-barl-69.70 S = 9.0 Hypothesis Testing At the significance level, α-500, do you have enough evidence for the population that the average course grade would be lower than the average first exam score Show the following items: 1. Critical value - 2. Test Statistic Value 3. Reiection Rule states: Decision: Reject the null hypothesis - or Do Not Reject 4.

Answer 1

Solution:

Here, we have to use paired t test.

H₀: µ_d = 0 versus H_a: µ_d < 0

We assume D = Y – X for the given scenario.

We are given α = 0.05

Dbar = Ybar – Xbar = 69.70 – 74.98 = -5.28

S_d = 9

n = 9

df = n – 1 = 8

α = 0.05

Q.(1)

Critical value = -1.8595

(by using t-table)

Q.(2)

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

t = (-5.28 – 0)/[9/sqrt(9)]

t = -5.28/3

t = -1.76

Test statistic value = -1.76

P-value = 0.0582

Q.(3)

Rejection rule:

Reject H₀ if test statistic t value < critical value -1.8595

We have test statistic > critical value, so we do not reject the null hypothesis

P-value > α = 0.05

Q.(4)

So, we do not reject the null hypothesis