Question

In order to determine whether or not a motorcycle education course improves the scores on a driving exam, a sample of 6 drivers were given the exam before and after taking the course. The results are shown below. Assume the population of differences is normally distributed.

Let d = Score After - Score Before.

Driver	Score Before the Course	Score After the Course
1	83	87
2	89	89
3	93	91
4	77	77
5	86	93
6	79	83

​a. Compute the test statistic.

b. Using α = .05 and the p-value approach, test to see if taking the course actually increased scores on the driving exam.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis: μ(d) = 0
Alternative Hypothesis: μ(d) > 0

a)

Test statistic,
t = (dbar - 0)/(s(d)/sqrt(n))
t = (2.17 - 0)/(3.3714/sqrt(6))
t = 1.577

b)

P-value Approach
P-value = 0.0878
As P-value >= 0.05, fail to reject null hypothesis.

Answer 2

To test whether taking the course actually increased scores on the driving exam, we need to perform a paired samples t-test. The null hypothesis (H0) states that there is no difference in scores before and after taking the course, while the alternative hypothesis (Ha) states that there is a significant difference (increase) in scores after taking the course.

Given data:

mathematicaCopy codeDriver   Score Before   Score After1        83             872        89             893        93             914        77             775        86             936        79             83

a. Compute the test statistic:

First, we need to calculate the differences for each driver (d = Score After - Score Before):

makefileCopy coded1 = 87 - 83 = 4
d2 = 89 - 89 = 0
d3 = 91 - 93 = -2
d4 = 77 - 77 = 0
d5 = 93 - 86 = 7
d6 = 83 - 79 = 4

Next, we compute the sample mean (Σd / n) and the sample standard deviation (s):

scssCopy codeSample mean (x̄) = (4 + 0 + (-2) + 0 + 7 + 4) / 6 = 13 / 6 ≈ 2.17Sample standard deviation (s) = √[( (4-2.17)^2 + (0-2.17)^2 + (-2-2.17)^2 + (0-2.17)^2 + (7-2.17)^2 + (4-2.17)^2 ) / 5] ≈ 3.59

The test statistic (t) is given by:

scssCopy codet = (x̄ - μ) / (s / √n)

Where μ is the hypothesized population mean difference (in this case, 0), s is the sample standard deviation, and n is the sample size.

scssCopy codet = (2.17 - 0) / (3.59 / √6) ≈ 2.39

b. Using α = 0.05 and the p-value approach, we will now find the critical value and compare it with the test statistic.

Degrees of freedom (df) = n - 1 = 6 - 1 = 5

Using a t-table or statistical software, we find the critical value at α = 0.05 with df = 5 to be approximately ±2.571.

The p-value is the probability of obtaining a test statistic as extreme as the one calculated, assuming the null hypothesis is true.

Since the absolute value of the test statistic (|2.39|) is less than the critical value (2.571), and the p-value is greater than the significance level (0.05), we fail to reject the null hypothesis.

Conclusion: Based on the data and the results of the test, there is not enough evidence to conclude that taking the course significantly increased scores on the driving exam.