Question

Answer in Excel way.
1.        

          In order to determine whether or not a driver's education course improves the scores on a driving exam, a sample of 6 students were given the exam before and after taking the course. The results are shown below.

Let d = Score After - Score Before.

Student	Score Before the Course	Score After the Course
1	83	87
2	89	88
3	93	91
4	77	77
5	86	93
6	79	83

a.            State the null and alternative hypothesis.

b.	Compute the test statistic.
c.	At alpha of 0.05 using the p-value approach, test to see if taking the course actually increased scores on the driving exam.

           

d.	Construct a 95 % confidence interval for the mean difference score for after and before the driver education course.

Answer 1

a)

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

b)

Test statistic,
t = (dbar - 0)/(s(d)/sqrt(n))
t = (2 - 0)/(3.5214/sqrt(6))
t = 1.391

c)

P-value Approach
P-value = 0.1115
As P-value >= 0.05, fail to reject null hypothesis.

d)

sample mean, xbar = 2
sample standard deviation, s = 3.5214
sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.571

ME = tc * s/sqrt(n)
ME = 2.571 * 3.5214/sqrt(6)
ME = 3.696

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (2 - 2.571 * 3.5214/sqrt(6) , 2 + 2.571 * 3.5214/sqrt(6))
CI = (-1.7 , 5.7)