Question

In order to determine whether or not a driver's 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	88
3	93	91
4	77	77
5	86	93
6	79	83

Complete the following table (round numbers to two decimal places):

	After	Before
Mean
Variance
t-Stat (test statistic)
p-value (one tail)
t critical value (one tail)

Using α = 0.05, test to see if taking the course actually increased scores on the driving exam. What do you conclude?

Answer 1

Solution-:

Let, x=Score driver before the course and y= score driver after thecourse

d=score after-score before

Here the situation is suitable for paired t- test

$\mu_{d}:$ Average difference in score

Hypothesis: We want to test, $HOM=0$ Vs   (Two tailed test)

H:Hd 70

Given data:

x	y
83	87
89	88
93	91
77	77
86	93
79	83

By using MS-Excel :

t-Test: Paired Two Sample for Means
	x	y
Mean	84.5000	86.5000
Variance	36.7000	33.5000
Observations	6.0000	6.0000
Pearson Correlation	0.8242
Hypothesized Mean Difference	0.0000
df	5.0000
t Stat	-1.3912
P(T<=t) one-tail	0.1114
t Critical one-tail	2.0150
P(T<=t) two-tail	0.2229
t Critical two-tail	2.5706

Here, t-stat$=13912$, P-value (One tailed)
=0.1114
, Critical value (One tailed)

Decision : Here, t-stat$=13912$ < Critical value (One tailed) hence, we accept $H_{0}$ .

Conclusion: We conclude that taking the course actually same (equal) scores on the driving exam.

OR

By using R-software:

> x=c( 83,89,93,77,86,79);x

[1] 83 89 93 77 86 79
> y=c(87,88,91,77,93,83);y
[1] 87 88 91 77 93 83
> t.test(y,x, paired=T)

Paired t-test

data: y and x
t = 1.3912, df = 5, p-value = 0.2229
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.695444 5.695444
sample estimates:
mean of the differences
2

Decision : Here, P-value =0.2229> l.o.s. =0.05 hence we accept $H_{0}$ .

Conclusion: We conclude that taking the course actually same (equal) scores on the driving exam.