Question

1. A tutoring company has offered students a program of 10 weekly sessions to help them improve their test score on a particular final exam. The average score of all the students who have taken this exam so far is 4.6. To test whether this tutoring company works, 5 willing students were sent to complete the 10 week program and take the test afterward. The data are presented below. Compute the appropriate t-test. Sample Scores X-M (X-M)2 12 SS = X = s? n = M =

Answer 1

Solution:

The formulas for mean and standard deviation are given as below:

Mean = M = ∑X/n

Variance = S^2 = ∑(X - M)^2 / (n – 1)

Standard deviation = S = Sqrt[∑(X - M)^2 / (n – 1)]

SS = ∑(X - M)^2

The calculation table is given as below:

Sample scores	X - M	(X - M)^2
6	-1.8	3.24
6	-1.8	3.24
12	4.2	17.64
7	-0.8	0.64
8	0.2	0.04
∑X = 39		SS = 24.8
n = 5		S^2 = 6.2
M = 7.8		S = 2.48997992

Now, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: The average exam score is 4.6.

Alternative hypothesis: Ha: The average exam score is more than 4.6.

H0: µ = 4.6 versus Ha: µ > 4.6

This is an upper tailed or right tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 4.6

Xbar = 7.8

S = 2.48997992

n = 5

df = n – 1 = 4

We assume α = 0.05

Critical value = 2.1318

(by using t-table or excel)

t = (7.8 – 4.6)/[ 2.48997992/sqrt(5)]

t = 2.8737

P-value = 0.0227

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis.

There is sufficient evidence to conclude that the tutoring company works.