Question

Do SAT coaching classes work? Do they help students to improve their test scores? Four students were selected randomly from all of the students that completed an SAT coaching class. For each student, we recorded their first SAT score (before the coaching class) and their second SAT score (after the coaching class).

Student

1

2

3

4

First SAT score

920

830

960

910

Second SAT score

1010

800

1000

980


To analyze these data, we should use

A.

the one-sample t test., one answer Yahoo,

B.

the matched pairs t test.

C.

the two-sample t test.

D.) Any of the above are valid. It just needs to be a t since ? is unknown.

Answer 1

Answer:

B.  the matched pairs t test.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

Let put some notation  

x=first test value , y = second test value

x: 920, 830, 960, 910

y: 1010, 800, 1000, 980

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \leq 0$

Alternative hypothesis: $\mu_y -\mu_x >0$

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: 90, -30, 40, 70

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{170}{4}=42.5$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =52.520$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{42.5 -0}{\frac{52.520}{\sqrt{4}}}=1.618$

The next step is calculate the degrees of freedom given by:

$df=n-1=4-1=3$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(3)}>1.618) =0.1024$

So the p value is higher than any significance level assumed (0.05), so then we can conclude that we reject the null hypothesis. So we can conclude that the difference between the after and the initial score it's not significantly higher at 5% of significance.