Question

Use the traditional method of hypothesis testing to test the given claim about the means of two populations. Assume that two dependent samples have been randomly selected from normally distributed populations. 37) Five students took a math test before and after tutoring. Their scores were as follows.
Subject A B C D E
Before 71 66 67 77 75
After 75 75 65 80 87

Using a 0.01 level of significance, test the claim that the tutoring has an effect on the math scores

Answer 1

Answer:

The appropriate hypothesis for the test model are as follows:

Null hypothesis:

$H_o = The \ tutoring \ has \ no \ effect \ on \ the \ math \ scores$

Alternative Hypothesis :

$H_a : The \ tutoring \ has \ an \ effect \ on \ the \ math \ scores$

The mean of the difference = -5.2

The standard deviation of the difference = 5.45

The value of the test statistic t is = -2.13

Step-by-step explanation:

The appropriate hypothesis for the test model are as follows:

Null hypothesis:

$H_o = The \ tutoring \ has \ no \ effect \ on \ the \ math \ scores$

Alternative Hypothesis :

$H_a : The \ tutoring \ has \ an \ effect \ on \ the \ math \ scores$

We use  MINITAB to obtain the p-value as follows:

Step 1 : Choose Stat>Basic Statistics> Paired t

Step 2 : Choose Samples in columns.

Step 3 : In First sample , enter column as Before

Step 4 : In Second sample, enter the column as After

MINITAB output:

Paired T-Test and CI : Before , After

Paired T for Before  -   After

                     N          Mean                StDev                   SE   Mean

Before           5          71.2000            4.8166                   2.1541

After               5          76.4000           8.0498                 3.6000

Difference      5        -5.20000           5.44977               2.43721

99% CI for mean difference :  (-16.42115, 6.02115)

T-Test of mean difference = 0 (vs  not = 0)

T-value = -2.13

P-Value = 0.100

However, from the MINITAB output ; it is obvious that the mean and the standard deviation of the difference are -5.2 and 5.45.

Therefore, the mean of the difference = -5.2

The standard deviation of the difference = 5.45

Also, we can see that the T-value = -2.13 ;

Thus, the value of the test statistic t is = -2.13