Question

A study randomly assigned students to take notes either longhand or using a laptop. The resulting scores of the students on a test of the material are given in the table below. Does the data provide evidence that it is more effective to take noteslong hand rather than on a laptop? Use the pooled procedure. The units here are points. (SHOW YOUR WORK)

Note-taking   n         x             s
Longhand    42     27.6          9.8
Laptop        40     18.3          9.0

(a) (1 point) Why is the pooled procedure more appropriate in this situation?
(b) (2 points) State your hypotheses in words and using proper mathematical symbols and notation.
(c) (4 points) Compute your test statistic and state the degrees of freedom. Round your test statistic to two
decimal places.
(d) (3 points) Find the p-value and give a p-value range.
(e) (2 points) State your conclusion in context using = 0:05.

Answer 1

We want to test "wether the data provide evidence that it is more effective to take noteslong hand rather than on a laptop?"

So null hypothesis(H₀ ) and alternative hypothesis (Ha ) are as follow:

Null hyypothesis :

$H_{0}:\mu _{1} -\mu_{2}=0$

Alternative hypothesis :

$H_{1}:\mu _{1} -\mu_{2}>0$

Where $\mu _{1}$ is the population mean of the variable Longhand

and $\mu _{2}$ is the population mean of the variable Laptop

First we need to say that why is the pooled procedure more appropriate in this situation?

This can be done by using F test

Let's use minitab:

Step 1) Click on Stat >>>Basic Statistics >>>2-Variances ...

Fill the necessary information and then click on Option again fill the necessary information.

Look the following image:

Then click on OK again Click on OK , so we get the following output:

p-value = 0.595

Since p-value > 0.05 we fail to reject the equality assumption of two populations.

So that  the pooled procedure more appropriate in this situation

Now lets test the equality of two means using pooled t test:

Level of significance = $\alpha$ = 0.05

therfore level of confidence = 100 - 5 = 95%

Let's used minitab:

Steps 1) Click on Stat>>>Basic Statistics>>>2-Sample t...

Steps 1) Click on summarized data and then fill the required information in the boxes : look the following picture.

step 3) Click on Option, Look the following image :

then click on OK again click on OK

So we get the following output

From the above output p-value = 0.000

Test statistic = T value = 4.47

Degrees of freedom = 42 + 40 - 2 = 80

P-range = (0 , 0.0005)

Because for one tailed test and 80 degrees of freedom t test statistic value 4.47 > 3.416 with corresponding area 0.0005 and less.

therefore range of P-value is ( 0 , 0.0005)

e) Decision rule: 1) If p-value <= level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.000< 0.05 so we used first rule.

That is we reject null hypothesis

Conclusion: At 5% level of significance the data provide evidence that it is more effective to take noteslong hand rather than on a laptop