Question

An adviser is testing out a new online learning module for a placement test. They wish to test the claim that on average the new online learning module increased placement scores at a significance level of α = 0.05. For the context of this problem, μD=μnew–μold where the first data set represents the new test scores and the second data set represents old test scores. Assume the population is normally distributed.H0: μD = 0H1: μD < 0You obtain the following paired sample of 19 students that took the placement test before and after the learning module: New Old LM LM 57.1 55.8 58.3 51.7 83.6 76.6 49.5 47.5 51.1 48.6 20.6 11.4 35.2 30.6 46.7 53 22.5 21 47.7 58.5 51.5 42.6 76.6 61.2 28.6 26.8 14.5 11.4 43.7 56.3 57 46.1 66.1 72.8 38.1 42.2 42.4 51.3a) Find the p-value. Round answer to 4 decimal places.b) Choose the correct decision and summary.Reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores.Do not reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores.Do not reject H0, there is enough evidence to support the claim that on average the new online learning module increased placement scores.Reject H0, there is enough evidence to support the claim that on average the new online learning module increased placement scores.

Answer 1

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-1.337 -0}{\frac{7.755}{\sqrt{19}}}=-0.751$

$p_v =P(t_{(18)}$

So the p value is higher than the $\apha=0.05$ used, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is higher or equal than 0.

And the best option for this case would be:

Do not reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores.

Step-by-step explanation:

Previous concepts

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=test value old , y = test value new

x: 55.8,51.7,76.6,47.5,48.6,11.4,30.6,53,21,58.5,42.6,61.2,26.8,11.4,56.3,46.1,72.8,42.2,51.3

y: 57.1,58.3,83.6,49.5,51.1,20.6,35.2,46.7,22.5,47.7,51.5,76.6,28.6,14.5,43.7,57,66.1,38.1,42.4

The system of hypothesis for this case are:

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

Alternative hypothesis: $\mu_x -\mu_y$

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

d: -1.3  -6.6  -7.0  -2.0  -2.5  -9.2  -4.6   6.3   -1.5  10.8  -8.9 -15.4  -1.8  -3.1  12.6 -10.9  6.7   4.1   8.9

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=-1.337$

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

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =7.755$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-1.337 -0}{\frac{7.755}{\sqrt{19}}}=-0.751$

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

$df=n-1=19-1=18$

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

$p_v =P(t_{(18)}$

So the p value is higher than the $\apha=0.05$ used, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is higher or equal than 0.

And the best option for this case would be:

Do not reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores.