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 CL , y = test value CO

x: 57.8 35.7 54.5 56.8 51.1 70.8 77.3 51.6 54.7 63.6 59.2 59.2 55.8 38.5

y: 44.7 52.1 60.2 52.7 47.2 65.6 71.4 48.8 53.1 66.3 59.8 47.5 64.5 34.4

The system of hypothesis for this case are:

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

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

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

d: 13.1, -16.4,-5.7,4.1,3.9,5.2,5.9,2.8,1.6,-2.7,-0.6,11.7,-8.7,4.1

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{29}{8}=1.307$

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

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =7.798$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1.307 -0}{\frac{7.798}{\sqrt{14}}}=0.6271$

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

$df=n-1=14-1=13$

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

$p_v =P(t_{(13)}>0.6271) =0.2707$

So the p value is higher than the significance level provided of 0.01, 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. So we can't conclude that the mean neck rotation is greater in the clockwise direction than in the counterclockwise direction at 1% of significance.

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3 years ago