Question

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is µd = 0. Compute the value of the t test statistic. Round intermediate calculations to four decimal places as needed and final answers to three decimal places as needed.x: 34 39 28 33 27 23 35 33y: 32 35 34 33 28 28 35 32a. t = -0.523b. t = -1.480c. t = 0.690d. t = -0.185

Answer 1

Answer:

d: 2, 4, -6, 0, -1, -5, 0, 1

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{29}{8}=-0.625$

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

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.378$

The next step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.625 -0}{\frac{3.378}{\sqrt{8}}}=-0.523$

And the correct option would be:

a. t = -0.523

Step-by-step explanation:

We assume the following notation:

x=test value after , y = test value before

x: 34 39 28 33 27 23 35 33

y: 32 35 34 33 28 28 35 32

The system of hypothesis for this case is given by:

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

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

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

d: 2, 4, -6, 0, -1, -5, 0, 1

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{29}{8}=-0.625$

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

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.378$

The next step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.625 -0}{\frac{3.378}{\sqrt{8}}}=-0.523$

And the correct option would be:

a. t = -0.523