Respuesta :
Answer:
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{33.17 -15}{\frac{37.420}{\sqrt{10}}}=1.536[/tex]
[tex]p_v P(t_{(9)}>1.536)= 0.079[/tex]
So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis and we can't conclude that the difference between the means after and before is higher than 15 at 1% of significance
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 before , y = test value after
x: 439 451 435 431 451 454 453 463 491 455
y: 511 526 529 473 493 440 466 481 482 459
Solution to the problem
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x \leq 15[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x >15[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: 72, 75, 94, 42, 42, -14, 13, 18, -9, 4
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=33.7[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =37.420[/tex]
The 4 step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{33.17 -15}{\frac{37.420}{\sqrt{10}}}=1.536[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=10-1=9[/tex]
Now we can calculate the p value, since we have a left tailed test the p value is given by:
[tex]p_v P(t_{(9)}>1.536)= 0.079[/tex]
So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis and we can't conclude that the difference between the means after and before is higher than 15 at 1% of significance
