Respuesta :
Answer:
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369[/tex]
[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/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 that the difference mean between Drug A and Drug B is equal to 0.
Step-by-step explanation:
Data given
1 2 3 4 5 6 7
Drug A 6 3 4 5 7 1 4
Drug B 5 1 5 5 5 2 2
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 for A , y = test value for B
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: -1, -2, 1, 0, -2, 1, -2
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.714[/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} =1.38[/tex]
The 4 step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=7-1=6[/tex]
Now we can calculate the p value, since we have a two tailed test the p value is given by:
[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/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 that the difference mean between Drug A and Drug B is equal to 0.
