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]df=n-1=7-1=6[/tex]
[tex]p_v =2*P(t_{(6)}<-1.369) =0.220[/tex]
We see that the p value is higher than the ususal significance levels commonly used of 1% or 5% so then we can conclude that we FAIL to reject the null hypothesis, and there is not enough evidence to conclude that we have a different response between the two drugs
Step-by-step explanation:
We have the following info given by the problem
Subject 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
x=value for drug A , y = value for drug B
x: 6 3 4 5 7 1 4
y: 5 1 5 5 5 2 2
We want to verify if the mean response differs between the two drugs then 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]
We can begin calculating the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: -1, -2, 1, 0, -2, 1, -2
Now we can calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.714[/tex]
Now we can find the the standard deviation for the differences, and we got:
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}}=1.38[/tex]
And now we can 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]
Now we can find the degrees of freedom given by:
[tex]df=n-1=7-1=6[/tex]
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]
We see that the p value is higher than the ususal significance levels commonly used of 1% or 5% so then we can conclude that we FAIL to reject the null hypothesis, and there is not enough evidence to conclude that we have a different response between the two drugs
