Answer:
![t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%20d%20-0%7D%7B%5Cfrac%7Bs_d%7D%7B%5Csqrt%7Bn%7D%7D%7D%3D%5Cfrac%7B-0.714%20-0%7D%7B%5Cfrac%7B1.38%7D%7B%5Csqrt%7B7%7D%7D%7D%3D-1.369)
![df=n-1=7-1=6](https://tex.z-dn.net/?f=df%3Dn-1%3D7-1%3D6)
![p_v =2*P(t_{(6)}](https://tex.z-dn.net/?f=p_v%20%3D2%2AP%28t_%7B%286%29%7D%3C-1.369%29%20%3D0.220)
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: ![\mu_y- \mu_x = 0](https://tex.z-dn.net/?f=%5Cmu_y-%20%5Cmu_x%20%3D%200)
Alternative hypothesis: ![\mu_y -\mu_x \neq 0](https://tex.z-dn.net/?f=%5Cmu_y%20-%5Cmu_x%20%5Cneq%200)
We can begin calculating the difference
and we obtain this:
d: -1, -2, 1, 0, -2, 1, -2
Now we can calculate the mean difference
![\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.714](https://tex.z-dn.net/?f=%5Cbar%20d%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20d_i%7D%7Bn%7D%3D-0.714)
Now we can find the the standard deviation for the differences, and we got:
![s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}}=1.38](https://tex.z-dn.net/?f=s_d%20%3D%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28d_i%20-%5Cbar%20d%29%5E2%7D%7Bn-1%7D%7D%3D1.38)
And now we can calculate the statistic given by :
![t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%20d%20-0%7D%7B%5Cfrac%7Bs_d%7D%7B%5Csqrt%7Bn%7D%7D%7D%3D%5Cfrac%7B-0.714%20-0%7D%7B%5Cfrac%7B1.38%7D%7B%5Csqrt%7B7%7D%7D%7D%3D-1.369)
Now we can find the degrees of freedom given by:
![df=n-1=7-1=6](https://tex.z-dn.net/?f=df%3Dn-1%3D7-1%3D6)
We can calculate the p value, since we have a two tailed test the p value is given by:
![p_v =2*P(t_{(6)}](https://tex.z-dn.net/?f=p_v%20%3D2%2AP%28t_%7B%286%29%7D%3C-1.369%29%20%3D0.220)
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