Respuesta :
Answer:
The 90% confidence interval is (-1.987, -1.320).
Step-by-step explanation:
The data provided is:
Sleep without Sleep using
the drug the drug d=without-with
2.5 4.3 -1.8
2.7 4.2 -1.5
3.8 5.0 -1.2
3.8 6.2 -2.4
1.8 2.5 -0.7
4.7 7.1 -2.4
4.8 7.1 -2.3
2.7 3.3 -0.6
1.8 3.5 -1.7
2.2 4.3 -2.1
3.6 4.3 -0.7
5.3 7.0 -1.7
5.2 7.6 -2.4
Compute the sample mean and sample standard deviation:
[tex]\bar d=\frac{1}{n}\sum d=\frac{1}{13}\times [(-1.8)+(-1.5)+(-1.2)+...+(-2.4)]=-1.654\\\\S_{d}=\sqrt{\frac{1}{n-1}\sum [d-\bar d]^{2}}=0.6753[/tex]
The degrees of freedom of the test is:
df = n - 1 = 13 - 1 = 12
Compute the critical value of t for 90% confidence level and 12 degrees of freedom as follows:
[tex]t_{\alpha/2, (n-1)}=t_{0.10/2, 12}=1.782[/tex]
Compute the 90% confidence interval as follows:
[tex]\bar d-t_{\alpha/2, (n-1)}\cdot\frac{S_{d}}{\sqrt{n}}<\mu_{d}<\bar d+t_{\alpha/2, (n-1)}\cdot\frac{S_{d}}{\sqrt{n}}[/tex]
[tex]-1.645-[1.782\cdot\frac{0.6753}{\sqrt{13}}]<\mu_{d}<-1.645+[1.782\cdot\frac{0.6753}{\sqrt{13}}]\\\\-1.645-0.334<\mu_{d}<-1.654+0.334\\\\-1.988<\mu_{d}<-1.320[/tex]
Thus, the 90% confidence interval is (-1.987, -1.320).
