Question

Construct a​ 90% confidence​ interval, using the inequality . To test the effectiveness of a new drug that is reported to increase the number of hours of sleep patients get during the​ night, researchers randomly select 13 patients and record the number of hours of sleep each gets with and without the new drug.

Answer 1

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:

$\bar d=\frac{1}{n}\sum d=\frac{1}{13}\times [(-1.8)+(-1.5)+(-1.2)+...+(-2.4)]=-1.654\\\\S_{d}=\sqrt{\frac{1}{n-1}\sum [d-\bar d]^{2}}=0.6753$

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:

$t_{\alpha/2, (n-1)}=t_{0.10/2, 12}=1.782$

Compute the 90% confidence​ interval as follows:

$\bar d-t_{\alpha/2, (n-1)}\cdot\frac{S_{d}}{\sqrt{n}}$

$-1.645-[1.782\cdot\frac{0.6753}{\sqrt{13}}]$

Thus, the 90% confidence​ interval is (-1.987, -1.320).