Question

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data AND determine the margin of error EBM.
DATA: 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5

Answer 1

Answer:

Confidence interval

$8.98-2.2\frac{1.29}{\sqrt{12}}=8.16$    

$8.98+2.2\frac{1.29}{\sqrt{12}}=9.80$  

And the margin of error would be:

$ME=2.2\frac{1.29}{\sqrt{12}}=0.819$

Step-by-step explanation:

For this case we have the followig dataset:

DATA: 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5

We can calculate the mean and the deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$\bar X= 8.98[/tex[tex]s = 1.29$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The degrees of freedom are given by:

$df=n-1=12-1=11$

The Confidence level is 0.95 or 95%, the value of significance is $\alpha=0.05$ and $\alpha/2 =0.025$, and the critical value would be $t_{\alpha/2}=2.20$

Repplacing the info we got:

$8.98-2.2\frac{1.29}{\sqrt{12}}=8.16$    

$8.98+2.2\frac{1.29}{\sqrt{12}}=9.80$    And the margin of error would be:

$ME=2.2\frac{1.29}{\sqrt{12}}=0.819$