Question

The following data show the number of hours per day 12 adults spent in front of screens watching​ television-related content. 1.8 4.7 4.1 5.2 7.6 7.3 5.7 3.2 5.4 1.9 2.7 8.1 a. Construct a 99 ​% confidence interval to estimate the average number of hours per day adults spend in front of screens watching​ television-related content.

Answer 1

Answer:

99% Confidence interval:  (2.9,6.7)    

Step-by-step explanation:

We are given the following data set:

1.8, 4.7, 4.1, 5.2, 7.6, 7.3, 5.7, 3.2, 5.4, 1.9, 2.7, 8.1

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.  

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$\bar{x} =\displaystyle\frac{57.7}{12} = 4.8$

Sum of squares of differences = 51.18

$s = \sqrt{\dfrac{51.18}{11}} = 2.16$

99% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 11 and}~\alpha_{0.01} = \pm 3.106$  

$4.8 \pm 3.106(\dfrac{2.16}{\sqrt{12}} )\\\\ = 4.8 \pm 1.937 \\\\= (2.863,6.737)\approx (2.9,6.7)$  

(2.9,6.7) is the required 99% confidence interval for average number of hours per day adults spend in front of screens watching​ television-related content.