Respuesta :
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:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]\bar{x} =\displaystyle\frac{57.7}{12} = 4.8[/tex]
Sum of squares of differences = 51.18
[tex]s = \sqrt{\dfrac{51.18}{11}} = 2.16[/tex]
99% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 11 and}~\alpha_{0.01} = \pm 3.106[/tex]
[tex]4.8 \pm 3.106(\dfrac{2.16}{\sqrt{12}} )\\\\ = 4.8 \pm 1.937 \\\\= (2.863,6.737)\approx (2.9,6.7)[/tex]
(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.
