Respuesta :
Answer:
99% Confidence interval: (183,246)
Step-by-step explanation:
We are given the following data set:
141, 144, 150, 161, 169, 179, 186, 194, 199, 209, 219, 220, 226, 237, 254, 261, 275, 278, 286, 295
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]Mean =\displaystyle\frac{4283}{20} = 214.15[/tex]
Sum of squares of differences = 45922.55
[tex]S.D = \sqrt{\frac{45922.55}{19}} = 49.16[/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 19 and}~\alpha_{0.01} = \pm 2.86[/tex]
[tex]214.15 \pm 2.86(\frac{49.16}{\sqrt{20}} ) = 214.15 \pm 31.44 =(182.71,245.59) \approx (183,246)[/tex]
