Respuesta :
Answer:
98% Confidence interval: (31.74,38.4)
Step-by-step explanation:
We are given the following data set:
29, 38, 38, 33, 38, 21, 45, 34, 40, 37, 37, 42, 30, 29, 35
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{526}{15} = 35.07[/tex]
Sum of squares of differences = 506.93
[tex]S.D = \sqrt{\displaystyle\frac{506.93}{14}} = 6.02[/tex]
98% 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 14 and}~\alpha_{0.02} = \pm 2.145[/tex]
[tex]35.07 \pm 2.145(\frac{6.02}{\sqrt{15}} ) = 35.07 \pm 3.33 = (31.74,38.4)[/tex]
Thus, there is 98% confidence that the population mean number of minutes spent traveling by workers is between 31.74 mins and 38.40
