Respuesta :
Answer:
(0.0772,0.1688)
Step-by-step explanation:
We are given the following data set:
0.12, 0.08, 0.18, 0.07, 0.12, 0.14, 0.15
n = 7
a) 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{0.86}{7} = 0.123[/tex]
Sum of squares of differences = 0.000008163265305 + 0.001836734694 + 0.003265306122 + 0.002793877551 + 0.000008163265305 + 0.0002938775509 + 0.0007367346937 = 0.008942857142
[tex]S.D = \sqrt{\frac{0.008942857142}{6}} = 0.0386[/tex]
Confidence interval:
[tex]\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at}~\alpha_{0.02}\text{ and degree of freedom 6} = \pm 3.142[/tex]
[tex]0.123 \pm 3.142(\frac{0.0386}{\sqrt{7}} ) = 0.123 \pm 0.0458 = (0.0772,0.1688)[/tex]
b) No, this requirement is not met because the interval contains value greater than 0.165 g as well.
