Respuesta :
Answer:
a) Sample Mean = 79.4
Sample standard deviation = 30.62
b) 90% Confidence interval: (67.56 ,91.24)
Step-by-step explanation:
We are given the following in the question:
Prices for sleeping bags has an approximately normal distribution.
We are given the following sample:
35, 85, 105, 40, 100, 50, 30, 23, 100, 110, 105, 95, 105, 60, 110, 120, 95, 90, 60, 70
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{1588}{20} = 79.4[/tex]
Sum of squares of differences = 1971.36 + 31.36 + 655.36 + 1552.36 + 424.36 + 864.36 + 2440.36 + 3180.96 + 424.36 + 936.36 + 655.36 + 243.36 + 655.36 + 376.36 + 936.36 + 1648.36 + 243.36 + 112.36 + 376.36 + 88.36 = 17816.8
[tex]S.D = \sqrt{\displaystyle\frac{17816.8}{19}} = 30.62[/tex]
b) 90% 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_{0.10} = \pm 1.7291[/tex]
[tex]79.4 \pm 1.73(\displaystyle\frac{ 30.62}{\sqrt{20}} ) = 79.4 \pm 11.84 = (67.56 ,91.24)[/tex]
