MATH SOLVE

3 months ago

Q:
# Given values: 54, 65, 76, 43, 45, 76, 87 12, 23, 34, 67,65, 78, 45, 34. Given 95% confidence and sigma (population standard deviation) (12) Compute the confidence interval. Round to one decimal place. (A) (47.5, 59.7)(B) (12, 87) (C) (48.5, 58.7) (D) None of the above.

Accepted Solution

A:

Answer:Option (D) None of the above.Step-by-step explanation:We are given the following data:54, 65, 76, 43, 45, 76, 87, 12, 23, 34, 67,65, 78, 45, 34Formula:
[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{804}{15} = 53.6[/tex]
Sum of squares of differences = 0.16 + 129.96 + 501.76 + 112.36 + 73.96 + 501.76 + 1115.56 + 1730.56 + 936.36 + 384.16 + 179.56 + 129.96 + 595.36 + 73.96 + 384.16 = 6849.6[tex]S.D = \sqrt{\frac{6849.6}{15}} = 21.4[/tex]
Confidence interval:
[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]
[tex]53.6 \pm 1.96(\frac{21.4}{\sqrt{15}} ) = 53.6 \pm 10.83 = (42.8,64.4)[/tex]