Question

GL Stats: lviixea conrldence Intervals Practice

On each problem, verify that the conditions for a confidence interval are met!
(1) Suppose the height of senior girls at Anytown High School is known to be normally distributed. A sample of
leights in inches of 23 randomly selected senior girls were: 63, 68, 60, 59, 68, 65, 67, 64, 69, 69, 61, 67, 61, 60,
66, 67, 68, 66, 70, 79, 76, 75, 65. Construct and interpret a 99% confidence interval for the true mean height​

Answer 1

99% Confidence interval: (63.65,69.65

Explanation:

We are given the following data:

63, 68, 60, 59, 68, 65, 67, 64, 69, 69, 61, 67, 61, 60, 66, 67, 68, 66, 70, 79, 76, 75, 65

Formula:

`\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}`

where
`x_i` are data points,
`\bar{x}` is the mean and n is the number of observations.

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(1533)/(23) = 66.65`

Sum of squares of differences = 575.217

`S.D = \sqrt{(575.217)/(22)} = 5.11`

99% Confidence interval:

`\bar{x} \pm t_(critical)\displaystyle(s)/(√(n))`

Putting the values, we get,

`t_(critical)\text{ at degree of freedom 22 and}~\alpha_(0.01) = \pm 2.818`

`66.65 \pm 2.818((5.11)/(√(23)) ) = 66.65 \pm 3.002 = (63.65,69.65)`