Question

14. A job candidate with an offer from a prominent investment bank wanted to estimate how many hours she would have to work per week during her first year at the bank. She took a sample of six first-year analysts, asking how many hours they worked in the last week. Construct a 95% confidence interval with her results: 64, 82, 74, 73, 78, and 87 hours.

Answer 1

95% confidence interval: (67.97,84.69)

Explanation:

We are given the following data set:

64, 82, 74, 73, 78, 87

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}}`

`\bar{x} =\displaystyle(458)/(6) = 76.33`

Sum of squares of differences = 317.33

`s = \sqrt{(317.33)/(5)} = 7.97`

95% Confidence interval:

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

Putting the values, we get,

`t_(critical)\text{ at degree of freedom 5 and}~\alpha_(0.05) = \pm 2.5705`

`76.33 \pm 2.5705((7.97)/(√(6)) )\\\\ = 76.33 \pm 8.3637\\\\ = (67.9663 ,84.6937)\approx (67.97,84.69)`

(67.97,84.69) is the required 95% confidence interval.