Mathematics: 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.

The product of a constant and a variables...

14. A job candidate with an offer from a prominent

anabelleacunamu

23.01.2022

95% confidence interval: (67.97,84.69)

Step-by-step 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.

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

$\bar{x} =\displaystyle\frac{458}{6} = 76.33$

Sum of squares of differences = 317.33

$s = \sqrt{\dfrac{317.33}{5}} = 7.97$

95% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

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

$76.33 \pm 2.5705(\dfrac{7.97}{\sqrt{6}} )\\\\ = 76.33 \pm 8.3637\\\\ = (67.9663 ,84.6937)\approx (67.97,84.69)$

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

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