Answered: Dr. Pat Jones is an industrial…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

14.

Dr. Pat Jones is an industrial psychologist and is currently studying stress amount internet company executives. A questionnaire was developed to measure stress. A score above 80 indicates stress at a dangerous level. A random sample of 15 executives revealed the stress scores in the attached file.

homework 3 - 2.xlsx ( Stress 94, 78, 83, 90, 78, 99, 97, 90, 97, 90, 93, 94, 100, 75, 84 )

Construct a 95% confidence interval for the population mean.

What is the upper boundary? (answer should be in the following format xx.xxx)

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Expert Solution

Step 1

Given data,

94, 78, 83, 90, 78, 99, 97, 90, 97, 90, 93, 94, 100, 75, 84

Mean:

$\begin{array}{rcl} x & = & \frac{\sum_{}^{} x_{i}}{n} \\ = & \frac{94 + 78 + 83 + . . . . + 75 + 84}{15} \\ = & 89.267 \end{array}$

Standard deviation:

$\begin{array}{rcl} s & = & \sqrt{\frac{\sum_{}^{} {(x_{i} - x)}^{2}}{n - 1}} \\ = & \sqrt{\frac{{(94 - 89.267)}^{2} + {(78 - 89.267)}^{2} + . . . {(75 - 89.267)}^{2} + {(84 - 89.267)}^{2}}{15 - 1}} \\ = & 8.022 \end{array}$

Here, sample size is 15. Population standard deviation is not known. It is appropriate to use t interval.

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