Answered: A corporation that maintains a large…

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

A corporation that maintains a large fleet of company cars for the use of its sales staff is interested in the mean distance driven monthly per sales person. The following list gives the monthly distances in miles driven by a random sample of

10 sales persons.

2255 , 2610 , 2566 , 2645 , 2671, 1979, 2434, 2485, 2290, 2580

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Based on this sample, find a

95% confidence interval for the mean number of miles driven monthly by members of the sales staff, assuming that monthly driving distances are normally distributed. Then find the lower limit and upper limit of the

95% confidence interval.

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

Lower limit:	?
Upper limit:	?

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

$S a m p l e s i z e (n) = 10$

Here sample data is given hence we use T Test to find confidence interval

$\begin{array}{rcl} D e g r e e s o f f r e e d o m (d f) & = & n - 1 \\ = & 10 - 1 \\ = & 9 \end{array}$

$C o n f i d e n c e l e v e l = 0.95 (95 %)$

$\begin{array}{rcl} α & = & 1 - c o n f i d e n c e l e v e l \\ = & 1 - 0.95 \\ = & 0.05 \end{array}$

Using T table or Excel command =T.INV.2T(0.05,9) we find critical value

$C r i t i c a l v a l u e (t c) = 2.262$

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