The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Chapter 8.1, Problem 6E

(a)

To determine

To calculate:The shape, center and the spread of the sampling distribution of x¯ .

(a)

Expert Solution
Check Mark

Answer to Problem 6E

The shape of sampling distribution of the sample mean x¯ is approximately normally distributed with mean μ and standard deviation 0.057.

Explanation of Solution

Given information:

Number of trucks tested for NOX emission test is n = 50

Standard deviation σ = 0.04

Formula Used:

  Populationstandarddeviation=StandarddeviationSquarerootofsamplesize

Given, number of trucks tested for NOX emission test is n = 50

Standard deviation σ = 0.04

For the shape of the sampling distribution, use the Central Limit Theorem (CLT).

The shape of sampling distribution of the sample mean x¯ is approximately normally distributed.

For the center, from given information of the sampling distribution of the sample mean is μ .

And for the spread, use the formula.

  Populationstandarddeviation=StandarddeviationSquarerootofsamplesizeσx¯=σn=0.4500.057

Hence, the shape is approximately normally distributed with center μ and spread is 0.057.

(b)

To determine

To draw:The sampling distribution of the mean and mark three standard deviation value on each side of the mean.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information:

Number of trucks tested for NOX emission test is n = 50

Standard deviation σ = 0.04

Given, number of trucks tested for NOX emission test is n = 50

Standard deviation σ = 0.04

  The Practice of Statistics for AP - 4th Edition, Chapter 8.1, Problem 6E , additional homework tip  1

Below figure shows the sampling distribution of x¯ with mean μ as center and three standard deviation values on each sides.

  The Practice of Statistics for AP - 4th Edition, Chapter 8.1, Problem 6E , additional homework tip  2

(c)

To determine

Tocalculate: The distance m of the mean of the sampling distribution.

(c)

Expert Solution
Check Mark

Answer to Problem 6E

Distance m of the mean of the sampling distribution is 0.114.

Explanation of Solution

Given information:

Number of trucks tested for NOX emission test is n = 50

Standard deviation σ = 0.04 Population standard deviation σx=0.057

Given, number of trucks tested for NOX emission test is n = 50

Standard deviation σ = 0.04 Population standard deviation σx=0.057

By 68-95-00.7% rule, about 95% of all values of x¯ lie within a distance m of the mean i.e. twice the population standard deviations of the mean.

  m=2σ=2×0.057=0.114

  The Practice of Statistics for AP - 4th Edition, Chapter 8.1, Problem 6E , additional homework tip  3

Hence, distance m of the mean of the sampling distribution is 0.114.

(d)

To determine

The percent captured μ by the sample intervals.

(d)

Expert Solution
Check Mark

Answer to Problem 6E

Sample intervals capture 95% of μ .

Explanation of Solution

Given information:

Number of trucks tested for NOX emission test is n = 50

Standard deviation σ = 0.04

From 6c, 95% of all values of x¯ lie within a distance m of the mean of the sampling distribution.

Hence 95% of all possible sample capture μ .

Chapter 8 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 8.1 - Prob. 8ECh. 8.1 - Prob. 9ECh. 8.1 - Prob. 10ECh. 8.1 - Prob. 11ECh. 8.1 - Prob. 12ECh. 8.1 - Prob. 13ECh. 8.1 - Prob. 14ECh. 8.1 - Prob. 15ECh. 8.1 - Prob. 16ECh. 8.1 - Prob. 17ECh. 8.1 - Prob. 18ECh. 8.1 - Prob. 19ECh. 8.1 - Prob. 20ECh. 8.1 - Prob. 21ECh. 8.1 - Prob. 22ECh. 8.1 - Prob. 23ECh. 8.1 - Prob. 24ECh. 8.1 - Prob. 25ECh. 8.1 - Prob. 26ECh. 8.2 - Prob. 1.1CYUCh. 8.2 - Prob. 1.2CYUCh. 8.2 - Prob. 2.1CYUCh. 8.2 - Prob. 2.2CYUCh. 8.2 - Prob. 2.3CYUCh. 8.2 - Prob. 2.4CYUCh. 8.2 - Prob. 3.1CYUCh. 8.2 - Prob. 3.2CYUCh. 8.2 - Prob. 27ECh. 8.2 - Prob. 28ECh. 8.2 - Prob. 29ECh. 8.2 - Prob. 30ECh. 8.2 - Prob. 31ECh. 8.2 - Prob. 32ECh. 8.2 - Prob. 33ECh. 8.2 - Prob. 34ECh. 8.2 - Prob. 35ECh. 8.2 - Prob. 36ECh. 8.2 - Prob. 37ECh. 8.2 - Prob. 38ECh. 8.2 - Prob. 39ECh. 8.2 - Prob. 40ECh. 8.2 - Prob. 41ECh. 8.2 - Prob. 42ECh. 8.2 - Prob. 43ECh. 8.2 - Prob. 44ECh. 8.2 - Prob. 45ECh. 8.2 - Prob. 46ECh. 8.2 - Prob. 47ECh. 8.2 - Prob. 48ECh. 8.2 - Prob. 49ECh. 8.2 - Prob. 50ECh. 8.2 - Prob. 51ECh. 8.2 - Prob. 52ECh. 8.2 - Prob. 53ECh. 8.2 - Prob. 54ECh. 8.3 - Prob. 1.1CYUCh. 8.3 - Prob. 2.1CYUCh. 8.3 - Prob. 3.1CYUCh. 8.3 - Prob. 3.2CYUCh. 8.3 - Prob. 3.3CYUCh. 8.3 - Prob. 3.4CYUCh. 8.3 - Prob. 55ECh. 8.3 - Prob. 56ECh. 8.3 - Prob. 57ECh. 8.3 - Prob. 58ECh. 8.3 - Prob. 59ECh. 8.3 - Prob. 60ECh. 8.3 - Prob. 61ECh. 8.3 - Prob. 62ECh. 8.3 - Prob. 63ECh. 8.3 - Prob. 64ECh. 8.3 - Prob. 65ECh. 8.3 - Prob. 66ECh. 8.3 - Prob. 67ECh. 8.3 - Prob. 68ECh. 8.3 - Prob. 69ECh. 8.3 - Prob. 70ECh. 8.3 - Prob. 71ECh. 8.3 - Prob. 72ECh. 8.3 - Prob. 73ECh. 8.3 - Prob. 74ECh. 8.3 - Prob. 75ECh. 8.3 - Prob. 76ECh. 8.3 - Prob. 77ECh. 8.3 - Prob. 78ECh. 8.3 - Prob. 79ECh. 8.3 - Prob. 80ECh. 8 - Prob. 1CRECh. 8 - Prob. 2CRECh. 8 - Prob. 3CRECh. 8 - Prob. 4CRECh. 8 - Prob. 5CRECh. 8 - Prob. 6CRECh. 8 - Prob. 7CRECh. 8 - Prob. 8CRECh. 8 - Prob. 9CRECh. 8 - Prob. 10CRECh. 8 - Prob. 1PTCh. 8 - Prob. 2PTCh. 8 - Prob. 3PTCh. 8 - Prob. 4PTCh. 8 - Prob. 5PTCh. 8 - Prob. 6PTCh. 8 - Prob. 7PTCh. 8 - Prob. 8PTCh. 8 - Prob. 9PTCh. 8 - Prob. 10PTCh. 8 - Prob. 11PTCh. 8 - Prob. 12PTCh. 8 - Prob. 13PT
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