EBK INTRODUCTION TO THE PRACTICE OF STA
EBK INTRODUCTION TO THE PRACTICE OF STA
9th Edition
ISBN: 8220103674638
Author: Moore
Publisher: YUZU
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6.4, Problem 115E

(a)

To determine

To find: The power by using Statistical Power applet.

(a)

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The power is 0.885.

Explanation of Solution

To obtain the power by using “Statistical Power applet”, follow the steps below:

Step 1: Go to the “Statistical power” on the website. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  1

Step 2: Specify “ H0:μ=0 ” and select Ha as “ Ha:μ0 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  2

Step 3: Specify σ as “ σ=1 ” and specify n as “ n=10 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  3

Step 4: Specify Alpha (α) as “ α=0.05 ” and specify μ as “ μ=1 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  4

The obtained result is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  5

The obtained power is 0.885.

To determine

To find: The power for different values of μ.

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The table providing μ and corresponding the power is shown below:

μ

Power

0.1

0.062

0.2

0.097

0.3

0.158

0.4

0.244

0.5

0.353

0.6

0.475

0.7

0.600

0.8

0.716

0.9

0.812

Explanation of Solution

Calculation:

To obtain the power by using Statistical Power applet, follow the steps below:

Step1: Go to “Statistical power” on the website. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  6

Step2: Specify “ H0:μ=0 ” and select Ha as “ Ha:μ0 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  7

Step3: Specify σ as “ σ=1 ” and specify n as “ n=10 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  8

Step4: Specify Alpha (α) as “ α=0.05 ” and specify μ as “ μ=0.1 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  9

The obtained result is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  10

The obtained power is 0.062. Now, repeat this process for the other provided values of μ. Thus, the table given μ and the power is shown below:

μ

Power

0.1

0.062

0.2

0.097

0.3

0.158

0.4

0.244

0.5

0.353

0.6

0.475

0.7

0.600

0.8

0.716

0.9

0.812

To determine

To explain: The obtained power.

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The power decreases as the alternative changes from one-sided alternative to the two-sided alternative.

Explanation of Solution

Power of a test is the capability to discard the null hypothesis when it is not true. It can be seen from the obtained table in exercise 6.114 that as the value of μ increases, the power also increases. In the above table when one-sided alternative changes to two sided alternative, the power gets decreases.

(b)

To determine

To find: The power.

(b)

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The power is calculated as 1.00.

Explanation of Solution

Calculation:

To obtain the power by using the Statistical Power applet”, follow the steps below:

Step1: Go to the “Statistical power” on the website. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  11

Step2: Specify “ H0:μ=0 ” and select Ha as “ Ha:μ>0 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  12

Step3: Specify σ as “ σ=0.5 ” and specify n as “ n=10 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  13

Step4: Specify Alpha (α) as “ α=0.05 ” and specify μ as “ μ=1 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  14

The obtained result is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  15

The obtained power is 1.00.

To determine

To find: The power for the different values of μ.

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The table providing μ and the power is shown below:

μ

Power

0.1

0.156

0.2

0.352

0.3

0.600

0.4

0.812

0.5

0.935

0.6

0.984

0.7

0.997

0.8

1.00

0.9

1.00

Explanation of Solution

Calculation:

To obtain the power by using the Statistical Power applet, follow the steps below:

Step1: Go to the “Statistical power” on the website. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  16

Step2: Specify “ H0:μ=0 ” and select Ha as Ha:μ>0. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  17

Step3: Specify σ as “ σ=0.5 ” and specify n as “ n=10 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  18

Step4: Specify Alpha (α) as “ α=0.05 ” and specify μ as “ μ=0.1 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  19

The obtained result is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  20

The obtained power is 0.156. Now, repeat this process for the other provided values of μ. Thus, the table given μ and the power is shown below:

μ

Power

0.1

0.156

0.2

0.352

0.3

0.600

0.4

0.812

0.5

0.935

0.6

0.984

0.7

0.997

0.8

1.00

0.9

1.00

To determine

To explain: The obtained power.

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The power increases as σ decreases to 0.5..

Explanation of Solution

Power of a test is the capability to discard the null hypothesis when it is not true. It can be seen from the obtained table in the exercise 6.114 that as the value of μ increases, the power decreases. In the above table when the value of σ decreases, the power gets increases.

(c)

To determine

To find: The power.

(c)

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The power is calculated as 1.00.

Explanation of Solution

Calculation:

To obtain the power by using the Statistical Power applet”, follow the steps below:

Step1: Go to the “Statistical power” on the website. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  21

Step2: Specify “ H0:μ=0 ” and select Ha as “ Ha:μ0 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  22

Step3: Specify σ as “ σ=1 ” and specify n as “ n=30 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  23

Step4: Specify Alpha (α) as “ α=0.05 ” and specify μ as “ μ=1 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  24

The obtained result is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  25

The obtained power is 1.00.

To determine

To find: The power for the different values of μ.

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The table providing μ and the power is shown below:

μ

Power

0.1

0.136

0.2

0.291

0.3

0.499

0.4

0.707

0.5

0.863

0.6

0.950

0.7

0.986

0.8

0.997

0.9

0.999

Explanation of Solution

Calculation:

To obtain the power by using the Statistical Power applet, follow the steps below:

Step1: Go to the “Statistical power” on the website. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  26

Step2: Specify “ H0:μ=0 ” and select Ha as Ha:μ>0. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  27

Step3: Specify σ as “ σ=1 ” and specify n as “ n=30 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  28

Step4: Specify Alpha (α) as “ α=0.05 ” and specify μ as “ μ=0.1 ”. The screenshot is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  29

The obtained result is shown below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 6.4, Problem 115E , additional homework tip  30

The obtained power is 0.136. Now, repeat this process for the other provided values of μ. Thus, the table given μ and the power is shown below:

μ

Power

0.1

0.136

0.2

0.291

0.3

0.499

0.4

0.707

0.5

0.863

0.6

0.950

0.7

0.986

0.8

0.997

0.9

0.999

To determine

To explain: The obtained power.

Expert Solution
Check Mark

Answer to Problem 115E

Solution: The power increases as the ‘n’ changes 10 to 30.

Explanation of Solution

Power of a test is the capability to discard the null hypothesis when it is not true. It can be seen from the obtained table in the exercise 6.114 that as the value of μ increases, the power increases. In the above table when the value of n increases, the power gets increases.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
(b) In various places in this module, data on the silver content of coins minted in the reign of the twelfth-century Byzantine king Manuel I Comnenus have been considered. The full dataset is in the Minitab file coins.mwx. The dataset includes, among others, the values of the silver content of nine coins from the first coinage (variable Coin1) and seven from the fourth coinage (variable Coin4) which was produced a number of years later. (For the purposes of this question, you can ignore the variables Coin2 and Coin3.) In particular, in Activity 8 and Exercise 2 of Computer Book B, it was argued that the silver contents in both the first and the fourth coinages can be assumed to be normally distributed. The question of interest is whether there were differences in the silver content of coins minted early and late in Manuel’s reign. You are about to investigate this question using a two-sample t-interval. (i) Using Minitab, find either the sample standard deviations of the two variables…
Homework Let X1, X2, Xn be a random sample from f(x;0) where f(x; 0) = (-), 0 < x < ∞,0 € R Using Basu's theorem, show that Y = min{X} and Z =Σ(XY) are indep. -
Homework Let X1, X2, Xn be a random sample from f(x; 0) where f(x; 0) = e−(2-0), 0 < x < ∞,0 € R Using Basu's theorem, show that Y = min{X} and Z =Σ(XY) are indep.

Chapter 6 Solutions

EBK INTRODUCTION TO THE PRACTICE OF STA

Ch. 6.1 - Prob. 11UYKCh. 6.1 - Prob. 12ECh. 6.1 - Prob. 13ECh. 6.1 - Prob. 14ECh. 6.1 - Prob. 15ECh. 6.1 - Prob. 16ECh. 6.1 - Prob. 17ECh. 6.1 - Prob. 18ECh. 6.1 - Prob. 19ECh. 6.1 - Prob. 20ECh. 6.1 - Prob. 21ECh. 6.1 - Prob. 22ECh. 6.1 - Prob. 23ECh. 6.1 - Prob. 24ECh. 6.1 - Prob. 25ECh. 6.1 - Prob. 26ECh. 6.1 - Prob. 27ECh. 6.1 - Prob. 28ECh. 6.1 - Prob. 29ECh. 6.1 - Prob. 30ECh. 6.1 - Prob. 31ECh. 6.1 - Prob. 32ECh. 6.1 - Prob. 33ECh. 6.1 - Prob. 34ECh. 6.1 - Prob. 35ECh. 6.1 - Prob. 36ECh. 6.1 - Prob. 37ECh. 6.2 - Prob. 38UYKCh. 6.2 - Prob. 39UYKCh. 6.2 - Prob. 40UYKCh. 6.2 - Prob. 41UYKCh. 6.2 - Prob. 42UYKCh. 6.2 - Prob. 43UYKCh. 6.2 - Prob. 44UYKCh. 6.2 - Prob. 45UYKCh. 6.2 - Prob. 46UYKCh. 6.2 - Prob. 47UYKCh. 6.2 - Prob. 48UYKCh. 6.2 - Prob. 49UYKCh. 6.2 - Prob. 50UYKCh. 6.2 - Prob. 51UYKCh. 6.2 - Prob. 52ECh. 6.2 - Prob. 53ECh. 6.2 - Prob. 54ECh. 6.2 - Prob. 55ECh. 6.2 - Prob. 56ECh. 6.2 - Prob. 57ECh. 6.2 - Prob. 58ECh. 6.2 - Prob. 59ECh. 6.2 - Prob. 60ECh. 6.2 - Prob. 61ECh. 6.2 - Prob. 62ECh. 6.2 - Prob. 63ECh. 6.2 - Prob. 64ECh. 6.2 - Prob. 65ECh. 6.2 - Prob. 66ECh. 6.2 - Prob. 67ECh. 6.2 - Prob. 68ECh. 6.2 - Prob. 69ECh. 6.2 - Prob. 70ECh. 6.2 - Prob. 71ECh. 6.2 - Prob. 72ECh. 6.2 - Prob. 73ECh. 6.2 - Prob. 74ECh. 6.2 - Prob. 75ECh. 6.2 - Prob. 76ECh. 6.2 - Prob. 77ECh. 6.2 - Prob. 78ECh. 6.2 - Prob. 79ECh. 6.2 - Prob. 80ECh. 6.2 - Prob. 81ECh. 6.2 - Prob. 82ECh. 6.2 - Prob. 83ECh. 6.2 - Prob. 84ECh. 6.2 - Prob. 85ECh. 6.2 - Prob. 86ECh. 6.2 - Prob. 87ECh. 6.2 - Prob. 88ECh. 6.2 - Prob. 89ECh. 6.3 - Prob. 90UYKCh. 6.3 - Prob. 91UYKCh. 6.3 - Prob. 92ECh. 6.3 - Prob. 93ECh. 6.3 - Prob. 94ECh. 6.3 - Prob. 95ECh. 6.3 - Prob. 96ECh. 6.3 - Prob. 97ECh. 6.3 - Prob. 98ECh. 6.3 - Prob. 99ECh. 6.3 - Prob. 100ECh. 6.3 - Prob. 101ECh. 6.3 - Prob. 102ECh. 6.3 - Prob. 103ECh. 6.3 - Prob. 104ECh. 6.3 - Prob. 105ECh. 6.3 - Prob. 106ECh. 6.3 - Prob. 107ECh. 6.3 - Prob. 108ECh. 6.3 - Prob. 109ECh. 6.4 - Prob. 110ECh. 6.4 - Prob. 111ECh. 6.4 - Prob. 112ECh. 6.4 - Prob. 113ECh. 6.4 - Prob. 114ECh. 6.4 - Prob. 115ECh. 6.4 - Prob. 116ECh. 6.4 - Prob. 117ECh. 6.4 - Prob. 118ECh. 6.4 - Prob. 119ECh. 6.4 - Prob. 120ECh. 6.4 - Prob. 121ECh. 6.4 - Prob. 122ECh. 6.4 - Prob. 123ECh. 6 - Prob. 124ECh. 6 - Prob. 125ECh. 6 - Prob. 126ECh. 6 - Prob. 127ECh. 6 - Prob. 128ECh. 6 - Prob. 129ECh. 6 - Prob. 130ECh. 6 - Prob. 131ECh. 6 - Prob. 132ECh. 6 - Prob. 133ECh. 6 - Prob. 134ECh. 6 - Prob. 135ECh. 6 - Prob. 136ECh. 6 - Prob. 137ECh. 6 - Prob. 138ECh. 6 - Prob. 139ECh. 6 - Prob. 140ECh. 6 - Prob. 141ECh. 6 - Prob. 142ECh. 6 - Prob. 143ECh. 6 - Prob. 144E
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
What is a Linear Equation in One Variable?; Author: Don't Memorise;https://www.youtube.com/watch?v=lDOYdBgtnjY;License: Standard YouTube License, CC-BY
Linear Equation | Solving Linear Equations | What is Linear Equation in one variable ?; Author: Najam Academy;https://www.youtube.com/watch?v=tHm3X_Ta_iE;License: Standard YouTube License, CC-BY