PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 12.2, Problem 38E

(a)

To determine

To find: that is it reasonable to use a power model to explain the relation between the period of pendulum and the length.

(a)

Expert Solution
Check Mark

Answer to Problem 38E

From the equation, it is noticed that the scatter plot of log (volume) and log (pressure) is linear and the residual figure represents no obvious leftover patterns.

Explanation of Solution

Given:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  1

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  2

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  3

From the equation, it is noticed that the scatter plot of log (volume) and log (pressure) is linear and the residual figure represents no obvious leftover patterns.

(b)

To determine

To find: the equation of the least squares of regression line.

(b)

Expert Solution
Check Mark

Answer to Problem 38E

  log(pressure)=1.111160.81344log(volume)

Explanation of Solution

Given:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  4

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  5

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  6

Calculation:

On the basis of given information the general regression equation is

  log(pressure)=α+βlog(volume)

Calculate the α (constant) is mention in the row “constant” and in the column “Coef” of the mention output from the computer

  α =1.11116

Calculate the β (constant) is mention in the row “ log(volume) ” and in the column “Coef” of the mention output from the computer

  β =-0.81344

Putting the value of α and β in the equation

  log(pressure)=α+βlog(volume)log(pressure)=1.111160.81344log(volume)

(c)

To determine

To Explain: the prediction of the pressure in the syringe if the volume is 17 cubic cm with using the part (a)

(c)

Expert Solution
Check Mark

Answer to Problem 38E

1.28914 atm

Explanation of Solution

Given:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  7

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  8

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 12.2, Problem 38E , additional homework tip  9

Calculation:

From the part (b)

  log(pressure)=1.111160.81344log(volume)

Putting the value of volume by 17 and calculate

  log(pressure)=1.111160.81344log(17)log(pressure)=0.1103

Taking the exponential with 10

  pressure=10log(pressure)=100.1103=1.28914

Therefore the expected pressure is 1.28914 atm.

Chapter 12 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 12.1 - Prob. 11ECh. 12.1 - Prob. 12ECh. 12.1 - Prob. 13ECh. 12.1 - Prob. 14ECh. 12.1 - Prob. 15ECh. 12.1 - Prob. 16ECh. 12.1 - Prob. 17ECh. 12.1 - Prob. 18ECh. 12.1 - Prob. 19ECh. 12.1 - Prob. 20ECh. 12.1 - Prob. 21ECh. 12.1 - Prob. 22ECh. 12.1 - Prob. 23ECh. 12.1 - Prob. 24ECh. 12.1 - Prob. 25ECh. 12.1 - Prob. 26ECh. 12.1 - Prob. 27ECh. 12.1 - Prob. 28ECh. 12.1 - Prob. 29ECh. 12.1 - Prob. 30ECh. 12.1 - Prob. 31ECh. 12.1 - Prob. 32ECh. 12.2 - Prob. 33ECh. 12.2 - Prob. 34ECh. 12.2 - Prob. 35ECh. 12.2 - Prob. 36ECh. 12.2 - Prob. 37ECh. 12.2 - Prob. 38ECh. 12.2 - Prob. 39ECh. 12.2 - Prob. 40ECh. 12.2 - Prob. 41ECh. 12.2 - Prob. 42ECh. 12.2 - Prob. 43ECh. 12.2 - Prob. 44ECh. 12.2 - Prob. 45ECh. 12.2 - Prob. 46ECh. 12.2 - Prob. 47ECh. 12.2 - Prob. 48ECh. 12.2 - Prob. 49ECh. 12.2 - Prob. 50ECh. 12.2 - Prob. 51ECh. 12.2 - Prob. 52ECh. 12.2 - Prob. 53ECh. 12.2 - Prob. 54ECh. 12.2 - Prob. 55ECh. 12.2 - Prob. 56ECh. 12.2 - Prob. 57ECh. 12.2 - Prob. 58ECh. 12 - Prob. R12.1RECh. 12 - Prob. R12.2RECh. 12 - Prob. R12.3RECh. 12 - Prob. R12.4RECh. 12 - Prob. R12.5RECh. 12 - Prob. R12.6RECh. 12 - Prob. T12.1SPTCh. 12 - Prob. T12.2SPTCh. 12 - Prob. T12.3SPTCh. 12 - Prob. T12.4SPTCh. 12 - Prob. T12.5SPTCh. 12 - Prob. T12.6SPTCh. 12 - Prob. T12.7SPTCh. 12 - Prob. T12.8SPTCh. 12 - Prob. T12.9SPTCh. 12 - Prob. T12.10SPTCh. 12 - Prob. T12.11SPTCh. 12 - Prob. T12.12SPTCh. 12 - Prob. AP4.1CPTCh. 12 - Prob. AP4.2CPTCh. 12 - Prob. AP4.3CPTCh. 12 - Prob. AP4.4CPTCh. 12 - Prob. AP4.5CPTCh. 12 - Prob. AP4.6CPTCh. 12 - Prob. AP4.7CPTCh. 12 - Prob. AP4.8CPTCh. 12 - Prob. AP4.9CPTCh. 12 - Prob. AP4.10CPTCh. 12 - Prob. AP4.11CPTCh. 12 - Prob. AP4.12CPTCh. 12 - Prob. AP4.13CPTCh. 12 - Prob. AP4.14CPTCh. 12 - Prob. AP4.15CPTCh. 12 - Prob. AP4.16CPTCh. 12 - Prob. AP4.17CPTCh. 12 - Prob. AP4.18CPTCh. 12 - Prob. AP4.19CPTCh. 12 - Prob. AP4.20CPTCh. 12 - Prob. AP4.21CPTCh. 12 - Prob. AP4.22CPTCh. 12 - Prob. AP4.23CPTCh. 12 - Prob. AP4.24CPTCh. 12 - Prob. AP4.25CPTCh. 12 - Prob. AP4.26CPTCh. 12 - Prob. AP4.27CPTCh. 12 - Prob. AP4.28CPTCh. 12 - Prob. AP4.29CPTCh. 12 - Prob. AP4.30CPTCh. 12 - Prob. AP4.31CPTCh. 12 - Prob. AP4.32CPTCh. 12 - Prob. AP4.33CPTCh. 12 - Prob. AP4.34CPTCh. 12 - Prob. AP4.35CPTCh. 12 - Prob. AP4.36CPTCh. 12 - Prob. AP4.37CPTCh. 12 - Prob. AP4.38CPTCh. 12 - Prob. AP4.39CPTCh. 12 - Prob. AP4.40CPTCh. 12 - Prob. AP4.41CPTCh. 12 - Prob. AP4.42CPTCh. 12 - Prob. AP4.43CPTCh. 12 - Prob. AP4.44CPTCh. 12 - Prob. AP4.45CPTCh. 12 - Prob. AP4.46CPT
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