Statistical Techniques in Business and Economics
Statistical Techniques in Business and Economics
16th Edition
ISBN: 9780077639723
Author: Lind
Publisher: Mcgraw-Hill Course Content Delivery
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 13, Problem 62DE

a.

To determine

Find the regression equation.

Find the selling price of a home with an area of 2,200 square feet.

Construct a 95% confidence interval for all 2,200 square foot homes.

Construct a 95% prediction interval for the selling price of a home with 2,200 square feet.

a.

Expert Solution
Check Mark

Answer to Problem 62DE

The regression equation is y^=64.7931+0.0703x.

The selling price of a home with an area of 2,200 square feet is $219,429.30.

The 95% confidence interval for all 2,200 square foot homes is (210,882.70,227,975.90)

The 95% prediction interval for the selling price of a home with 2,200 square feet is (131,837.60,307,021.0)

Explanation of Solution

Here, the selling price is the dependent variable and size of the home is the independent variable.

Step-by-step procedure to obtain the ‘regression equation’ using MegaStat software:

  • In an EXCEL sheet enter the data values of x and y.
  • Go to Add-Ins > MegaStat > Correlation/Regression > Regression Analysis.
  • Select input range as ‘Sheet1!$B$2:$B$106’ under Y/Dependent variable.
  • Select input range ‘Sheet1!$A$2:$A$106’ under X/Independent variables.
  • Select ‘Type in predictor values’.
  • Enter 2,200 as ‘predictor values’ and 95% as ‘confidence level’.
  • Click on OK.

Output obtained using MegaStat software is given below:

Statistical Techniques in Business and Economics, Chapter 13, Problem 62DE , additional homework tip  1

From the regression output, the regression equation is as follows.

y^=64.7931+0.0703x.

Where y is the selling price and x is the size of the home.

The selling price of a home with an area of 2,200 square feet is $219,429.30.

The 95% confidence interval for all 2,200 square foot homes is (210,882.70,227,975.90)

The 95% prediction interval for the selling price of a home with 2,200 square feet is (131,837.60,307,021.0)

b.

To determine

Find the regression equation.

Find the selling price of a home with 20 miles from the center of the city.

Construct a 95% confidence interval for homes with 20 miles from the center of the city.

Construct a 95% prediction interval a home with 20 miles from the center of the city.

b.

Expert Solution
Check Mark

Answer to Problem 62DE

The regression equation is y^=270.16703.3540(x).

The estimated selling price of a home with 20 miles from the center of the city is $203087.10.

The 95% confidence interval for homes with 20 miles from the center of the city is

 (190,267.40,215,906.80)

The 95% prediction interval for homes with 20 miles from the center of the city is

(114,117.8,292,056.5)

Explanation of Solution

Step-by-step procedure to obtain the ‘regression equation’ using MegaStat software:

  • In an EXCEL sheet enter the data values of x and y.
  • Go to Add-Ins > MegaStat > Correlation/Regression > Regression Analysis.
  • Select input range as ‘Sheet1!$C$2:$C$106’ under Y/Dependent variable.
  • Select input range ‘Sheet1!$B$2:$B$106’ under X/Independent variables.
  • Select ‘Type in predictor values’.
  • Enter 20 as ‘predictor values’ and 95% as ‘confidence level’.
  • Click on OK.

Output obtained using MegaStat software is given below:

Statistical Techniques in Business and Economics, Chapter 13, Problem 62DE , additional homework tip  2

From the above output, the regression equation is as follows:

y^=270.16703.3540(x)

Where y is the selling price of a home and x is the distance from the center of the city.

The estimated selling price of a home with 20 miles from the center of the city is $203087.10.

The 95% confidence interval for homes with 20 miles from the center of the city is (190,267.40,215,906.80)

The 95% prediction interval for a home with 20 miles from the center of the city is

(114,117.8,292,056.5)

c.

To determine

Check whether there is a negative correlation between “distance from the center of the city” and “selling price” using 0.05 significance level.

Check whether there is a positive correlation between the “area of the home” and “selling price” using 0.05 significance level.

Report the p-value of the test and summarize the results.

c.

Expert Solution
Check Mark

Answer to Problem 62DE

There is a negative association between “distance from the center of the city” and “selling price”.

There is a positive association between “selling price” and “area of the home”.

Explanation of Solution

Denote the population correlation as ρ.

The null and alternative hypotheses are stated below:

Null hypothesis:

H0:ρ0

That is, the correlation between “distance from the center of the city” and “selling price” is greater than or equal to zero.

Alternative hypothesis:

H1:ρ<0

That is, the correlation between “distance from the center of the city” and “selling price” is less than zero.

Here, the sample size is 105 and the correlation coefficient between “distance from the center of the city” and “selling price” is –0.347.

The test statistic is as follows:

t=0.34710521(0.347)2=0.347×1030.9378=3.75

Thus, the test statistic value is –3.75.

The degrees of freedom is as follows:

df=n2=1052=103

The level of significance is 0.05. Therefore, 1α=0.95.

Critical value:

Step-by-step software procedure to obtain the critical value using EXCEL software:

  • Open an EXCEL file.
  • In cell A1, enter the formula “=T.INV (0.95, 103)”.

Output obtained using EXCEL is given as follows:

Statistical Techniques in Business and Economics, Chapter 13, Problem 62DE , additional homework tip  3

Decision rule:

Reject the null hypothesis H0, if |t|-calculated>|t|-critical value. Otherwise, fail to reject H0.

Conclusion:

The value of test statistic is –3.75 and the critical value is 1.659.

Here, |t|-calculated(=3.75)>|t|-critical value(1.659).

By the rejection rule, reject the null hypothesis.

Thus, it can be concluded that there is a negative association between“distance from the center of the city” and “selling price”.

Calculation of p-value:

Using excel formula = T.DIST (–3.75,103,TRUE)

Statistical Techniques in Business and Economics, Chapter 13, Problem 62DE , additional homework tip  4

The p-value of the test is 0.000.

Check the correlation between independent variables “area of the home” and “selling price” is positive:

The hypotheses are given below:

Null hypothesis:

H0:ρ0

That is, the correlation between “size of the home” and “selling price” is less than or equal to zero.

Alternative hypothesis:

H1:ρ>0

That is, the correlation between “size of the home” and “selling price” is positive.

Here, the sample size is 105 and the correlation coefficient is 0.371.

The test statistic is as follows:

t=0.371105210.3712=0.371×1030.9286=3.76

Thus, the t-test statistic value is 3.76.

Conclusion:

The value of test statistic is 3.76 and the critical value is 1.659.

Here, t-calculated(=3.76)>t-critical value(=1.659).

By the rejection rule, reject the null hypothesis.

Thus, it can be concluded that there is a positive association between “size of the home” and “selling price”.

Calculation of p-value:

Using excel formula = T.DIST (3.76,103,TRUE)

Statistical Techniques in Business and Economics, Chapter 13, Problem 62DE , additional homework tip  5

The p-value of the test is 0.000141.

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
Problem 3. Pricing a multi-stock option the Margrabe formula The purpose of this problem is to price a swap option in a 2-stock model, similarly as what we did in the example in the lectures. We consider a two-dimensional Brownian motion given by W₁ = (W(¹), W(2)) on a probability space (Q, F,P). Two stock prices are modeled by the following equations: dX = dY₁ = X₁ (rdt+ rdt+0₁dW!) (²)), Y₁ (rdt+dW+0zdW!"), with Xo xo and Yo =yo. This corresponds to the multi-stock model studied in class, but with notation (X+, Y₁) instead of (S(1), S(2)). Given the model above, the measure P is already the risk-neutral measure (Both stocks have rate of return r). We write σ = 0₁+0%. We consider a swap option, which gives you the right, at time T, to exchange one share of X for one share of Y. That is, the option has payoff F=(Yr-XT). (a) We first assume that r = 0 (for questions (a)-(f)). Write an explicit expression for the process Xt. Reminder before proceeding to question (b): Girsanov's theorem…
Problem 1. Multi-stock model We consider a 2-stock model similar to the one studied in class. Namely, we consider = S(1) S(2) = S(¹) exp (σ1B(1) + (M1 - 0/1 ) S(²) exp (02B(2) + (H₂- M2 where (B(¹) ) +20 and (B(2) ) +≥o are two Brownian motions, with t≥0 Cov (B(¹), B(2)) = p min{t, s}. " The purpose of this problem is to prove that there indeed exists a 2-dimensional Brownian motion (W+)+20 (W(1), W(2))+20 such that = S(1) S(2) = = S(¹) exp (011W(¹) + (μ₁ - 01/1) t) 롱) S(²) exp (021W (1) + 022W(2) + (112 - 03/01/12) t). where σ11, 21, 22 are constants to be determined (as functions of σ1, σ2, p). Hint: The constants will follow the formulas developed in the lectures. (a) To show existence of (Ŵ+), first write the expression for both W. (¹) and W (2) functions of (B(1), B(²)). as (b) Using the formulas obtained in (a), show that the process (WA) is actually a 2- dimensional standard Brownian motion (i.e. show that each component is normal, with mean 0, variance t, and that their…
The scores of 8 students on the midterm exam and final exam were as follows.   Student Midterm Final Anderson 98 89 Bailey 88 74 Cruz 87 97 DeSana 85 79 Erickson 85 94 Francis 83 71 Gray 74 98 Harris 70 91   Find the value of the (Spearman's) rank correlation coefficient test statistic that would be used to test the claim of no correlation between midterm score and final exam score. Round your answer to 3 places after the decimal point, if necessary. Test statistic: rs =

Chapter 13 Solutions

Statistical Techniques in Business and Economics

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
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Text book image
Functions and Change: A Modeling Approach to Coll...
Algebra
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Cengage Learning
Text book image
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Text book image
College Algebra
Algebra
ISBN:9781337282291
Author:Ron Larson
Publisher:Cengage Learning
Mod-01 Lec-01 Discrete probability distributions (Part 1); Author: nptelhrd;https://www.youtube.com/watch?v=6x1pL9Yov1k;License: Standard YouTube License, CC-BY
Discrete Probability Distributions; Author: Learn Something;https://www.youtube.com/watch?v=m9U4UelWLFs;License: Standard YouTube License, CC-BY
Probability Distribution Functions (PMF, PDF, CDF); Author: zedstatistics;https://www.youtube.com/watch?v=YXLVjCKVP7U;License: Standard YouTube License, CC-BY
Discrete Distributions: Binomial, Poisson and Hypergeometric | Statistics for Data Science; Author: Dr. Bharatendra Rai;https://www.youtube.com/watch?v=lHhyy4JMigg;License: Standard Youtube License