
Concept explainers
(a)
To describe: The direction, form and strength of the relationship.
(a)

Answer to Problem 5CRE
There is a strong, negative, linear association between the variables Temperature and first blossom days in April.
Explanation of Solution
Given information:
Make a well-labelled
Formula used:
EXCEL software
Calculation:
Draw a scatter plot for the given data:
Given data represents the values of Temperature (0C) and the first blossom days in April.
Here, the dependent variable is y = first blossom days in April and the independent variable x = Temperature (0C).
EXCEL software can be used to draw a scatter plot for the given data.
Software Procedure:
The step-by-step procedure to draw a scatter plot using EXCEL software is given below:
- Open an EXCEL file.
- Enter the data values of Temperature (0C) in column A and name it as Temperature .
- Enter the data values of Days in column B and name it as Days .
- Select Data and go to Insert tab.
- Select Scatter or Bubble chart > Scatter plot.
- Click OK.
- Select any point on the Scatter plot and left click on it.
The output of scatter plot is shown below:
The relationship between two variables Temperature and first blossom days in April can be described on the basis of scatter plot.
Associated variables:
Two variables are associated or related if the value of one variable gives you information about the value of the other variable.
From the above obtained scatter plot, it is seen that the two variables Temperature and first blossom days in April are associated variables.
Direction of association:
If the increase in the values of one variable increases the values of another variable, then the direction is positive. If the increase in the values of one variable decreases the values of another variable, then the direction is negative.
Here, the values of days decrease with the increase in the values of temperature and the values of day’s increases with the decrease in the values of temperature.
Hence, the direction of the association is negative.
Form of the association between variable:
The form of the association describes whether the data points follow a linear pattern or some other complicated curves. For data if it appears that a line would do a reasonable job of summarizing the overall pattern in the data. Then, the association between two variables is linear.
From the scatter plot, it is observed that the pattern of the relationship between the variables Temperature and first blossom days in April is slightly linear.
Hence, there is a linear association between the variables Temperature and first blossom days in April.
Strength of the association:
The association is said to be strong if all the points are close to the straight line. It is said to be weak if all points are far away from the straight line and it is said to be moderate if the data points are moderately close to straight line.
From the scatter plot, it is observed that the variables will have perfect
Hence, the association between the variables is strong.
Hence, there is a strong, negative, linear association between the variables Temperature and first blossom days in April.
Conclusion:
There is a strong, negative, linear association between the variables Temperature and first blossom days in April.
(b)
To find: The equation of least square regression line.
(b)

Answer to Problem 5CRE
The intercept of the regression equation is 33.1203.
Explanation of Solution
Given information:
Use technology the equation of the least squares regression line is to be interpreted.
Formula used:
A simple linear regression model is given as y^ = b0 + bx + e
Calculation:
Regression equation to predict repair time given the repair type:
Simple linear regression model:
A simple linear regression model is given as y^ = b0 + bx + e
Where y^ is the predicted value of response variable, and x be the predictor variable. The quantity b is the estimated slope corresponding to x and b0 is the estimated intercept of the line, from the sample data.
Simple regression model can be build using EXCEL software:
Software Procedure:
The step-by-step procedure to obtain simple regression model using EXCEL software is given below:
- Open an EXCEL file.
- Enter the data values of Temperature (0C) in column A and name it as Temperature.
- Enter the data values of Days in column B and name it as Days.
- Click on Data > Data Analysis > Regression.
- In Input Y
Range enter $B$1:$B$25. - In Input X Range enter $A$1:$A$25.
- Click on
- Click OK.
The generated output is given below:
From the generate output, the intercept b0 = 33.1203 and slope coefficient b = -4.6855.
Thus, the simple linear regression equation to predict the first blossom days in April (y) given the temperature is Days = 33.1203 − 4.6855*Temperature.
Interpretation of slope:
The slope of the regression equation is -4.6855.
For any 1 unit increase in temperature, the number of days decreases by -4.6855 units.
Interpretation of intercept:
The intercept of the regression equation is 33.1203.
The number of days will be 33.1203 even when the temperature is 0.
Conclusion:
The intercept of the regression equation is 33.1203.
(c)
To analyse: The occurrence of the first cherry blossom.
(c)

Answer to Problem 5CRE
The predicted first cherry blossom day when the temperature is 3.50C is 17.
Explanation of Solution
Given information:
The average March temperate; this year was 3.5°C.
Formula used:
Calculation:
Predict the first cherry blossom day when the temperature is 3.50C:
The predicted first cherry blossom day when the temperature is 3.50C is obtained as 17 from the calculation given below:
Thus, the predicted first cherry blossom day when the temperature is 3.50C is 17.
Conclusion:
The predicted first cherry blossom day when the temperature is 3.50C is 17.
(d)
Tofind: The residual value for the year.
(d)

Answer to Problem 5CRE
The residual for the year when the average March temperature was 4. 50C is 2.
Explanation of Solution
Given information:
The average March temperate; this year was 4.5°C.
Formula used:
Calculation:
Predict the first cherry blossom day when the temperature is 4.50C:
The predicted first cherry blossom day when the temperature is obtained as given below:
Thus, the predicted first cherry blossom day when the temperature is 4.50C is 12.
Residual for the year when the average March temperature was 4. 50C :
The residual for the year when the average March temperature was 4. 50C is obtained as 2 from the calculation given below:
Thus, the residual for the year when the average March temperature was 4. 50C is 2.
Conclusion:
The residual for the year when the average March temperature was 4. 50C is 2.
(e)
To construct: The residual plot.
(e)

Answer to Problem 5CRE
From the obtained residual plot it is observed that, the data points are deviating from the straight line.
Explanation of Solution
Given information:
The residual plot needs to be constructed.
Formula used:
EXCEL software
Calculation:
Residual plot can be build using EXCEL software:
Software Procedure:
The step-by-step procedure to obtain simple regression model using EXCEL software is given below:
- Open an EXCEL file.
- Enter the data values of Temperature (0C) in column A and name it as Temperature .
- Enter the data values of Days in column B and name it as Days .
- Click on Data > Data Analysis > Regression.
- In Input Y Range enter $B$1:$B$25.
- In Input X Range enter $A$1:$A$25.
- Click on
- Click on Residual plots.
- Click OK.
The generated output is given below:
Observation: From the obtained residual plot it is observed that, the data points are deviating from the straight line. Therefore, the regression model does not seem of doing a very good job in predicting the days in April until the first blossom.
Conclusion:
From the obtained residual plot it is observed that, the data points are deviating from the straight line.
(f)
To find: The interpret value of r2
(f)

Answer to Problem 5CRE
The standard error in the regression model is 3.0216.
Explanation of Solution
Given information:
The values of r2 and s need to analyse.
Formula used:
The percentage of variation in the observed values of dependent variable
Calculation:
From the obtained regression output, the value of R-square is 0.7243.
R-squared:
The percentage of variation in the observed values of dependent variable that is explained by the regression is 72.43%, which indicates that 72.43% of the variability in the variable “First blossom days in April(Y)” is explained by the variability in independent variable “Temperature (X)” using the linear regression model.
From the obtained regression output, the value of standard error is 3.0216.
The standard error in the regression model is 3.0216.
Conclusion:
The standard error in the regression model is 3.0216.
Chapter 3 Solutions
The Practice of Statistics for AP - 4th Edition
Additional Math Textbook Solutions
Elementary Statistics: Picturing the World (7th Edition)
Calculus: Early Transcendentals (2nd Edition)
Algebra and Trigonometry (6th Edition)
Elementary Statistics (13th Edition)
A First Course in Probability (10th Edition)
University Calculus: Early Transcendentals (4th Edition)
- The managing director of a consulting group has the accompanying monthly data on total overhead costs and professional labor hours to bill to clients. Complete parts a through c Overhead Costs Billable Hours345000 3000385000 4000410000 5000462000 6000530000 7000545000 8000arrow_forwardUsing the accompanying Home Market Value data and associated regression line, Market ValueMarket Valueequals=$28,416plus+$37.066×Square Feet, compute the errors associated with each observation using the formula e Subscript ieiequals=Upper Y Subscript iYiminus−ModifyingAbove Upper Y with caret Subscript iYi and construct a frequency distribution and histogram. Square Feet Market Value1813 911001916 1043001842 934001814 909001836 1020002030 1085001731 877001852 960001793 893001665 884001852 1009001619 967001690 876002370 1139002373 1131001666 875002122 1161001619 946001729 863001667 871001522 833001484 798001589 814001600 871001484 825001483 787001522 877001703 942001485 820001468 881001519 882001518 885001483 765001522 844001668 909001587 810001782 912001483 812001519 1007001522 872001684 966001581 86200arrow_forwarda. Find the value of A.b. Find pX(x) and py(y).c. Find pX|y(x|y) and py|X(y|x)d. Are x and y independent? Why or why not?arrow_forward
- The PDF of an amplitude X of a Gaussian signal x(t) is given by:arrow_forwardThe PDF of a random variable X is given by the equation in the picture.arrow_forwardFor a binary asymmetric channel with Py|X(0|1) = 0.1 and Py|X(1|0) = 0.2; PX(0) = 0.4 isthe probability of a bit of “0” being transmitted. X is the transmitted digit, and Y is the received digit.a. Find the values of Py(0) and Py(1).b. What is the probability that only 0s will be received for a sequence of 10 digits transmitted?c. What is the probability that 8 1s and 2 0s will be received for the same sequence of 10 digits?d. What is the probability that at least 5 0s will be received for the same sequence of 10 digits?arrow_forward
- V2 360 Step down + I₁ = I2 10KVA 120V 10KVA 1₂ = 360-120 or 2nd Ratio's V₂ m 120 Ratio= 360 √2 H I2 I, + I2 120arrow_forwardQ2. [20 points] An amplitude X of a Gaussian signal x(t) has a mean value of 2 and an RMS value of √(10), i.e. square root of 10. Determine the PDF of x(t).arrow_forwardIn a network with 12 links, one of the links has failed. The failed link is randomlylocated. An electrical engineer tests the links one by one until the failed link is found.a. What is the probability that the engineer will find the failed link in the first test?b. What is the probability that the engineer will find the failed link in five tests?Note: You should assume that for Part b, the five tests are done consecutively.arrow_forward
- 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…arrow_forwardProblem 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…arrow_forwardThe 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 =arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman





