ACHIEVE F/PRACT OF STAT IN LIFE-ACCESS
ACHIEVE F/PRACT OF STAT IN LIFE-ACCESS
4th Edition
ISBN: 9781319509286
Author: BALDI
Publisher: MAC HIGHER
Question
Book Icon
Chapter 23, Problem 23.38E

(a)

To determine

To explain:

The scatterplot of white matter change and the explanatory of the given data.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The table of the exposure and change.

Concept used:

Formula

The binomial distribution

  p[X=x]=(nx)pxqnx

Calculation:

Scatter plot is given as the table

The weak is negative relationship exists.

Because the value of r=0.30

Where r is negative

Let us perform the regression

Regressing sub concussion exposure on white matter change with below mentioned steps in excel

  X And Y are two variables.

SUMMARY OUTPUT

Regression Statistics

Multiple R =0.368868063

R Square =0.136063648

Adjusted R Square =0.09286683

Standard Error = 3.467466463

Observations =22

Draw the table

    DegreeSSMSFSignificant
    Regression 137.8717083337.87173.1498540.091157025
    Residual 20240.466473512.02332
    Total 21278.3381818

Draw the second table

    CoefficientStandard Errort StatP-valueLower 95%Upper 95%Lower 95.0%Upper 95.0%
    Intercept2.2720152571.2015559641.8908940.073215-0.2343865574.778417071-0.234386557 4.778417071
    X-6.912959936 3.895102237-1.774780.091157-15.03800081.212080931-15.03800081.212080931

(b)

To determine

To find:

Theregression standard error of the given data.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The table of the exposure and change.

Concept used:

Formula

The binomial distribution

  p[X=x]=(nx)pxqnx

Calculation:

Scatter plot is given as the table

The weak is negative relationship exists.

Because the value of r=0.30

Where r is negative

Let us perform the regression

Regressing sub concussion exposure on white matter change with below mentioned steps in excel

  X And Y are two variables.

SUMMARY OUTPUT

Regression Statistics

Multiple R =0.368868063

R Square =0.136063648

Adjusted R Square =0.09286683

Standard Error = 3.467466463

Observations =22

Draw the table

    DegreeSSMSFSignificant
    Regression 137.8717083337.87173.1498540.091157025
    Residual 20240.466473512.02332
    Total 21278.3381818

Draw the second table

    CoefficientStandard Errort StatP-valueLower 95%Upper 95%Lower 95.0%Upper 95.0%
    Intercept2.2720152571.2015559641.8908940.073215-0.2343865574.778417071-0.234386557 4.778417071
    X-6.912959936 3.895102237-1.774780.091157-15.03800081.212080931-15.03800081.212080931

Here X= exposure

  Y= Change as variable

So least squares regression line

Change ( Y ) = 6.913 Exposure (X) + 2.272

Regression standard error = 3.467

The low p-value of 0.0912 of the Exposure coefficient, indicates that a confidence level of 90% (where p-value of 0.1 ) that the Exposure is a significant predictor of white matter Change.

Hence, the null hypothesis of no relationship between exposure and white matter change can be rejected in favor of the alternative hypothesis at a confidence level of 10%.

(c)

To determine

To explain:

Theequation for the least-squares regression line without data point and test null hypothesis of the given data.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The table of the exposure and change.

Concept used:

Formula

The binomial distribution

  p[X=x]=(nx)pxqnx

Calculation:

Scatter plot is given as the table

The weak is negative relationship exists.

Because the value of r=0.30

Where r is negative

Let us perform the regression

Regressing sub concussion exposure on white matter change with below mentioned steps in excel

  X And Y are two variables.

SUMMARY OUTPUT

Regression Statistics

Multiple R =0.368868063

R Square =0.136063648

Adjusted R Square =0.09286683

Standard Error = 3.467466463

Observations =22

Draw the table

    DegreeSSMSFSignificant
    Regression 137.8717083337.87173.1498540.091157025
    Residual 20240.466473512.02332
    Total 21278.3381818

Draw the second table

    CoefficientStandard Errort StatP-valueLower 95%Upper 95%Lower 95.0%Upper 95.0%
    Intercept2.2720152571.2015559641.8908940.073215-0.2343865574.778417071-0.234386557 4.778417071
    X-6.912959936 3.895102237-1.774780.091157-15.03800081.212080931-15.03800081.212080931

Here X= exposure

  Y= Change as variable

So least squares regression line

Change ( Y ) = 6.913 Exposure (X) + 2.272

Regression standard error = 3.467

The low p-value of 0.0912 of the Exposure coefficient, indicates that a confidence level of 90% (where p-value of 0.1 ) that the Exposure is a significant predictor of white matter Change.

Hence, the null hypothesis of no relationship between exposure and white matter change can be rejected in favor of the alternative hypothesis at a confidence level of 10%.

SUMMARY OUTPUT

Regression Statistics

Multiple R =0.104318418

R Square =0.010882332

Adjusted R Square =-0.041176492

Standard Error = 3.471079833

Observations =21

Draw the table

    DegreeSSMSFSignificant
    Regression 12.5185862442.5185860.2090390.652706126
    Residual 19228.91950912.0484
    Total 20231.4380952

Draw the second table

    CoefficientStandard Errort StatP-valueLower 95%Upper 95%Lower 95.0%Upper 95.0%
    Intercept1.475861.451931.016480.32216-1.563064.51480-1.563064.51480
    X-2.66665.83246-0.457210.65270-14.87419.5408-14.87419.54083

Standard error is same

But the p value is 0.65

Exposure is not good predictor of Change in white matter.

Also, the Exposure coefficient has come down from 6.91 to 2.67 ,

(d)

To determine

To explain:

Theimpact of the usual observation on the regression analysis of the given data.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given:

The table of the exposure and change.

Concept used:

Formula

The binomial distribution

  p[X=x]=(nx)pxqnx

Calculation:

Scatter plot is given as the table

The weak is negative relationship exists.

Because the value of r=0.30

Where r is negative

Let us perform the regression

Regressing sub concussion exposure on white matter change with below mentioned steps in excel

  X And Y are two variables.

SUMMARY OUTPUT

Regression Statistics

Multiple R =0.368868063

R Square =0.136063648

Adjusted R Square =0.09286683

Standard Error = 3.467466463

Observations =22

Draw the table

    DegreeSSMSFSignificant
    Regression 137.8717083337.87173.1498540.091157025
    Residual 20240.466473512.02332
    Total 21278.3381818

Draw the second table

    CoefficientStandard Errort StatP-valueLower 95%Upper 95%Lower 95.0%Upper 95.0%
    Intercept2.2720152571.2015559641.8908940.073215-0.2343865574.778417071-0.234386557 4.778417071
    X-6.912959936 3.895102237-1.774780.091157-15.03800081.212080931-15.03800081.212080931

Here X= exposure

  Y= Change as variable

So least squares regression line

Change ( Y ) = 6.913 Exposure (X) + 2.272

Regression standard error = 3.467

The low p-value of 0.0912 of the Exposure coefficient, indicates that a confidence level of 90% (where p-value of 0.1 ) that the Exposure is a significant predictor of white matter Change.

Hence, the null hypothesis of no relationship between exposure and white matter change can be rejected in favor of the alternative hypothesis at a confidence level of 10%.

SUMMARY OUTPUT

Regression Statistics

Multiple R =0.104318418

R Square =0.010882332

Adjusted R Square =-0.041176492

Standard Error = 3.471079833

Observations =21

Draw the table

    DegreeSSMSFSignificant
    Regression 12.5185862442.5185860.2090390.652706126
    Residual 19228.91950912.0484
    Total 20231.4380952

Draw the second table

    CoefficientStandard Errort StatP-valueLower 95%Upper 95%Lower 95.0%Upper 95.0%
    Intercept1.475861.451931.016480.32216-1.563064.51480-1.563064.51480
    X-2.66665.83246-0.457210.65270-14.87419.5408-14.87419.54083

Standard error is same

But the p value is 0.65

Exposure is not good predictor of Change in white matter.

Also, the Exposure coefficient has come down from 6.91 to 2.67 ,

The Change in white matter will not actually change as fast

So,eliminate of the outlier data points.

The Outlier data points tend to increase the correlation between the two variables.

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
The class will include a data exercise where students will be introduced to publicly available data sources. Students will gain experience in manipulating data from the web and applying it to understanding the economic and demographic conditions of regions in the U.S. Regions and topics of focus will be determined (by the student with instructor approval) prior to April. What data exercise can I do to fulfill this requirement? Please explain.
Consider the ceocomp dataset of compensation information for the CEO’s of 100 U.S. companies. We wish to fit aregression model to assess the relationship between CEO compensation in thousands of dollars (includes salary andbonus, but not stock gains) and the following variates:AGE: The CEOs age, in yearsEDUCATN: The CEO’s education level (1 = no college degree; 2 = college/undergrad. degree; 3 = grad. degree)BACKGRD: Background type(1= banking/financial; 2 = sales/marketing; 3 = technical; 4 = legal; 5 = other)TENURE: Number of years employed by the firmEXPER: Number of years as the firm CEOSALES: Sales revenues, in millions of dollarsVAL: Market value of the CEO's stock, in natural logarithm unitsPCNTOWN: Percentage of firm's market value owned by the CEOPROF: Profits of the firm, before taxes, in millions of dollars1) Create a scatterplot matrix for this dataset. Briefly comment on the observed relationships between compensationand the other variates.Note that companies with negative…
6 (Model Selection, Estimation and Prediction of GARCH) Consider the daily returns rt of General Electric Company stock (ticker: "GE") from "2021-01-01" to "2024-03-31", comprising a total of 813 daily returns. Using the "fGarch" package of R, outputs of fitting three GARCH models to the returns are given at the end of this question. Model 1 ARCH (1) with standard normal innovations; Model 2 Model 3 GARCH (1, 1) with Student-t innovations; GARCH (2, 2) with Student-t innovations; Based on the outputs, answer the following questions. (a) What can be inferred from the Standardized Residual Tests conducted on Model 1? (b) Which model do you recommend for prediction between Model 2 and Model 3? Why? (c) Write down the fitted model for the model that you recommended in Part (b). (d) Using the model recommended in Part (b), predict the conditional volatility in the next trading day, specifically trading day 814.
Knowledge Booster
Background pattern image
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