Compare the simple linear model in Question 2 and the exponential model in Question 3, do you observe any improvements after conducting the exponential model relative to the linear model? Support your conclusion with details (such as R-Square, homogeneity of variance and normality test) this is Q2 and my answer to it: Fit the simple linear regression model TIME = β0 + β1ENZ + ε. Write down the estimated regression function and examine the residual plot and normality test. Describe what you observed and make brief comments. Hint: you need to check the ANOVA table (that is the F-Statistic and its p-value on the last line of the summary(yourmodel) output)), parameter estimates tables, R-Square, residual plot and normality test. Coefficients: (Intercept) enz -108.716 3.967 TIME= -108.716 + 3.967 (ENZ) + ε By checking normality using the Shapiro-Wilk test, the p-value is 0.0000 less than 0.05. The null is rejected, and the data is not normally distributed. ----------------------------------------------- Test Statistic pvalue ----------------------------------------------- Shapiro-Wilk 0.866 0.0000 Kolmogorov-Smirnov 0.14 0.2190 Cramer-von Mises 5.1667 0.0000 Anderson-Darling 1.9562 0.0000 ----------------------------------------------- Residuals: Min 1Q Median 3Q Max -243.33 -64.74 -25.19 49.45 486.50 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -108.7161 61.7191 -1.761 0.084 . enz 3.9668 0.7721 5.137 4.25e-06 *** Residual standard error: 119.5 on 52 degrees of freedom Multiple R-squared: 0.3367, Adjusted R-squared: 0.3239 F-statistic: 26.39 on 1 and 52 DF, p-value: 4.25e-06 This is Q3 and my answer to it: Fit the exponential model logTIME = β0 + β1ENZ + ε. Write down the estimated regression function. Does the model fit well? Why? Hint: you need to check the ANOVA table (that is the F-Statistic and its p-value on the last line of the summary(yourmodel) output)), parameter estimates tables, RSquare, residual plot and normality test. lm(formula = log(time) ~ enz, data = patients) logTIME = β0 + β1ENZ + ε. = 3.558633 ~ 0.019727ENZ + ε Residuals: Min 1Q Median 3Q Max -1.19415 -0.29725 -0.02198 0.34125 1.01853 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 3.558633 0.245526 14.494 < 2e-16 *** enz 0.019727 0.003072 6.423 4.12e-08 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.4753 on 52 degrees of freedom Multiple R-squared: 0.4424, Adjusted R-squared: 0.4316 F-statistic: 41.25 on 1 and 52 DF, p-value: 4.118e-08 The model fit well, as the p-value is 4.118e-08, Normality test: W = 0.98822, p-value = 0.8716, more than 0.05. Fail to reject the null and conclude the data are normally distributed.

Compare the simple linear model in Question 2 and the exponential model in Question 3, do you observe any improvements after conducting the exponential model relative to the linear model? Support your conclusion with details (such as R-Square, homogeneity of variance and normality test) this is Q2 and my answer to it: Fit the simple linear regression model TIME = β0 + β1ENZ + ε. Write down the estimated regression function and examine the residual plot and normality test. Describe what you observed and make brief comments. Hint: you need to check the ANOVA table (that is the F-Statistic and its p-value on the last line of the summary(yourmodel) output)), parameter estimates tables, R-Square, residual plot and normality test. Coefficients: (Intercept) enz -108.716 3.967 TIME= -108.716 + 3.967 (ENZ) + ε By checking normality using the Shapiro-Wilk test, the p-value is 0.0000 less than 0.05. The null is rejected, and the data is not normally distributed. ----------------------------------------------- Test Statistic pvalue ----------------------------------------------- Shapiro-Wilk 0.866 0.0000 Kolmogorov-Smirnov 0.14 0.2190 Cramer-von Mises 5.1667 0.0000 Anderson-Darling 1.9562 0.0000 ----------------------------------------------- Residuals: Min 1Q Median 3Q Max -243.33 -64.74 -25.19 49.45 486.50 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -108.7161 61.7191 -1.761 0.084 . enz 3.9668 0.7721 5.137 4.25e-06 * Residual standard error: 119.5 on 52 degrees of freedom Multiple R-squared: 0.3367, Adjusted R-squared: 0.3239 F-statistic: 26.39 on 1 and 52 DF, p-value: 4.25e-06 This is Q3 and my answer to it: Fit the exponential model logTIME = β0 + β1ENZ + ε. Write down the estimated regression function. Does the model fit well? Why? Hint: you need to check the ANOVA table (that is the F-Statistic and its p-value on the last line of the summary(yourmodel) output)), parameter estimates tables, RSquare, residual plot and normality test. lm(formula = log(time) ~ enz, data = patients) logTIME = β0 + β1ENZ + ε. = 3.558633 ~ 0.019727ENZ + ε Residuals: Min 1Q Median 3Q Max -1.19415 -0.29725 -0.02198 0.34125 1.01853 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 3.558633 0.245526 14.494 < 2e-16 * enz 0.019727 0.003072 6.423 4.12e-08 * --- Signif. codes: 0 ‘*’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.4753 on 52 degrees of freedom Multiple R-squared: 0.4424, Adjusted R-squared: 0.4316 F-statistic: 41.25 on 1 and 52 DF, p-value: 4.118e-08 The model fit well, as the p-value is 4.118e-08, Normality test: W = 0.98822, p-value = 0.8716, more than 0.05. Fail to reject the null and conclude the data are normally distributed.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Compare the simple linear model in Question 2 and the exponential model in Question 3, do you observe any improvements after conducting the exponential model relative to the linear model? Support your conclusion with details (such as R-Square, homogeneity of variance and normality test)

this is Q2 and my answer to it:

Fit the simple linear regression model TIME = β0 + β1ENZ + ε. Write down the estimated regression function and examine the residual plot and normality test. Describe what you observed and make brief comments. Hint: you need to check the ANOVA table (that is the F-Statistic and its p-value on the last line of the summary(yourmodel) output)), parameter estimates tables, R-Square, residual plot and normality test.

Coefficients:

(Intercept) enz

-108.716 3.967

TIME= -108.716 + 3.967 (ENZ) + ε

By checking normality using the Shapiro-Wilk test, the p-value is 0.0000 less than 0.05. The null is rejected, and the data is not normally distributed.

-----------------------------------------------

Test Statistic pvalue

-----------------------------------------------

Shapiro-Wilk 0.866 0.0000

Kolmogorov-Smirnov 0.14 0.2190

Cramer-von Mises 5.1667 0.0000

Anderson-Darling 1.9562 0.0000

-----------------------------------------------

Residuals:

Min 1Q Median 3Q Max

-243.33 -64.74 -25.19 49.45 486.50

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) -108.7161 61.7191 -1.761 0.084 .

enz 3.9668 0.7721 5.137 4.25e-06 ***

Residual standard error: 119.5 on 52 degrees of freedom

Multiple R-squared: 0.3367, Adjusted R-squared: 0.3239

F-statistic: 26.39 on 1 and 52 DF, p-value: 4.25e-06

This is Q3 and my answer to it:

Fit the exponential model logTIME = β0 + β1ENZ + ε. Write down the estimated regression function. Does the model fit well? Why? Hint: you need to check the ANOVA table (that is the F-Statistic and its p-value on the last line of the summary(yourmodel) output)), parameter estimates tables, RSquare, residual plot and normality test.

lm(formula = log(time) ~ enz, data = patients)

logTIME = β0 + β1ENZ + ε.

= 3.558633 ~ 0.019727ENZ + ε

Residuals:

Min 1Q Median 3Q Max

-1.19415 -0.29725 -0.02198 0.34125 1.01853

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 3.558633 0.245526 14.494 < 2e-16 ***

enz 0.019727 0.003072 6.423 4.12e-08 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.4753 on 52 degrees of freedom

Multiple R-squared: 0.4424, Adjusted R-squared: 0.4316

F-statistic: 41.25 on 1 and 52 DF, p-value: 4.118e-08

The model fit well, as the p-value is 4.118e-08,

Normality test:

W = 0.98822, p-value = 0.8716, more than 0.05. Fail to reject the null and conclude the data are normally distributed.

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enz
log(time)
3.5 4.0 4.5 5.0 5.5 6.0 6.5
80

Residuals vs Fitted
540
053
ㅇㅇ
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Fitted values
Im(time - enz)
Residuals
-200
007

Expert Solution

Step 1

The coefficient of determination is defined as the proportion of total variability in the dependent variable, which is explained by the model. It is denoted by $R^{2}$ .

The value of $R^{2} = 0$ indicates that the dependent variable cannot be predicted using the independent variable. The value of $R^{2} = 1$ indicates the perfect fit. Hence, the closer the value of $R^{2}$ is to $1$ , the better the fit.

For the linear model, the multiple $R^{2}$ -squared and adjusted $R^{2}$ -squared both lies close to $33 %$ indicated not so the good fit of the model. Whereas the multiple $R^{2}$ -squared and adjusted $R^{2}$ -squared for exponential model lie close to $45 %$ indicating a comparatively better fit of the model.

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