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
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Residuals vs Fitted
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053
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Im(time - enz)
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Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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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