Provide a detailed explanation of how the Taylor-Russell Model applies as many statistics as possible from Statistics for the Behavioral Sciences Chapters 1-6 and 15.1-15.3. Systematically break down the model/method into its constituent statistics and explain in detail how the method/model and its constituent statistics map against the statistics learned in Statistics for the Behavioral Sciences Chapters 1-6 and 15.1-15.3.
Added by Ignacio G.
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Show that the covariance matrix ̑ = [1.0 .63 .45 .63 1.0 .35 .45 .35 1.0] for the p = 3 standardized random variables Z1, Z2, and Z3 can be generated by the m = 1 factor model Z1 = .9F1 + ̑1 Z2 = .7F1 + ̑2 Z3 = .5F1 + ̑3 where Var(F1) = 1, Cov(̑, F1) = 0, and ̑ = Cov(̑) = [.19 0 0 0 .51 0 0 0 .75] That is, write ̑ in the form ̑ = LL' + ̑. Use the information in Exercise 9.1. (a) Calculate communalities h², i = 1, 2, 3, and interpret these quantities. (b) Calculate Corr(Zi, F1) for i = 1, 2, 3. Which variable might carry the greatest weight in "naming" the common factor? Why? The eigenvalues and eigenvectors of the correlation matrix ̑ in Exercise 9.1 are ̑1 = 1.96, e'1 = [.625, .593, .507] ̑2 = .68, e'2 = [-.219, -.491, .843] ̑3 = .36, e'3 = [.749, -.638, -.177] (a) Assuming an m = 1 factor model, calculate the loading matrix L and matrix of specific variances ̑ using the principal component solution method. Compare the results with those in Exercise 9.1. (b) What proportion of the total population variance is explained by the first common factor?
Lien L.
****PLEASE SHOW WORK! INCLUDING CALCULATIONS FOR THE SHARED CONTRIBUTION!**** 2. To what extent does ATTIME2 (a student's attitude towards statistics at time 2), TEST2 (percentage score on the midterm test), ANXIETY2 score (student's anxiety towards statistics at time 2) uniquely contribute to the prediction of the grade a student is trying for at time 3 (TRYING3)? What is the shared contribution? R OUTPUT for QUESTION #2 > descriptive.table(vars = d(TEST2,ANXIETY2,ATTIME2,TRYING3),data= ASS2_12, + func.names =c("Mean","St. Deviation","Valid N","Minimum","Maximum")) Descriptive Statistics Mean St. Deviation Valid N Minimum Maximum TEST2 70.43 17.33 117 26.72 99.04 ANXIETY2 11.79 5.50 117 0.00 24.00 ATTIME2 3.31 8.59 117 -21.00 21.00 TRYING3 74.70 10.08 103 51.00 95.00 subA2Q2<-subset(ASS2_12,TEST2 >= 26.72 & ATTIME2 >= -21 & ANXIETY2 >= 0 & TRYING3 >= 51) > descriptive.table(vars = d(TEST2,ANXIETY2,ATTIME2,TRYING3),data= subA2Q2, + func.names =c("Mean","St. Deviation","Valid N","Minimum","Maximum")) Descriptive Statistics Mean St. Deviation Valid N Minimum Maximum TEST2 73.08 16.13 103 33.72 99.04 ANXIETY2 11.41 5.39 103 0.00 24.00 ATTIME2 3.82 8.10 103 -17.00 21.00 TRYING3 74.70 10.08 103 51.00 95.00 > corr.mat<-cor.matrix(variables=d(TRYING3), + with.variables=d(TEST2,ANXIETY2,ATTIME2), + data=subA2Q2, + test=cor.test, + method='pearson', + alternative="two.sided") > print(corr.mat,CI=FALSE,stat=FALSE) Correlation Pearson's product-moment correlation TRYING3 TEST2 cor 0.793 N 103 p-value* <0.001 ANXIETY2 cor -0.381 N 103 p-value* <0.001 ATTIME2 cor 0.261 N 103 p-value* 0.00786 Notes: H0: correlation = 0 *HA: two-sided > rm('corr.mat') > mod1 <- lm(formula=TRYING3 ~ TEST2 + ANXIETY2 + ATTIME2,data=subA2Q2,na.action=na.omit) > Anova(mod1,type='II') Anova Table (Type II tests) Response: TRYING3 Sum Sq Df F value Pr(>F) TEST2 5216.6 1 143.6575 < 2e-16 *** ANXIETY2 6.9 1 0.1896 0.66418 ATTIME2 146.2 1 4.0252 0.04755 * Residuals 3595.0 99 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > summarylm(mod1) Linear Model Summary Call: lm(formula = TRYING3 ~ TEST2 + ANXIETY2 + ATTIME2, data = subA2Q2, na.action = na.omit) Residuals: Min -18.6124 1Q -3.6376 Median -0.7338 3Q 3.8223 Max 12.1970 Coefficients: Estimate Std. Error t value Pr(>|t|) sig. (Intercept) 39.90517 3.89769 10.238 2e-16 *** TEST2 0.47642 0.03975 11.986 2e-16 *** ANXIETY2 -0.06019 0.13824 -0.435 0.6642 ATTIME2 0.17378 0.08662 2.006 0.0476 * Note: ***p<.001, **p<.01, *p<.05, .p<.10 Residual standard error: 6.026 on 99 degrees of freedom Multiple R-squared: 0.6532 Adjusted R-squared: 0.6427 F-statistic: 62.17 DF: ( 3 , 99 ) p-value: 2.2e-16 > tmp<-residuals(mod1) > subA2Q2[names(tmp),"Residuals"]<-tmp > lm.sumSquares(mod1) lm(formula = TRYING3 ~ TEST2 + ANXIETY2 + ATTIME2, data = subA2Q2, na.action = na.omit) Coefficients SSR df pEta-sqr dR-sqr (Intercept) 3806.3337 1 0.5143 NA TEST2 5216.6385 1 0.5920 0.5032 ANXIETY2 6.8855 1 0.0019 0.0007 ATTIME2 146.1656 1 0.0391 0.0141 Sum of squared errors (SSE): 3595.0 Sum of squared total (SST): 10367.7
Adi S.
In Problem $14.1$ data regarding voter turnout in five cities were presented. For the sake of convenience, the data for three of the variables are presented again here along with descriptive statistics and zero-order correlations. egin{tabular}{lccc} City & Turnout & Unemployment Rate & $%$ Negative Ads \ hline A & 55 & 5 & 60 \ B & 60 & 8 & 63 \ C & 65 & 9 & 55 \ D & 68 & 9 & 53 \ E & 70 & 10 & 48 \ Mean $=$ & $63.6$ & $8.2$ & $55.8$ \ s $=$ & $5.5$ & $1.7$ & $5.3$ end{tabular} egin{tabular}{lcc} & Unemployment Rate & $%$ Negative Ads \ hline Turnout & $0.95$ & $-0.87$ \ Unemployment Rate & & $-0.70$ end{tabular} a. Compute the partial correlation coefficient for the relationship between turnout $(Y)$ and unemployment $(X)$ while controlling for the effect of negative advertising $(Z)$. What effect does this control variable have on the bivariate relationship? Is the relationship between turnout and unemployment direct? (HINT: Use Formula 15.1 and see Section 15.2.) b. Compute the partial correlation coefficient for the relationship between turnout $(Y)$ and negative advertising $(X)$ while controlling for the effect of unemployment $(Z)$. What effect does this have on the bivariate relationship? Is the relationship between turnout and negative advertising direct? (HINT: Use Formula 15.1 and see Section 15.2. You will need this partial correlation to compute the multiple correlation coefficient.) c. Find the unstandardized multiple regression equation with unemployment $left(X_{1} ight)$ and negative ads $left(X_{2} ight)$ as the independent variables. What turnout would be expected in a city in which the unemployment rate was $10 %$ and $75 %$ of the campaign ads were negative? (HINT: Use Formulas $15.4$ and $15.5$ to compute the partial slopes and then use Formula $15.6$ to find a, the $Y$ intercept. The regression line is stated in Formula 15.3. Substitute 10 for $X_{1}$ and 75 for $X_{2}$ to compute predicted $Y$.) d. Compute beta-weights for each independent variable. Which has the stronger impact on turnout? (HINT: Use Formulas $15.7$ and $15.8$ to calculate the beta-weights.) e. Compute the coefficient of multiple determination $left(R^{2} ight)$. How much of the variance in voter turnout is explained by the two independent variables combined? (HINT: Use Formula 15.11. You calculated the partial correlation coefficient in part b of this problem.) f. Write a paragraph summarizing your conclusions about the relationships among these three variables.
Dominador T.
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