The number of crimes reported ( in millions ) and the number of arrests reported
ID: 3311521 • Letter: T
Question
The number of crimes reported ( in millions ) and the number of arrests reported ( in millions ) by the U.S department of justice for 14 years ( adapted from the national crime victimization served and uniform crime reports )- ontent × 0321jerry-Bing e/myb7 cmd display&ctoold-MyGradesonMyBb; MyGradesTorol Crystal Williams ction Lit tudent Self Service Student ID Card Info eForms Community Virtual Student Lounge & Question Completion Status: 12 points Save AnawerT The number of crimes reported (in millions) and the muimber of arrests reported (in millions) by the U.S. Departmeat of Justice for 14 years (Adapted from the National Crime Victimization Survey and Uniform Crime Reports) Crimes,X1 Lo | 1.95 | 1.44 1A0 . 1321.23 1.227 Armestsy 978 3 072 0.68 064 0 Crimes1.23 122 118 116 119 1.21 1.20 Arrests, 0.63 0621 aso 0.59 0.50.61 058 a) Find the regression equation that models the number of crimes as a function of the number of arrests is y = x + b) The correlation coefficient is R c) The linear correlation coefficient is R- d) SST- e) SSR, ROUND ALL ANSWERS TO 4 DECIMAL PLACES X.Xxxx 316 PM 11/28/2017
Explanation / Answer
> x=c(1.6,1.55,1.44,1.4,1.32,1.23,1.22,1.23,1.22,1.18,1.16,1.19,1.21,1.2)
> y=c(0.78,0.80,0.73,0.72,0.68,0.64,0.63,0.63,0.62,0.60,0.59,0.60,0.61,0.58)
> a=lm(y~x)
> summary(a)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-0.030455 -0.004227 0.000640 0.010345 0.017494
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.02057 0.03702 0.556 0.589
x 0.49157 0.02840 17.311 7.47e-10 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.01473 on 12 degrees of freedom
Multiple R-squared: 0.9615, Adjusted R-squared: 0.9583
F-statistic: 299.7 on 1 and 12 DF, p-value: 7.473e-10
(a)
y = 0.0206 + 0.4916 x
(b)
The correlation coefficient R2 = 0.9615
(c)
The linear correlation R = 0.9806
(d)
> sum((y-mean(y))^2)
[1] 0.06763571
SST = 0.0676
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