Exam 9: Bivariate Correlation and Regression

Correct Answer:

Verified

With every unit increase in IV (x), the DV (y) increases by 6 units.

Correct Answer:

Verified

a. H₀: There is no statistically significant relationship between visual consumption of crime shows (average per day) and levels of fear of crime (scale 0-30; treated as a continuous variable).
H₁: There is a statistically significant relationship between visual consumption of crime shows (average per day) and levels of fear of crime (scale 0-30; treated as a continuous variable).
b.
$$\begin{array} { | r | r | r | r | r | r | } \hline \text { Case } & x & y & x ^ { 2 } & y ^ { 2 } & x y \\\hline 1 & 0 & 12 & 0 & 144 & 0 \\\hline 2 & 3 & 11 & 9 & 121 & 33 \\\hline 3 & 2 & 11 & 4 & 121 & 22 \\\hline 4 & 9 & 18 & 81 & 324 & 162 \\\hline 5 & 4 & 19 & 16 & 361 & 76 \\\hline 6 & 5 & 20 & 25 & 400 & 100 \\\hline 7 & 7 & 11 & 49 & 121 & 77 \\\hline 8 & 9 & 16 & 81 & 256 & 144 \\\hline 9 & 8 & 18 & 64 & 324 & 144 \\\hline 10 & 9 & 25 & 81 & 625 & 225 \\\hline 11 & 11 & 24 & 121 & 576 & 264 \\\hline 12 & 1 & 9 & 1 & 81 & 9 \\\hline 13 & 3 & 11 & 9 & 121 & 33 \\\hline 14 & 2 & 11 & 4 & 121 & 22 \\\hline \text { Totals } & 73 & 216 & 545 & 3,696 & 1,311 \\\hline\end{array}$$
$$\text { i. } b = \frac { S S _ { y x } } { S S _ { x } } = \frac { N \Sigma X Y - ( \Sigma X ) ( \Sigma Y ) } { N \Sigma X ^ { 2 } - ( \Sigma X ) ^ { 2 } } = \frac { 14 \times 1311 - 73 \times 216 } { 14 \times 545 - 5,329 } = \frac { 2,586 } { 2,301 } = 1.1239$$
With every unit increase in visual consumption of TV crime shows (unit = 1 hour), the level of fear of crime increases by 1.1239 units.
ii.
$$\begin{array} { | l | r | r | } \hline & x & { y } \\\hline & \begin{array} { l } \text { Hours of } \\\text { crime } \\\text { shows }\end{array} & { \begin{array} { l } \text { Fear of } \\\text { crime }\end{array} } \\\hline & 0 & 12 \\\hline & 3 & 11 \\\hline & 2 & 11 \\\hline & 9 & 18 \\\hline & 4 & 19 \\\hline & 5 & 20 \\\hline & 7 & 11 \\\hline & 9 & 16 \\\hline & 8 & 18 \\\hline & 9 & 25 \\\hline & 11 & 24 \\\hline & 1 & 9 \\\hline & 3 & 11 \\\hline & 2 & 11 \\\hline \text { Totals } & 73 & 216 \\\hline \text { Means } & 5.214286 & 15.42857 \\\hline\end{array}$$
$$\text { Consant } ( a ) = \bar { y } - b \bar { x } = 15.4286 - 1.124 \times 5.2143 = 9.5677$$
It can be predicted that an individual who does not watch any crime shows (x = 0) would still have a level of fear of crime of 9.5677 (y). In other words, the regression line intersects with the x-axis at 9.5677.
$$\begin{array} { l } \text { iii. } \\r = \frac { N \Sigma X Y - ( \Sigma X ) ( \Sigma Y ) } { \sqrt { \left[ N \Sigma X ^ { 2 } - ( \Sigma X ) ^ { 2 } \right] \left[ N \Sigma Y ^ { 2 } - ( \Sigma Y ) ^ { 2 } \right] } } = \frac { 2,586 } { 3,421.6207 } = 0.7558 \\\end{array}$$
The correlation coefficient of 0.7558 indicates that there is a moderate to strong positive relationship between visual consumption of crime shows (hours/day) and the level of fear of crime (scale 0-30). That is, as the hours of crime shows watched by an individual increases, so does his/her level of fear of crime. Recall: this is a hypothetical example.
iv. r²= 0.7558² = 0.5712.
57.12% of the variance in level of fear of crime (scale 0-30) is accounted for by the average number of hours of crime shows watched per week. In other words, when knowledge of the distribution of hours of crime shows watched is taken into account, we improve our ability to predict values on the independent variable (level of fear of crime) by a factor of 57.12%.
.
This value represents the total amount of (squared) errors we could possibly make if we knew absolutely nothing about the distribution of hours of TV crime shows watched.
^v.
$$\begin{array} { | r | r | r | r | r | } \hline { \text { Case } } & x & y & y & y \\\hline 1 & 0 & 12 & - 3.4286 & 11.7553 \\\hline 2 & 3 & 11 & - 4.4286 & 19.6125 \\\hline 3 & 2 & 11 & - 4.4286 & 19.6125 \\\hline 4 & 9 & 18 & 2.5714 & 6.612098 \\\hline 5 & 4 & 19 & 3.5714 & 12.7549 \\\hline 6 & 5 & 20 & 4.5714 & 20.8977 \\\hline 7 & 7 & 11 & - 4.4286 & 19.6125 \\\hline 8 & 9 & 16 & 0.5714 & 0.326498 \\\hline 9 & 8 & 18 & 2.5714 & 6.612098 \\\hline 10 & 9 & 25 & 9.5714 & 91.6117 \\\hline 11 & 11 & 24 & 8.5714 & 73.4689 \\\hline 12 & 1 & 9 & - 6.4286 & 41.3269 \\\hline 13 & 3 & 11 & - 4.4286 & 19.6125 \\\hline 14 & 2 & 11 & - 4.4286 & 19.6125 \\\hline \text { Totals } & 73 & 216 & - 0.0004 & 363.4286 \\\hline\end{array}$$
vi. First you must compute y? for every case observed: y? = a + bx
$\Sigma$ (y - y?)² = 155.8357.
The number of possible errors we could make (155.8357) trying to predict y is substantially reduced by knowing the distribution of hours of TV crime shows watched.
$\begin{array}{|r|r|r|l|r|r|}\hline \text { case } & x & y & y^{\prime} & y-y^{\prime} & \left(y-y^{\prime}\right)^{2} \\\hline 1 & 0 & 12 & 9.5677 & 2.4323 & 5.916083 \\\hline 2 & 3 & 11 & 12.9394 & -1.9394 & 3.761272 \\\hline 3 & 2 & 11 & 11.8155 & -0.8155 & 0.66504 \\\hline 4 & 9 & 18 & 19.6828 & -1.6828 & 2.831816 \\\hline 5 & 4 & 19 & 14.0633 & 4.9367 & 24.37101 \\\hline 6 & 5 & 20 & 15.1872 & 4.8128 & 23.16304 \\\hline 7 & 7 & 11 & 17.435 & -6.435 & 41.40923 \\\hline 8 & 9 & 16 & 19.6828 & -3.6828 & 13.56302 \\\hline 9 & 8 & 18 & 18.5589 & -0.5589 & 0.312369 \\\hline 10 & 9 & 25 & 19.6828 & 5.3172 & 28.27262 \\\hline 11 & 11 & 24 & 21.9306 & 2.0694 & 4.282416 \\\hline 12 & 1 & 9 & 10.6916 & -1.6916 & 2.861511 \\\hline 13 & 3 & 11 & 12.9394 & -1.9394 & 3.761272 \\\hline 14 & 2 & 11 & 11.8155 & -0.8155 & 0.66504 \\\hline \text { Totals } & 73 & 216 & & & 155.8357 \\\hline\end{array}$

vii. SS_total = SS_explained + SS_error
SS_explained = 363.4286 - 155.8357 = 207.5929
With the knowledge of the distribution of hours of TV crime shows (x) watched, we can explain 207.5929 of the variance within the dependent variable (numbers of violent crimes/month/100 inhabitants).
c. .
We can reject the null hypothesis if t exceeds ±2.228 (df = 14 - 2 = 10).
d. We use the t distribution to determine significance.
$$t = r \sqrt { \frac { n - 2 } { 1 - r ^ { 2 } } } = 0.7558 \sqrt { \frac { 14 - 2 } { 1 - 0.5712 } } = 0.7558 \times 5.2901 = 3.9983$$
We can reject the null hypothesis because t exceeds ±2.228 (t(12)=3.9983; p < 0.05). This means there is a statistically significant relationship between hours of TV crime shows watched and levels of fear of crime.
e.
$$S _ { y . x } = \sqrt { \frac { \Sigma \left( y - y ^ { \prime } \right) ^ { 2 } } { n - 2 } } = \sqrt { \frac { 155.8357 } { 12 } } = 3.6037$$