Exam 8: Regression, Associations, and Predictive Modeling

Correct Answer:

Verified

a. $P$ (neither ham nor turkey)
$= 1 - P ($ ham or turkey $)$
$= 1 - [ P ($ ham $) + P$ (turkey) $- P ($ ham and turkey $) ]$
$= 1 - [ 0.44 + 0.58 - 0.16 ] = 1 - 0.86 = 0.14$
Or, using the Venn diagram, 14\%
b. $P ($ ham only $) = P ($ ham $) - P ($ ham and turkey $)$
$$= 0.44 - 0.16$$
$$= 0.28$$
Or, using the Venn diagram, $28 \%$
c. $P ($ ham I turkey $) = \frac { P ( \text { ham and turkey } ) } { P ( \text { turkey } ) } = \frac { 0.16 } { 0.58 } = 0.2759$
d. No, the events are not disjoint, since some families $( 16 \% )$ have both ham and turkey at their holiday meals.

Correct Answer:

Verified

a. Temperature and Rent Control
b. For cities with similar values for the other explanatory variables (poverty, unemployment, etc.), on average, each additional degree of average yearly temperature is associated with an increase of $0.14$ homeless people per 10,000 residents of the city.
c. For cities with similar values of the other explanatory variables (poverty, unemployment, etc.), cities with rent control laws have, on average, $2.94$ more homeless people per 10,000 residents than cities without rent control laws. d. No. This is an observational study, so the results cannot be used to establish cause-and-effect. There is evidence that cities with rent control laws are associated with higher levels of homelessness, but the cause may not be the rent control laws themselves.
e. The adjusted $R ^ { 2 }$ should be used to compare regression models because the models have different numbers of predictor variables.
f. * Straight enough condition: OK. The plot of residuals vs. predicted values shows no obvious curvature. (We should also check plots of Homeless vs. each X variable.) * Does the plot thicken? condition: OK. The plot of residuals vs. predicted values is approximately the same width throughout.
* Randomization condition: Caution. We do not know whether or not the 50 U.S. cities in our sample were chosen randomly, or whether they are representative of all large U.S. cities.
* Nearly Normal condition: Violated. The normal probability plot is curved, which indicates that the regression errors do not follow a Normal model. (It may be possible to re-express the $y$ -variable to fix this problem.)

Correct Answer:

Verified

C.
$\begin{array}{|c|c|c|c|}\hline X=\text { amount Alphonso wins } & \$ 8 & \$ 2 & -\$ 3 \\\hline P(X) & \frac{6}{36} & \frac{10}{36} & \frac{20}{36} \\\hline\end{array}$

$\begin{array}{|c|c|c|c|c|c|c|}\hline & 8 & 2 & 2 & 2 & 2 & 2 \\\hline 3 & \mathrm{~A} 8 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 \\\hline 3 & \mathrm{~A} 8 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 \\\hline 3 & \mathrm{~A} 8 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 \\\hline 3 & \mathrm{~A} 8 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 & \mathrm{~B} 3 \\\hline \mathrm{l} & \mathrm{A} 8 & \mathrm{~A} 2 & \mathrm{~A} 2 & \mathrm{~A} 2 & \mathrm{~A} 2 & \mathrm{~A} 2 \\\hline \mathrm{l} & \mathrm{A} 8 & \mathrm{~A} 2 & \mathrm{~A} 2 & \mathrm{~A} 2 & \mathrm{~A} 2 & \mathrm{~A} 2 \\\hline\end{array}$
d. $E ( X ) = \$ 8 \left( \frac { 6 } { 36 } \right) + \$ 2 \left( \frac { 10 } { 36 } \right) + ( - \$ 3 ) \left( \frac { 20 } { 36 } \right) = \$ 0.22$
Standard deviation $= \sqrt { ( 8 - 0.22 ) ^ { 2 } \left( \frac { 6 } { 36 } \right) + ( 2 - 0.22 ) ^ { 2 } \left( \frac { 10 } { 36 } \right) + ( - 3 - 0.22 ) ^ { 2 } \left( \frac { 20 } { 36 } \right) } = \sqrt { 16.7284 }$ $= 4.09$
e. Alphonso has the advantage, because his expected value is positive. He expects to win an average of $\$ 0.22$ for each time they play the game.