The strength of correlation is measured on an absolute value scale of $\displaystyle 0$ to $\displaystyle 1.0$ with $\displaystyle 0$ being the least correlated and $\displaystyle 1.0$ being the most correlated. The positive or negative in front of the correlation integer simply determines whether or not there is a positive or negative correlation between the variables. 

A correlation of $\displaystyle 0$ means that there is no correlation at all between two variables.

A high correlation coefficient regardless of sign implies a stronger relationship. In this case $\displaystyle -0.9$ has a stronger negative relationship than the positive relationship described by a value of $\displaystyle 0.8$

i is correct because there is a positive correlation between the number of texts sent each day and average credit card debt. 

ii is incorrect because the word "cause" was used in the statement. Correlation does not mean causation. There is a relationship between the number of texts sent each day and the number of books that a person reads each month. However, the number of texts sent each day does not cause a person to read a certain number of books each month. 

iii is correct because the absolute values of the correlations indicate which correlation is stronger. $\displaystyle .35$ is a stronger correlation than $\displaystyle -.2$.

Linear regression is the best option for determining whether the value of one variable predicts the value of a second variable.  Since that is exactly what the coach is trying to do, he should use linear regression.

Regression tests seek to determine one variable's ability to predict another variable.  In this analysis, one variable is dependent (the one predicted), and the other is independent (the variable that predicts).  Therefore, the dependent variable is the y-variable and the independent variable is the x-variable.

The key here is to utilize

$\displaystyle b_{1} = r\frac{s_{y}}{s_{x}}$.

"Perfect negative linear correlation" means $\displaystyle r = -1$, while the rest of the problem indicates $\displaystyle b_{1} = -7$ and $\displaystyle s_{x} = 2$. This enables us to solve for $\displaystyle s_{y}$.

$\displaystyle b_{1} = r\frac{s_{y}}{s_{x}}$
$\displaystyle \frac{b_{1}}{r} = \frac{s_{y}}{s_{x}}$
$\displaystyle s_{y} = \frac{b_{1}}{r}{s_{x}} = \frac{-7}{-1}2 = 7\cdot 2 = 14$

Use the formula $\displaystyle \beta _1 = r\frac{s_y}{s_x}$.

Plug in the given values for $\displaystyle r$ and $\displaystyle s_x$ and this becomes an algebra problem.

$\displaystyle 0.64 = 0.8\frac{s_y}{5}$

$\displaystyle 0.8 = \frac{s_y}{5}$

$\displaystyle s_y = 0.8\cdot 5 = 4$

An observation is an outlier if it falls more than $\displaystyle 1.5(IQR)$ above the upper quartile or more than $\displaystyle 1.5(IQR)$ below the lower quartile.

$\displaystyle IQR= Q3-Q1= 12-6=6$

$\displaystyle 1.5(IQR)= 11..5(6)= 9$

$\displaystyle Q1-9= 6-9= -3$. The minimum value is $\displaystyle 3$ so there are no outliers in the low end of the distribution.

$\displaystyle Q3+9= 12+9= 21$. The maximum value is $\displaystyle 20$ so there are no outliers in the high end of the distribution.

Use the $\displaystyle 1.5 \cdot IQR$ criteria:

This states that anything less than $\displaystyle Q1-1.5IQR$ or greater than $\displaystyle Q3+1.5IQR$ will be an outlier.

Thus, we want to find

$\displaystyle Q3+1.5IQR$  where $\displaystyle IQR =Q3-Q1$.

$\displaystyle 1.5 \cdot (70 - 40) =1.5\cdot 30=45$

$\displaystyle Q_3 + (1.5\cdot IQR) = 70 + 45 = 115$

Therefore, any new observation greater than 115 can be considered an outlier.

In order to find the outliers, we can use the $\displaystyle Q1-1.5*IQR$ and $\displaystyle Q3+1.5IQR$ formulas.

$\displaystyle IQR = Q3-Q1 = 86-45 = 41$

$\displaystyle 1.5*IQR = 61.5$

$\displaystyle Q1-1.5*IQR = 45-61.5 = -16.5$

$\displaystyle Q3+1.5*IQR = 86+61.5 = 147.5$

Only two numbers are outside of the calculated range and therefore are outliers: $\displaystyle 150$ and $\displaystyle 200$.