Chapter ISG, Problem 4.6.2.4P

a.

To determine

To graph: The scatter plot for the given data and determine the type of correlation.

Answer to Problem 4.6.2.4P

The given scatter plot has a weak negative correlation.

Explanation of Solution

Given: The given data which shows the number of visitors to a ski park and the outside temperature is as follows-

Temperature (In Fahrenheit)	Number of visitors
20	600
25	620
30	650
35	585
40	502
45	565
50	380
55	362
60	403

Graph: The scatter plot of the given data is as follows-

  

Interpretation: The line of best fit has a negative slope. Also, the scatter points are not clustered together near the line. Hence, the given scatter plot depicts a weak negative correlation.

b.

To determine

The equation of the line of regression.

Answer to Problem 4.6.2.4P

The equation of the line of regression is-

  $y=\frac{-118x}{15}+\frac{2450}{3}$

Explanation of Solution

Given: The given data which shows the number of visitors to a ski park and the outside temperature is as follows-

Temperature (In Fahrenheit)	Number of visitors
20	600
25	620
30	650
35	585
40	502
45	565
50	380
55	362
60	403

Concept Used: Any line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ has the equations-

  $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)$.

Calculation: From the graph it can be considered that the line of best fit or the regression line passes through the points (25,620) and (40,502). Thus, the equation of the regression line will be-

  $y-620=\left(\frac{502-620}{40-25}\right)(x-25)$

  $\Rightarrow y-620=\left(\frac{-118}{15}\right)(x-25)$

  $\Rightarrow 15(y-620)=-118(x-25)$

  $\Rightarrow 15y-9300=-118x+2950$

  $\Rightarrow 15y=-118x+9300+2950$

  $\Rightarrow y=\frac{-118x+12250}{15}$

  $\Rightarrow y=\frac{-118x}{15}+\frac{2450}{3}$ , which is the approximate equation of the regression line.

c.

To determine

To verify: The interpretation of the slope and y-intercept of the regression line.

Explanation of Solution

Given: The equation of the line of regression as determined earlier is-

  $y=\frac{-118x}{15}+\frac{2450}{3}$

Comparing the equation of the regression line with the general equation of a straight line, i.e.- $y=mx+b$ , the values of the slope and y -intercept are-

  $m=\frac{-118}{15}=-7.86667$

  $b=\frac{2450}{3}=816.6667$

The negative slope implies a negative correlation, i.e.- the values of y decreases with an increase in the values of x . Thus, it shows that with increase in outside temperature the number of visitors at the ski park decreases.

The y -intercept gives an assumed value of the y variable when the value of the x variable is zero, i.e.- it gives a statistical value of the number of visitors at the ski park when the outside temperature is 0 degrees Fahrenheit.

d.

To determine

The reasonable outputs of y for given values of x in the domain of the equation of the regression line.

Explanation of Solution

Given: The regression line has the equation-

  $y=\frac{-118x}{15}+\frac{2450}{3}$

No, the regression line gives a statistical interpretation of a given data which can’t be used to determine the values of y for any given value of x for real work scenarios. For example, the y -intercept of the regression line is 816.66667 (approx.) which suggests that when the outside temperature is 0 degrees Fahrenheit, the number of visitors in the ski park is about 817. At such low temperature people prefer to stay indoors and it is an unreasonable number of visitors for a ski park to accommodate.

e.

To determine

The number of visitors if the outside temperature is 32 degrees Fahrenheit.

Answer to Problem 4.6.2.4P

The ski park can expect 565 visitors when the outside temperature is 32 degrees.

Explanation of Solution

Given: The regression line has the equation-

  $y=\frac{-118x}{15}+\frac{2450}{3}$

Concept Used: The regression line gives a statistical interpretation of the given data which can be used to find the values of one variable if the other variable is known.

Calculations: The regression line is-

  $y=\frac{-118x}{15}+\frac{2450}{3}$

When the outside temperature is 32 degrees, i.e.- when x = 32, the number of expected visitors will be-

  $y=\frac{-118}{15}\times 32+\frac{2450}{3}$

  $=\frac{-3776}{15}+\frac{2450}{3}$

  $=-251.7333+816.6667$

  $=564.9334$

  $=565$ (approx.)