Question

In: Statistics and Probability

Consider the following regression model. Weekend is whether or not the visit was on a weekend....

Consider the following regression model. Weekend is whether or not the visit was on a weekend. Distance is how far the guests have to travel to get to the amusement park. Rides and Games are the number of rides and games, respectively. Clean is a cleanliness score from 1-10. Num.Child is the number of children with the guest. Wait is the average wait time for the rides.

Multiple R-squared: 0.8632,

Adjusted R-squared: 0.8787

F-statistic: 151.6 on 7 and 492 DF, p-value: .00000000022

Coeffiients:

Estimate Std. Error t value Pr(>ItI)

(Intercept) -140.61254 7.15405 -19.655 0.0000016

wekend -0.71573 0.80870 -0.885 0.376572

distance 0.04494 0.01219 3.686 0.000253

rides 0.61361 0.01219 5.072 0.0000059

games 0.13833 0.05872 2.356 0.18882

clean 0.92725 0.13593 6.821 0.000061

num.child 3.61602 0.26980 13.403 0.000025

wait 0.56476 0.04064 13.896 0.000031

a) do you think that this is a good regression model? Why or why not?

b)should all of the input variables in the model be included? If not, which variables should be removed from the model and why?

c) Generate a point estimate for the satisfaction level of an amusement park visit that is on a Friday, to an amusement park that is 63 miles away, that has 20 rides and 15 games. The park has a cleanliness score of 8, and an average wait time for each ride of 10 minutes. The guest has 3 children with them.

Solutions

Expert Solution

a) There are two statistic, which can be useful to analysis the above output which is R-squared = 0.8632, and hence 86.32% variability is explained by the model and for overall model the p-value is 00000000022 < 0.05 which shows that the above is a good model.

b) From the above all input variables, the p-value which is less than 0.05, which significantly effect the model. Hence, Intercept, distance, rides, clean, num.child and wait are significantly effect the model.

c ) The regression line of best fit is

y =-140.61254 -0.71573* wekend + 0.04494*distance + 0.61361 * rides + 0.13833 *games + 0.92725 * clean + 3.61602 *num.child + 0.56476 *wait


y= -140.61254 -0.71573* 0 + 0.04494* 63 + 0.61361 *  20 + 0.13833 *15 +   0.92725 * 8 + 3.61602 * 3 +  0.56476 *10 =  -99.52051

A a point estimate for the satisfaction level of an amusement park is -99.52051

orchestra answered 2 years ago
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