Chef Lim Chef Lim is a Chinese restaurant that takes delivery orders. Most of it

ID: 3247855 • Letter: C

Question

Chef Lim

Chef Lim is a Chinese restaurant that takes delivery orders. Most of its deliveries are within a 10-mile radius, but it occasionally delivers to customers more than 10 miles away. Chef Lim employs a number of delivery people, four of whom are relatively new hires. The restaurant has recently been receiving customer complaints about excessively long delivery times. Therefore, SnowPea has collected data on a random sample of deliveries by its four new delivery people during the peak dinner time. Chef Lim is concerned that one or more of the new delivery people might be slower than others.

Managerial report: Using a word processor, write a managerial report about the recommendation and the reason why you make such a recommendation. You must write the report in a clear and concise manner with supporting graphs and tables, equations. Your report will be graded by the rubric. When you are developing a conclusion, consider the following guideline.

***Please need help with the following***:

A: Using the regression analysis and develop a regression equation to predict delivery time.

-Make four separate regression models: one for each deliverer. For each deliverer, what is the predicted delivery time of the customer who lives 15 miles away from the restaurant?

-Make one regression model ignoring deliverer information, what is the predicted delivery time of the customer who lives 15 miles away from the restaurant?

-What is the 95% confidence interval of the prediction in part b.

Do other analyses that you think are necessary or beneficial to make recommendations.

The variables in the dataset are as follows:

Deliverer: which person made the delivery

PrepTime: time from when order was placed until delivery person started driving it to the customer

TravelTime: time to drive from Chef Lim to customer

Distance: distance (miles) from Chef Lim to customer

Order Deliverer PrepTime TravelTime Distance 1 3 13.0 30.0 13.3 2 1 10.9 17.8 8.6 3 1 9.2 9.2 4.7 4 3 7.2 14.7 4.3 5 2 14.5 21.9 12.2 6 2 9.5 8.1 3.6 7 2 7.1 31.6 11.3 8 3 8.5 29.0 10.3 9 2 9.8 4.9 2.6 10 1 14.3 13.4 5.3 11 1 12.0 12.9 4.5 12 2 13.0 8.8 4.1 13 1 14.9 8.7 4.2 14 1 9.7 29.9 11.7 15 2 10.0 18.2 9.5 16 3 10.8 22.9 8.8 17 1 12.1 5.0 1.7 18 1 9.9 9.9 4.5 19 1 7.9 9.0 3.7 20 2 17.2 6.5 3.4 21 4 10.2 29.3 3.4 22 3 15.0 24.0 9.8 23 4 12.5 19.5 5.4 24 1 11.0 10.5 5.0 25 4 15.7 29.0 7.9 26 2 10.1 2.6 1.4 27 3 8.2 19.0 7.5 28 1 9.2 13.4 7.0 29 2 18.7 8.9 5.6 30 4 11.0 9.3 2.2 31 1 9.3 22.8 8.2 32 2 10.1 14.5 7.7 33 1 11.4 26.1 12.9 34 3 18.0 10.8 3.8 35 4 14.0 28.8 11.7 36 2 15.4 21.2 10.7 37 4 6.1 6.9 2.8 38 3 13.9 21.9 8.4 39 4 9.5 17.8 4.6 40 1 10.3 9.5 4.3 41 3 14.4 29.7 6.9 42 2 9.4 23.5 12.2 43 3 6.6 17.0 6.3 44 4 15.6 18.5 5.9 45 2 9.7 13.8 9.2 46 1 14.0 5.3 2.5 47 1 11.7 3.3 1.4 48 4 12.7 0.7 0.2 49 1 14.6 1.9 0.6 50 1 9.5 12.6 6.5 51 1 7.1 7.2 3.3 52 3 8.5 7.3 3.0 53 4 9.9 26.2 5.2 54 4 14.3 28.7 8.9 55 4 12.0 23.1 7.7 56 1 13.0 12.6 7.2 57 4 16.2 28.3 8.2 58 4 10.3 9.4 4.1 59 4 8.5 29.8 13.0 60 1 9.5 21.6 11.6 61 3 13.1 10.1 3.1 62 3 8.4 24.1 5.8 63 3 17.0 6.0 2.8 64 4 14.2 18.6 2.4 65 3 8.8 13.6 4.6 66 2 12.9 29.5 14.0 67 2 10.9 17.8 6.9 68 3 9.7 23.8 6.4 69 2 13.4 18.9 10.1 70 2 8.7 7.0 2.9 71 1 17.3 16.6 9.3 72 3 7.0 31.3 8.6 73 1 9.0 3.5 1.5 74 1 13.1 11.9 4.4 75 1 11.0 11.6 3.6 76 2 12.0 7.5 3.3

Explanation / Answer

We will run the linear regression in R. Following is the code to import the data in R and run the logistic regression.

Code to import the data:

delvr<-c(3,
1,
1,
3,
2,
2,
2,
3,
2,
1,
1,
2,
1,
1,
2,
3,
1,
1,
1,
2,
4,
3,
4,
1,
4,
2,
3,
1,
2,
4,
1,
2,
1,
3,
4,
2,
4,
3,
4,
1,
3,
2,
3,
4,
2,
1,
1,
4,
1,
1,
1,
3,
4,
4,
4,
1,
4,
4,
4,
1,
3,
3,
3,
4,
3,
2,
2,
3,
2,
2,
1,
3,
1,
1,
1,
2
)
travtime<-c(30,
17.8,
9.2,
14.7,
21.9,
8.1,
31.6,
29,
4.9,
13.4,
12.9,
8.8,
8.7,
29.9,
18.2,
22.9,
5,
9.9,
9,
6.5,
29.3,
24,
19.5,
10.5,
29,
2.6,
19,
13.4,
8.9,
9.3,
22.8,
14.5,
26.1,
10.8,
28.8,
21.2,
6.9,
21.9,
17.8,
9.5,
29.7,
23.5,
17,
18.5,
13.8,
5.3,
3.3,
0.7,
1.9,
12.6,
7.2,
7.3,
26.2,
28.7,
23.1,
12.6,
28.3,
9.4,
29.8,
21.6,
10.1,
24.1,
6,
18.6,
13.6,
29.5,
17.8,
23.8,
18.9,
7,
16.6,
31.3,
3.5,
11.9,
11.6,
7.5
)
distance<-c(13.3,
8.6,
4.7,
4.3,
12.2,
3.6,
11.3,
10.3,
2.6,
5.3,
4.5,
4.1,
4.2,
11.7,
9.5,
8.8,
1.7,
4.5,
3.7,
3.4,
3.4,
9.8,
5.4,
5,
7.9,
1.4,
7.5,
7,
5.6,
2.2,
8.2,
7.7,
12.9,
3.8,
11.7,
10.7,
2.8,
8.4,
4.6,
4.3,
6.9,
12.2,
6.3,
5.9,
9.2,
2.5,
1.4,
0.2,
0.6,
6.5,
3.3,
3,
5.2,
8.9,
7.7,
7.2,
8.2,
4.1,
13,
11.6,
3.1,
5.8,
2.8,
2.4,
4.6,
14,
6.9,
6.4,
10.1,
2.9,
9.3,
8.6,
1.5,
4.4,
3.6,
3.3
)
data<-data.frame(delvr,travtime,distance)

Code to run the linear regression by each deliverer.

Deliverer --1:

lg1<-lm(travtime~distance, data= subset(data,delvr==1))
summary(lg1)

Ouput:

Call:

lm(formula = travtime ~ distance, data = subset(data, delvr ==

1))

Residuals:

Min 1Q Median 3Q Max

-3.2168 -0.9405 -0.6122 0.4332 5.2674

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 1.1557 0.8939 1.293 0.209

distance 2.0066 0.1392 14.415 5.25e-13 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.275 on 23 degrees of freedom

Multiple R-squared: 0.9003, Adjusted R-squared: 0.896

F-statistic: 207.8 on 1 and 23 DF, p-value: 5.246e-13

So, the regression equation for deliverer-1 is :

Travtime = 1.1557 + 2.0066* Distance

Predicted Travtime for distance = 15 miles = 1.1557 + 2.0066*15 =31.25 minutes.

For deliverer -2:

Code:

lg2<-lm(travtime~distance, data= subset(data,delvr==2))
summary(lg2)

Output:

Call:

lm(formula = travtime ~ distance, data = subset(data, delvr ==

2))

Residuals:

Min 1Q Median 3Q Max

-4.8561 -1.1996 -0.4122 0.7791 8.6951

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 0.04258 1.47137 0.029 0.977

distance 2.02321 0.17869 11.322 4.75e-09 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.944 on 16 degrees of freedom

Multiple R-squared: 0.889, Adjusted R-squared: 0.8821

F-statistic: 128.2 on 1 and 16 DF, p-value: 4.755e-09

So, the regression equation for deliverer-2 is :

Travtime = 0.0426 + 2.0232* Distance

Predicted Travtime for distance = 15 miles = 0.0426 + 2.0232*15 =30.39 minutes.

For deliverer -3:

Code:

lg3<-lm(travtime~distance, data= subset(data,delvr==3))
summary(lg3)

Output:

Call:

lm(formula = travtime ~ distance, data = subset(data, delvr ==

3))

Residuals:

Min 1Q Median 3Q Max

-5.3480 -2.6367 -1.7999 0.7441 9.4817

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 3.9065 2.7649 1.413 0.178

distance 2.3640 0.3801 6.220 1.64e-05 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.484 on 15 degrees of freedom

Multiple R-squared: 0.7206, Adjusted R-squared: 0.702

F-statistic: 38.69 on 1 and 15 DF, p-value: 1.64e-05

So, the regression equation for deliverer-3 is :

Travtime = 3.9065 + 2.3640* Distance

Predicted Travtime for distance = 15 miles = 3.9065 + 2.3640*15 =39.37 minutes.

For deliverer -4:

Code:

lg4<-lm(travtime~distance, data= subset(data,delvr==4))
summary(lg4)

Output:

Call:

lm(formula = travtime ~ distance, data = subset(data, delvr ==

4))

Residuals:

Min 1Q Median 3Q Max

-7.9911 -3.8763 -0.4072 3.5795 14.0658

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 8.2822 3.0789 2.690 0.017600 *

distance 2.0447 0.4549 4.495 0.000504 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 6.195 on 14 degrees of freedom

Multiple R-squared: 0.5907, Adjusted R-squared: 0.5615

F-statistic: 20.21 on 1 and 14 DF, p-value: 0.0005039

So, the regression equation for deliverer-4 is :

Travtime = 8.2822 + 2.0447* Distance

Predicted Travtime for distance = 15 miles = 8.2822 + 2.0447*15 =38.95 minutes.

Now, we will conduct the overall regression without the deliverer information.

Code:

lg<-lm(travtime~distance, data)
summary(lg)

Output:

Call:

lm(formula = travtime ~ distance, data = data)

Residuals:

Min 1Q Median 3Q Max

-8.382 -3.220 -2.083 1.782 18.960

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 3.3981 1.2337 2.754 0.0074 **

distance 2.0417 0.1726 11.831 <2e-16 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 5.178 on 74 degrees of freedom

Multiple R-squared: 0.6542, Adjusted R-squared: 0.6495

F-statistic: 140 on 1 and 74 DF, p-value: < 2.2e-16

The regression equation is

Travel time = 3.3981+2.0417*Distance.

now, to get the 95% confidence interval for the predicted tarveltime when the distance is 15 miles, use the following code.

newdata<-data.frame(distance=15)
predict(lg,newdata, interval="confidence",conf.level=0.95)

Output:

> predict(lg,newdata, interval="confidence",conf.level=0.95)
fit lwr upr
1 34.02321 30.7951 37.25132

So, the predicted delivery time when distance is 15 miles = 3.3981+2.0417*15 = 34.02 minutes and

the 95% confidence interval for the predicted value is (30.80,37.25) minutes

Navigate

Chef Lim Chef Lim is a Chinese restaurant that takes delivery orders. Most of it

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.

Chef Lim Chef Lim is a Chinese restaurant that takes delivery orders. Most of it

Question

Explanation / Answer

Related Questions

Navigate