Question

Total Labor Hours = Beta0 + Beta1 Number of Cases Shipped + Beta2 Indirect Costs of Total Labor Hours + Beta3 Holiday + E where E is assumed normal with mean 0 and constant variance. Assume that the regression model here is appropriate. A new shipment is to be received with x1=250,000; x2=6.70; x3=0. Obtain a 95% confidence interval for the mean response for this shipment. Interpret this interval.

y	x1	x2	x3
4325	301995	6.88	0
4110	269334	7.23	0
4111	267631	6.27	0
4161	296350	6.49	0
4560	277223	6.37	0
4401	269189	7.05	0
4251	277133	6.34	0
4222	282892	6.94	0
4063	306639	8.56	0
4343	328405	6.71	0
4833	321773	5.82	1
4453	272319	6.82	0
4195	293880	8.38	0
4394	300867	7.72	0
4099	296872	7.67	0
4816	245674	7.72	1
4867	211944	6.45	1
4114	227996	7.22	0
4314	248328	8.5	0
4289	249894	8.08	0
4269	302660	7.26	0
4347	273848	7.39	0
4178	245743	8.12	0
4333	267673	6.75	0
4226	256506	7.79	0
4121	271854	7.89	0
3998	293225	9.01	0
4475	269121	8.01	0
4545	322812	7.21	0
4016	252225	7.85	0
4207	261365	6.14	0
4148	287645	6.76	0
4562	289666	7.92	0
4146	270051	8.19	0
4555	265239	7.55	0
4365	352466	6.94	0
4471	426908	7.25	0
5045	369989	9.65	1
4469	472476	8.2	0
4408	414102	8.02	0
4219	302507	6.72	0
4211	382686	7.23	0
4993	442782	7.61	1
4309	322303	7.39	0
4499	290455	7.99	0
4186	411750	7.83	0
4342	292087	7.77	0

Answer 1

I have coplete this problem by using R software as well as Manually

first i will give the r codes

data=read.csv(file.choose())
> data
y x1 x2 x3
1 4325 301995 6.88 0
2 4110 269334 7.23 0
3 4111 267631 6.27 0
4 4161 296350 6.49 0
5 4560 277223 6.37 0
6 4401 269189 7.05 0
7 4251 277133 6.34 0
8 4222 282892 6.94 0
9 4063 306639 8.56 0
10 4343 328405 6.71 0
11 4833 321773 5.82 1
12 4453 272319 6.82 0
13 4195 293880 8.38 0
14 4394 300867 7.72 0
15 4099 296872 7.67 0
16 4816 245674 7.72 1
17 4867 211944 6.45 1
18 4114 227996 7.22 0
19 4314 248328 8.50 0
20 4289 249894 8.08 0
21 4269 302660 7.26 0
22 4347 273848 7.39 0
23 4178 245743 8.12 0
24 4333 267673 6.75 0
25 4226 256506 7.79 0
26 4121 271854 7.89 0
27 3998 293225 9.01 0
28 4475 269121 8.01 0
29 4545 322812 7.21 0
30 4016 252225 7.85 0
31 4207 261365 6.14 0
32 4148 287645 6.76 0
33 4562 289666 7.92 0
34 4146 270051 8.19 0
35 4555 265239 7.55 0
36 4365 352466 6.94 0
37 4471 426908 7.25 0
38 5045 369989 9.65 1
39 4469 472476 8.20 0
40 4408 414102 8.02 0
41 4219 302507 6.72 0
42 4211 382686 7.23 0
43 4993 442782 7.61 1
44 4309 322303 7.39 0
45 4499 290455 7.99 0
46 4186 411750 7.83 0
47 4342 292087 7.77 0

y=data$y
> y
[1] 4325 4110 4111 4161 4560 4401 4251 4222 4063 4343 4833 4453 4195 4394 4099 4816 4867 4114 4314 4289 4269 4347 4178 4333 4226
[26] 4121 3998 4475 4545 4016 4207 4148 4562 4146 4555 4365 4471 5045 4469 4408 4219 4211 4993 4309 4499 4186 4342
> x1=data$x1
> x1
[1] 301995 269334 267631 296350 277223 269189 277133 282892 306639 328405 321773 272319 293880 300867 296872 245674 211944 227996
[19] 248328 249894 302660 273848 245743 267673 256506 271854 293225 269121 322812 252225 261365 287645 289666 270051 265239 352466
[37] 426908 369989 472476 414102 302507 382686 442782 322303 290455 411750 292087
> x2=data$x2
> x2
[1] 6.88 7.23 6.27 6.49 6.37 7.05 6.34 6.94 8.56 6.71 5.82 6.82 8.38 7.72 7.67 7.72 6.45 7.22 8.50 8.08 7.26 7.39 8.12 6.75 7.79
[26] 7.89 9.01 8.01 7.21 7.85 6.14 6.76 7.92 8.19 7.55 6.94 7.25 9.65 8.20 8.02 6.72 7.23 7.61 7.39 7.99 7.83 7.77
> x3=data$x3
> x3
[1] 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0

model=lm(y~x1+x2+x3)
> model

Call:
lm(formula = y ~ x1 + x2 + x3)

Coefficients:
(Intercept) x1 x2 x3
4.137e+03 8.077e-04 -1.242e+01 6.094e+02

> summary(model)

Call:
lm(formula = y ~ x1 + x2 + x3)

Residuals:
Min 1Q Median 3Q Max
-263.65 -112.15 -16.26 80.20 297.83

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.137e+03 2.236e+02 18.503 < 2e-16 ***
x1 8.077e-04 3.917e-04 2.062 0.0453 *
x2 -1.242e+01 2.827e+01 -0.439 0.6627
x3 6.094e+02 7.055e+01 8.639 6.03e-11 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 148.3 on 43 degrees of freedom
Multiple R-squared: 0.6604, Adjusted R-squared: 0.6367
F-statistic: 27.87 on 3 and 43 DF, p-value: 3.611e-10

> x=matrix(c(x1,x2,x3),byrow=F,ncol=3)
> x
[,1] [,2] [,3]
[1,] 301995 6.88 0
[2,] 269334 7.23 0
[3,] 267631 6.27 0
[4,] 296350 6.49 0
[5,] 277223 6.37 0
[6,] 269189 7.05 0
[7,] 277133 6.34 0
[8,] 282892 6.94 0
[9,] 306639 8.56 0
[10,] 328405 6.71 0
[11,] 321773 5.82 1
[12,] 272319 6.82 0
[13,] 293880 8.38 0
[14,] 300867 7.72 0
[15,] 296872 7.67 0
[16,] 245674 7.72 1
[17,] 211944 6.45 1
[18,] 227996 7.22 0
[19,] 248328 8.50 0
[20,] 249894 8.08 0
[21,] 302660 7.26 0
[22,] 273848 7.39 0
[23,] 245743 8.12 0
[24,] 267673 6.75 0
[25,] 256506 7.79 0
[26,] 271854 7.89 0
[27,] 293225 9.01 0
[28,] 269121 8.01 0
[29,] 322812 7.21 0
[30,] 252225 7.85 0
[31,] 261365 6.14 0
[32,] 287645 6.76 0
[33,] 289666 7.92 0
[34,] 270051 8.19 0
[35,] 265239 7.55 0
[36,] 352466 6.94 0
[37,] 426908 7.25 0
[38,] 369989 9.65 1
[39,] 472476 8.20 0
[40,] 414102 8.02 0
[41,] 302507 6.72 0
[42,] 382686 7.23 0
[43,] 442782 7.61 1
[44,] 322303 7.39 0
[45,] 290455 7.99 0
[46,] 411750 7.83 0
[47,] 292087 7.77 0

>a=solve(t(x)%*%x)

>a

[,1] [,2] [,3]
[1,] 6.065369e-12 -2.419043e-07 -1.292230e-07
[2,] -2.419043e-07 1.007298e-02 1.986474e-03
[3,] -1.292230e-07 1.986474e-03 2.263496e-01

[1] 25000.0 6.7 0.0
> x0=c(250000,6.7,0)
> x0

ycap= 4.137e+03 + 8.077e-04 *250000 -1.242e+01 *6.7 + 6.094e+02 *0
> ycap

sqrt(t(x0)%*%a%*%x0)*148.3*qt(0.975,43)
[,1]
[1,] 43.2184

qt(0.975,43)
[1] 2.016692

ycap+sqrt(t(x0)%*%a%*%x0)*148.3*qt(0.975,43)
[,1]
[1,] 4298.929
> ycap-sqrt(t(x0)%*%a%*%x0)*148.3*qt(0.975,43)
[,1]
[1,] 4212.493