Question

A production plant cost-control engineer is responsible for cost reduction. One of the costly items in his plant is the amount of water used by the production facilities each month. He decided to investigate water usage by collecting seventeen observations on his plant's water usage and other variables.

Variable               Description

Temperature     Average monthly temperate (F)

Production          Amount of production (M pounds)

Days                      Number of plant operating days in the month

Persons                Number of persons on the monthly plant payroll

Water                   Monthly water usage (gallons)

Temperature	Production	Days	Persons	Water
58.8	7107	21	129	3067
65.2	6373	22	141	2828
70.9	6796	22	153	2891
77.4	9208	20	166	2994
79.3	14792	25	193	3082
81.0	14564	23	189	3898
71.9	11964	20	175	3502
63.9	13526	23	186	3060
54.5	12656	20	190	3211
39.5	14119	20	187	3286
44.5	16691	22	195	3542
43.6	14571	19	206	3125
56.0	13619	22	198	3022
64.7	14575	22	192	2922
73.0	14556	21	191	3950
78.9	18573	21	200	4488
79.4	15618	22	200	3295

Assume that you want to develop a linear model to predict the amount of “water” based on the monthly “production”.

i. Use the least squares method to estimate the regression coefficients b0 and b1 and state the regression equation.

ii. Which coefficients are statistically significant at the 5% level?

iii. Give the interpretation of the regression coefficients b0 and b1. What is the expected amount of “water”, if the “production” level is equal to 10000? Produce confidence interval for your estimate. Describe your finding in context.

iv. Interpret the value of the R2. Try to improve your model – Show an example

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
7107	3067	33564301.46	56029.67	1371348.57
6373	2828	42607872.28	226296.09	3105156.16
6796	2891	37264561.16	170326.15	2519350.92
9208	2994	13634339.04	95917.73	1143579.86
14792	3082	3577883.52	49153.50	-419363.20
14564	3898	2767330.10	353185.50	988625.74
11964	3502	876977.16	39320.56	-185696.61
13526	3060	391287.04	59392.56	-152445.20
12656	3211	59765.87	8594.38	22663.86
14119	3286	1484813.93	313.50	-21575.14
16691	3542	14368113.22	56784.09	903260.86
14571	3125	2790668.52	31935.79	-298533.43
13619	3022	516284.52	79358.20	-202413.96
14575	2922	2804048.75	145699.38	-639177.73
14556	3950	2740777.63	417696.09	1069958.92
18573	4488	32177589.93	1402552.56	6717943.21
15618	3295	7384966.10	75.79	-23658.49

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	219308	56163	199011580.2	3192631.5	15899024.35
mean	12900.47	3303.71	SSxx	SSyy	SSxy

1)

sample size ,   n =   17          
here, x̅ = Σx / n=   12900.47   ,     ȳ = Σy/n =   3303.71  
                  
SSxx =    Σ(x-x̅)² =    199011580.2353          
SSxy=   Σ(x-x̅)(y-ȳ) =   15899024.4          
                  
estimated slope , ß1 = SSxy/SSxx =   15899024.4   /   199011580.235   =   0.0799
                  
intercept,   ß0 = y̅-ß1* x̄ =   2273.0880          
                  
so, regression line is   Ŷ =   2273.0880   +   0.0799   *x

2)slope hypothesis test               tail=   2
Ho:   ß1=   0          
H1:   ß1╪   0          
n=   17              
alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    358.000   /√   199011580.24   =   0.0254
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.0799   /   0.0254   =   3.1481
                  
p-value =    0.0066              
decison :    p-value<α , reject Ho              
Conclusion:   Reject Ho and conclude that slope is significantly different from zero              

3)

Ŷ =   2273.0880   +   0.0799   *x

slope interpretation

for every 1 unit increase in production , water uses will increase by 0.0799 gallons

intercept interpretation

when amount of production is 0 ,water consumption will be 2273.08 gallons.

X Value=   10000
Confidence Level=   95%

Predicted Y at X=   10000   is                  
Ŷ =   2273.08799   +   0.079890   *   10000   =   3071.987

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    113.828              
margin of error,E=t*Std error=t* S(ŷ) =   2.1314   *   113.8283   =   242.6192
                  
Confidence Lower Limit=Ŷ +E =    3071.987   -   242.619   =   2829.368
Confidence Upper Limit=Ŷ +E =   3071.987   +   242.619   =   3314.607

4)

R² =    (Sxy)²/(Sx.Sy) =    0.3978

only 39.78% of variation is explained by regression model