Question

1) The owner of a restaurant with a large outdoor patio has noticed that sales of cold beverages on the patio increase on hotter days. The owner would like to develop a predictive model for sales as a function of temperature to use for short term scheduling. To develop the model, the owner collects data for 21 days. The data appear in the Beverage worksheet of the HW7 data workbook on Moodle.

a) Draw a scatter plot of the data using Sales as the Y variable and Temperature as the X variable. Does there appear to be a linear association between a month’s electric consumption and average daily temperature?

b) Fit the simple linear regression model using Sales as the dependent or Y variable and Temperature as the dependent or X variable. Is the model significant at  = 0.05?

c) What percentage of the variation in Beverage Sales has been explained by the linear relationship with Temperature?

d) Draw a scatter plot of the residuals versus Temperature. Do the residuals Satisfy the assumption that they are independent of Temperature?

f) Fit a multiple regression model using Beverage Sales as the Y or dependent variable with independent variables of Temperature, and the (Temperature)^2 quadratic term. Is the model significant at  = 0.05? Are both independent variables significant at  = 0.05?

g) Using the model in part f), estimate the change in Beverage Sales when the Temperature increases from 75 to 80 and from 85 to 90.

Temperature	Sales
85	$    1,810
90	$    4,825
79	$       438
82	$       775
84	$    1,213
96	$    8,692
88	$    2,356
76	$       266
93	$    4,930
97	$    9,138
89	$    2,714
83	$    1,082
85	$    1,290
90	$    3,970
82	$       894
91	$    2,906
90	$    4,615
84	$    1,168
79	$       462
81	$    1,018
95	$    5,950

Answer 1

a)

there is a linear relationship because the correlation coefficient is = 0.9224

b)Y = -32511.2467 + 408.6026 X₁

Source	DF	Sum of Squares	Mean Square	F Statistic	P-value
Regression (between ŷi andyibarbar)	1	117362193.5706	117362193.5706	108.2876	2.761e-9
Residual (between yi and ŷi)	19	20592209.6675	1083800.5088
Total(between yi andyibar)	20	137954403.2381	6897720.1619

	Coeff	SE	t-stat	lower t0.025(19)	upper t0.975(19)	Stand Coeff	p-value	VIF
b	-32511.2467	3408.7235	-9.5377	-39645.7869	-25376.7065	0.000	1.122e-8
X1	408.6026	39.2656	10.4061	326.4189	490.7864	0.9224	2.761e-9	1.0000

R square (R²) equals 0.8507. It means that the predictors (X_i) explain 85.1% of the variance of Y.
Adjusted R square equals 0.8429.
The coefficient of correlation (R) equals 0.9224. It means that there is a very strong direct relationship between the predicted data (ŷ) and the observed data (y).

Overall regression: right-tailed, F_(1,19) = 108.2876, p-value = 2.761e-9. Since p-value < α (0.05), we reject the H₀.
The linear regression model, Y = b₀+ b₁X₁ provides a better fit

All the independent variables (Xi) are significant.

The Y-intercept (b): two-tailed, T = -9.5377, p-value = 1.122e-8. Hence b is significantly different from zero.

c) R^2 = 0.8507 means 85.07 percentage of the variation in Beverage Sales has been explained by the linear relationship with Temperature?

d)Residual
(between y_i and ŷ_i)

f)

y = 23.30035581 x² - 3643.171723 x + 142850.3406

Regression Polynomial:	y=23.3004x2−3643.1717x+142850.3406
R-squared:	r2=0.9474
Adjusted R-squared:	r2adj=0.9446
Residual Standard Error:	635.1365 on 1818 degrees of freedom

Coefficient	Estimate	Standard Error	tt-statistic	pp-value
intercept	142850.3406	30575.7015	4.672	0.0002
β1	-3643.1717	705.2304	-5.1659	0.0001
β2	23.3004	4.0532	5.7486	0

Analysis of Variance Table

Source	df	SS	MS	FF-statistic	pp-value
Regression	2	130693232.2314	65346616.1157	161.9903	0
Residual Error	18	7261171.0068	403398.3893
Total	20	137954403.2381	6897720.1619

R square (R²) equals 0.9474. It means that the predictors (X_i) explain 85.1% of the variance of Y.
Adjusted R square equals 0.9446
The coefficient of correlation (R) equals 0.9733. It means that there is a very very strong direct relationship between the predicted data (ŷ) and the observed data (y).

Overall regression: right-tailed, F_(1,19) =1, p-value=0.000 Since p-value < α (0.05), we reject the H₀.
The quadric regression model provides a better fit

All the independent variables (Xi) are significant.

The Y-intercept (b): two-tailed, T =4.672, p-value = 0.002 Hence intercept is significantly different from zero.

everything is significant and best fit than the simple linear regression model

g) y = 23.30035581 x² - 3643.171723 x + 142850.3406

  y = 23.30035581 (75)^2 - 3643.171723 *75 + 142850.3406

y = 676.962

y =  23.30035581 (80)^2 - 3643.171723 *80 + 142850.3406

y = 518.8799

change = 676.962- 518.8799 = 158.0821

y =  23.30035581 (85)^2 - 3643.171723 *85 + 142850.3406

y = 1525.81

y =  23.30035581 (90)^2 - 3643.171723 *90 + 142850.3406

y = 3697.77

change = 3697.77- 1525.81 = 2171.96