A local retailer in East Lansing believes that the sales volume of ice cream (Y)

Question

A local retailer in East Lansing believes that the sales volume of ice cream (Y) depends on the temperature (X) of that particular day. Data were gathered for the sales volume (in thousands of dollars) and temperature (degrees Fahrenheit) for seven days, The regression model is Y-hat=137.11-10.88X. s(bo)= 39.77 and s(b1) = 3.69. SST= 131.73 and SSR= 106.3. Use alpha = .05.

Question 31 (1 point)

If we want to test if there is actually a linear relationship between the sales of ice cream and the temperature of that day using regression, what is the correct alternative hypothesis for this test?

=0           

0         

1=0         

10

0=0

Question 32 (1 point)

The t-critical value for 1 is:

+/-2.571   

+/-2.015   

+2.015

-2.015       

+1.96

Question 33 (1 point)

The calculated value of the t-test statistic is:

-12.602

-2.949       

-3.448       

-10.778     

2.949

Question 34 (1 point)

What is the appropriate managerial conclusion?      

the sales volume increases as the temperature increases

there is a linear relationship between sales volume and temperature

there is no relationship between sales volume and temperature

the sales volume in summer is higher than in winter      

the sales volume decreases when the temperature decreases

Question 35 (1 point)

What is the value of r-square?

-0.193

0.193

-0.807

.807

1.313

Question 36 (1 point)

If we want to conduct an over all model test, what is the value of the F-test statistic?

3.54          

17.7          

26.6          

6.61          

20.9

Question 37 (1 point)

What is the F-critcal value?   

230.2        

5.59

234

6.61

3.89

Explanation / Answer

Ans:

1)df=n-2=7-2=5

critical t value(2 tailed)=tinv(0.05,5)=+/-2.571

2)Test statistic:

t=b1/sb1=-10.88/3.69=-2.949

3)As,t=-2.949<-2.571,we reject null hypothesis.

There is a linear relationship between sales volume and temperature

4)R^2=SSR/SST=106.3/131.73=0.807

5)MSR=106.3/1=106.3

MSE=(131.73-106.3)/5=5.086

F=MSR/MSE=106.3/5.086=20.9

6)critical F value=FINV(0.05,1,5)=6.61

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