Ice Cream Sales vs Temperature Temperature °C Ice cream sales $215 14.2 $325 16.

Question

Ice Cream Sales vs Temperature Temperature °C Ice cream sales $215 14.2 $325 16.4° $185 11.9 $332 15.2 $406 18.5° 22.1 $522 19.4 $412 $614 25.1 $544 23.4 $421 18.1 $445 22.6 $408 17.2 Determine the equation for the regression line: y -159.4742 Determine the correlation coefficient: Strong Positive Associat Use the equation to predict the sales when it's 20° C: I Use the equation to predict the temperature when the sales are $150: The slope of the line tells you that when the temperature Calculate the residual for the point (14.2, 215): x (round to 4 decimal places when necessary) 30.0879 (use 4 decimals if there are more than 4) then the sales will

Explanation / Answer

Result:

The local ice cream shop keeps track of how much ice cream they sell versus the temperature that day. Here are their figures for the last 12 days:

Regression Analysis

r²

0.917

n

12

r

0.958

k

1

Std. Error

38.127

Dep. Var.

sales

ANOVA table

Source

SS

df

MS

F

p-value

Regression

160,218.6163

1  

160,218.6163

110.22

1.02E-06

Residual

14,536.3004

10  

1,453.6300

Total

174,754.9167

11  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=10)

p-value

95% lower

95% upper

Intercept

-159.4742

54.6407

-2.919

.0153

-281.2213

-37.7270

Temp

30.0879

2.8659

10.499

1.02E-06

23.7022

36.4735

Predicted values for: sales

95% Confidence Interval

95% Prediction Interval

Temp

Predicted

lower

upper

lower

upper

Leverage

20

442.2831

416.341

468.225

353.459

531.107

0.093

Predicts sales when Temperature is 20 is 442.2831

Not applicable. We are predicting the sales from temperature not otherway.

The slope of the line tells you that when the temperature increases by 1 , then the sales will increases by 30.0879.

Calculate residual for the point (14.2, 215)

Predicted sales for Temperature 14.2 = -159.4742+30.0879*14.2 =267.77398

Residual =(215-267.77398) =-52.77398

=-52.7740 ( 4 decimals)

Regression Analysis

r²

0.917

n

12

r

0.958

k

1

Std. Error

38.127

Dep. Var.

sales

ANOVA table

Source

SS

df

MS

F

p-value

Regression

160,218.6163

1  

160,218.6163

110.22

1.02E-06

Residual

14,536.3004

10  

1,453.6300

Total

174,754.9167

11  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=10)

p-value

95% lower

95% upper

Intercept

-159.4742

54.6407

-2.919

.0153

-281.2213

-37.7270

Temp

30.0879

2.8659

10.499

1.02E-06

23.7022

36.4735

Predicted values for: sales

95% Confidence Interval

95% Prediction Interval

Temp

Predicted

lower

upper

lower

upper

Leverage

20

442.2831

416.341

468.225

353.459

531.107

0.093

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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