Question

In: Statistics and Probability

Jason believes that the sales of coffee at his coffee shop depend upon the weather. He...

Jason believes that the sales of coffee at his coffee shop depend upon the weather. He has taken a sample of 6 days. Below you are given the results of the sample.

Cups of Coffee Sold Temperature (in ⁰F)
350 50
200 60
210 70
100 80
60 90
40 100

a. Compute the least squares estimated regression equation.

b. Is there a significant relationship between the sales of coffee and temperature? Use a t test and a .05 level of significance. Be sure to state the null and alternative hypotheses.

c. Perform F test for the model with a .01 level of significance

Solutions

Expert Solution

X Y XY
350 50 17500 122500 2500
200 60 12000 40000 3600
210 70 14700 44100 4900
100 80 8000 10000 6400
60 90 5400 3600 8100
40 100 4000 1600 10000
Ʃx = Ʃy = Ʃxy = Ʃx² = Ʃy² =
960 450 61600 221800 35500
Sample size, n = 6
x̅ = Ʃx/n = 960/6 = 160
y̅ = Ʃy/n = 450/6 = 75
SSxx = Ʃx² - (Ʃx)²/n = 221800 - (960)²/6 = 68200
SSyy = Ʃy² - (Ʃy)²/n = 35500 - (450)²/6 = 1750
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 61600 - (960)(450)/6 = -10400

a)

Slope, b = SSxy/SSxx = -10400/68200 =    -0.152492669

y-intercept, a = y̅ -b* x̅ = 75 - (-0.15249)*160 =    99.39882698

Regression equation :   

ŷ = 99.3988 + (-0.1525) x  

b)

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 1750 - (-10400)²/68200 = 164.0762463

Standard error, se = √(SSE/(n-2)) = √(164.07625/(6-2)) = 6.4046

Null and alternative hypothesis:

Ho: β₁ = 0 ; Ha: β₁ ≠ 0

α = 0.05

Test statistic:

t = b/(se/√SSxx) = -6.2180

df = n-2 = 4

p-value = T.DIST.2T(ABS(-6.218), 4) = 0.0034

Conclusion:

p-value < α, Reject the null hypothesis.

There is a significant relationship between the sales of coffee and temperature.

c)

Null and alternative hypothesis:

Ho: β₁ = 0 ; Ha: β₁ ≠ 0

α = 0.01

SSE = SSyy -SSxy²/SSxx = 1750 - (-10400)²/68200 = 164.08

SSR = SSxy²/SSxx = (-10400)²/68200 = 1585.9238

df1 = 1

df2 = n-2 = 6-2 = 4

Test statistic:

F = SSR/(SSE/(n-2)) = 1585.9238/(164.0762/4) = 38.6631

p-value = F.DIST.RT(38.6631, 1, 4) = 0.0034

Conclusion:

p-value < α, Reject the null hypothesis.

The model is significant.


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