In: Statistics and Probability
Assume 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 |
340 |
50 |
200 |
60 |
210 |
70 |
100 |
80 |
60 |
90 |
40 |
100 |
a. | Which variable is the dependent variable? |
b. | Using the Excel spreadsheet, run the least squares model and write the summary output here. |
c. | Compute the least squares estimated line. |
d. | Compute the correlation coefficient between temperature and the sales of coffee. |
e. | Is there a significant relationship between the sales of coffee and temperature? Use a .05 level of significance. Be sure to state the null and alternative hypotheses. |
f. | Predict sales of a 90 degree day. |
a)
dependent variable : Cups of Coffee Sold
b)
SUMMARY OUTPUT | |||||
Regression Statistics | |||||
Multiple R | 0.9555 | ||||
R Square | 0.9129 | ||||
Adjusted R Square | 0.8912 | ||||
Standard Error | 37.4611 | ||||
Observations | 6.0000 | ||||
ANOVA | |||||
df | SS | MS | F | Significance F | |
Regression | 1 | 58870 | 58870 | 41.9501 | 0.0029 |
Residual | 4 | 5613.333333 | 1403.333333 | ||
Total | 5 | 64483.33333 | |||
Coefficients | Standard Error | t Stat | P-value | ||
Intercept | 593.3333 | 68.8811 | 8.6139 | 0.0010 | |
Temperature | -5.8000 | 0.8955 | -6.4769 | 0.0029 |
c) e least squares estimated line. :Yhat=593.3333-5.8000*temperature
d)
correlation coefficient =-0.9555
e)
null hypothesis: | β1 | = | 0 | |
Alternate Hypothesis: | β1 | ≠ | 0 |
as p value =0.0029 is less than 0.05 ; we reject null hypothesis and conclude that there is a significant relationship between the sales of coffee and temperature .
f)
predicted sales =593.3333-5.8000*90=71.3333