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