In: Statistics and Probability
A freshly brewed shot of espresso has three distinct components: the heart, body, and crema. The separation of these three components typically lasts only 10 to 20 seconds. To use the espresso shot in making a latte, a cappuccino, or another drink, the shot must be poured into the beverage during the separation of the heart, body, and crema. If the shot is used after the separation occurs, the drink becomes excessively bitter and acidic, ruining the final drink. Thus, a longer separation time allows the drink-maker more time to pour the shot and ensure that the beverage will meet expectations. An employee at a coffee shop hypothesized that the harder the espresso grounds were tamped down into the portafilter before brewing, the longer the separation time would be. An experiment using 24 observations was conducted to test this relationship. The independent variable Tamp measures the distance, in inches, between the espresso grounds and the top of the portafilter (i.e., the harder the tamp, the greater the distance). The dependent variable Time is the number of seconds after the heart, body, and crema are separated (i.e., the amount of time after the shot is poured before it must be used for the customer’s beverage). The data can be seen below:
Tamp Time
0.20 14
0.50 14
0.50 18
0.20 16
0.20 16
0.50 13
0.20 12
0.35 15
0.50 9
0.35 15
0.50 11
0.50 16
0.50 18
0.50 13
0.35 19
0.35 19
0.20 17
0.20 18
0.20 15
0.20 16
0.35 18
0.35 16
0.35 14
0.35 16
(a) Use the least-squares method to develop a simple regression equation with Time as the dependent variable and Tamp as the independent variable.
(b) Predict the separation time for a tamp distance of 0.50 inch.
(c) Plot the residuals versus the time order of experimentation. Are there any noticeable patterns?
(d) Compute the Durbin-Watson statistic. At the 0.05 level of significance, is there evidence of positive autocorrelation among the residuals?
(e) Based on the results of (c) and (d), is there reason to question the validity of the model?
(a) Use the least-squares method to develop a simple regression equation with Time as the dependent variable and Tamp as the independent variable.
The simple regression equation is:
Time = 17.0833 - 5*Tamp
(b) Predict the separation time for a tamp distance of 0.50 inch.
Time = 17.0833 - 5*0.50
Time = 14.583
(c) Plot the residuals versus the time order of experimentation. Are there any noticeable patterns?
The plot is:
Yes, there is a noticeable pattern.
(d) Compute the Durbin-Watson statistic. At the 0.05 level of significance, is there evidence of positive autocorrelation among the residuals?
The Durbin-Watson statistic is 1.72.
Yes, there is evidence of positive autocorrelation among the residuals.
(e) Based on the results of (c) and (d), is there reason to question the validity of the model?
Yes, the validity of the model can be questioned.
r² | 0.061 | |||||
r | -0.247 | |||||
Std. Error | 2.508 | |||||
n | 24 | |||||
k | 1 | |||||
Dep. Var. | Time | |||||
ANOVA table | ||||||
Source | SS | df | MS | F | p-value | |
Regression | 9.0000 | 1 | 9.0000 | 1.43 | .2443 | |
Residual | 138.3333 | 22 | 6.2879 | |||
Total | 147.3333 | 23 | ||||
Regression output | confidence interval | |||||
variables | coefficients | std. error | t (df=22) | p-value | 95% lower | 95% upper |
Intercept | 17.0833 | |||||
Tamp | -5.0000 | 4.1793 | -1.196 | .2443 | -13.6673 | 3.6673 |
Observation | Time | Predicted | Residual | |||
1 | 14.0 | 16.1 | -2.1 | |||
2 | 14.0 | 14.6 | -0.6 | |||
3 | 18.0 | 14.6 | 3.4 | |||
4 | 16.0 | 16.1 | -0.1 | |||
5 | 16.0 | 16.1 | -0.1 | |||
6 | 13.0 | 14.6 | -1.6 | |||
7 | 12.0 | 16.1 | -4.1 | |||
8 | 15.0 | 15.3 | -0.3 | |||
9 | 9.0 | 14.6 | -5.6 | |||
10 | 15.0 | 15.3 | -0.3 | |||
11 | 11.0 | 14.6 | -3.6 | |||
12 | 16.0 | 14.6 | 1.4 | |||
13 | 18.0 | 14.6 | 3.4 | |||
14 | 13.0 | 14.6 | -1.6 | |||
15 | 19.0 | 15.3 | 3.7 | |||
16 | 19.0 | 15.3 | 3.7 | |||
17 | 17.0 | 16.1 | 0.9 | |||
18 | 18.0 | 16.1 | 1.9 | |||
19 | 15.0 | 16.1 | -1.1 | |||
20 | 16.0 | 16.1 | -0.1 | |||
21 | 18.0 | 15.3 | 2.7 | |||
22 | 16.0 | 15.3 | 0.7 | |||
23 | 14.0 | 15.3 | -1.3 | |||
24 | 16.0 | 15.3 | 0.7 | |||
Durbin-Watson = | 1.72 | |||||
Predicted values for: Time | ||||||
95% Confidence Interval | 95% Prediction Interval | |||||
Tamp | Predicted | lower | upper | lower | upper | Leverage |
0.5 | 14.583 | 12.905 | 16.262 | 9.119 | 20.048 | 0.104 |