The Toluca Company
www.oswego.edu/~srp/stats/toluca.txt
A scatterplot, regression analysis and residual plots are shown.
550
450
6
400
5
350
4
Frequency
workhours
500
300
250
200
3
2
1
150
0
100
-80
20
30
40
50
60
70
80
90
100
110
-60
-40
-20
0
20
40
60
80
100
RESI1
120
lotsize
97.64
Normal Probability Plot of the Residuals
(response is workhour)
2
48.82
residual
Normal Score
1
0
0.00
-48.82
-1
-2
-97.64
-100
0
100
Residual
20
70
120
lotsize
The regression equation is
workhours = 62.4 + 3.57 lotsize
Predictor
Constant
lotsize
Coef
62.37
3.5702
S = 48.82
SE Coef
26.18
0.3470
R-Sq = 82.2%
T
2.38
10.29
P
0.026
0.000
R-Sq(adj) = 81.4%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
23
24
SS
252378
54825
307203
MS
252378
2384
F
105.88
P
0.000
Predicted Values for New Observations
New Obs
1
Fit
294.43
SE Fit
9.92
(
90.0% CI
277.43, 311.43)
(
90.0% PI
209.04, 379.81)
Values of Predictors for New Observations
New Obs
1
lotsize
65.0
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1. Here are three of the assumptions that need to be assessed before making inferences. In this
situation all three are met. For each, describe briefly how the data confirms the assumption.
Assumption 1: The relationship between the
mean of the response and the predictor is
linear.
Assumption 2: The standard deviation is
approximately the same no matter what the
predictor value is.
Assumption 3: Deviations from the
relationship have approximate normal
distribution.
2. Draw the estimated regression line on the scatterplot. (Use x = 20 first. Now, if x = 120 we
have a 100 lot increase, so the estimated time increases by 357.02. Using x = 20 and 120 you can
plot the line quite accurately.)
3. Consider the observation with 20 lots, taking 113 hours. What are the fit and residual for this
case?
4. Take a look at the plot of residuals vs. lotsize. What percent of the data falls within 1 standard
deviation of the line?
Within two standard deviations of the line?
5. Obtain a 90% confidence interval for the increase in hours due to an additional lot. Then
rewrite the interval to reflect the change in hours due to an additional 10 lots.
6. Consider testing H0: 1 = 0 HA: HA: 1  0. Justify a decision:
(All your decisions should be equivalent.)
a) using common sense
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b) based on an examination of the scatterplot
c) using your confidence interval
d) by computing a test statistic and P-value
7. Associated with the estimated intercept b0 = 62.37 you see a test statistic of t = 2.38 and a Pvalue of 0.000. Verify that the test statistic is correct for a null hypothesis of H0: 0 = 0. Use your
table to verify that the P-value is correct for the alternative HA: 0  0. (Minitab produces the Pvalue for the two-sided alternative.) What is the P-value for HA: 0 > 0 (which is a better
alternative as no one would ever believe that a negative intercept = start-up time could occur).
8. Toluca Company guidelines specify an average start-up time of 50 hours for this type of job.
The plant manager is concerned the start-up process is running too slowly (relative to the
standard set in the guidelines). State appropriate hypotheses and them using the data. Obtain a Pvalue and use it to determine how strong the evidence is that the start-up time of 50 is not met.
Hypotheses:
H0: __________
HA: __________
Test statistic t =
P-value =
9. Plug a lot size of x = 65 into the regression equation ŷ = 62.37 + 3.5702x to confirm the “fit”
given at the bottom of the output. (You obtain this in Minitab by filling in the “Predictions for
new observations” box in the “Options” dialog. There’s also a place to specify the confidence.)
10. Use the fit, its standard error, and the t-table to confirm the given 90% confidence interval.
Interpret this interval. Then draw the interval as a vertical line over x = 65 on the scatterplot.
Would you say there’s a 90% chance that a job of 65 lots would fall in this range? (It might be
wise also to compare the width of this interval with the residual standard deviation.)
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Solutions
1. The scatterplot and the residual vs. lotsize plot make it clear that a linear relationship is
appropriate. The normal probability plot suggests that the error distribution is at least
approximately normal. The residual vs. lotsiz plot confirms that the standard deviation of the
times is fairly constant over the different values for lotsize.
2. When x = 20, ŷ = 133.77. When x = 120 ŷ = 133.77 + 357.02 = 490.79.
3. Fit fit is 133.77. The residual is -20.77.
4. 16 of 25 = 68%. Then 24 of 25 = 96%.
5. The estimated value is 3.5702 with a standard error of 0.3470. For 90% confidence use
t = 1.714. The interval is 3.5702  1.714(0.3470) or 3.5702  0.5948. We’re 90% confident
that a job that is 1 lot bigger will take, on average, between 2.98 and 4.16 hours longer. For
10 additional lots the interval is from 29.8 to 41.6 hours.
6. a) Of course a bigger job takes longer. Not only is the null obviously false, even the
alternative is wrong (how could bigger jobs take less time?). b) The scatterplot makes it clear
that there’s a strong relationship. c) The confidence interval is nowhere near 0; so 0 can be
rejected at the 5% level (and much smaller levels). d) The test statistic is
(3.5702 – 0)/0.3470 = 10.29 which gives a P-value of 0.000+. Reject H0.
7. (62.37 – 0)/26.18 = 2.38. From the table the tail area is about 0.015, for a P-value of about
0.03. (This is what Minitab has, exact to 0.026.) For the right-sided alternative the P-value
would be 0.013.
8. H0: 0 = 50 HA: 0 > 50. The test statistic is t = (62.37 – 50)/26.18 = 0.47. The P-value for
such a test is then about 0.32.
9. The fit is, as shown in Minitab, is 62.37 + 3.5702(65) = 294.43.
10. 294.43  1.714(9.92) or 294.43  17.00 which is 277.43 to 311.43. This is a confidence
interval. Confidence intervals do not predict individual values, they estimate means. We’re
90% confident that the average job of 65 lots takes between 277 and 311 hours. Individual
jobs vary considerably more (the standard deviation is 49, which is far larger than the error
margin of 17 for this CI).
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