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ISCAM Chapter 6 Exercises Solutions
19. (a) An ANOVA would not be appropriate here since the distributions of rainfall amounts were strongly
skewed to the right and the standard deviations were very different from each other.
(b) After taking log 10(rainfall):
Both groups appear to be reasonably modeled by a normal distribution and the standard deviations are also
much more similar (.7130/.6947 < 2). Combined with the fact that this was a randomized experiment, the
conditions for ANOVA appear to be met.
Let d represent the treatment effect on the mean of log(rainfall) from the cloud seeding.
H0 : d=0 (there is no treatment effect)
Ha: d>0 (the cloud seeding increases the average log rainfall)
With a test statistic of F=6.47 and a p-value of .014 we have fairly strong evidence against the null
hypothesis. We would reject the null hypothesis at the 5% level, but not quite at the 1% level. If there was
no treatment effect, we would observe a difference in the sample means at least this large in about 1.4% of
randomizations.
(c) Minitab output for pooled two-sample two-sided t-test:
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The p-values are the same, and the test statistics are related by: t 2 = (2.54)2 = 6.47 = F.
20. (a) This is an experiment because the list given to the subjects was randomly assigned by the students.
(b) The observational units are the people.
(c) The explanatory variable is the type of words (categorical with 4 categories) and the response variable is
how many words were remembered (quantitative).
(d) Graphical and numerical summaries:
There is a lot of overlap in the four distributions of number of words memorized. There appears to be a
tendency for fewer words to be remembered with the abstract lists, and perhaps a slight tendency for more
words to be remembered with the short lists. The spread and shape of the data also appear similar (shape is
a little hard to discern, perhaps skewed to the left).
(e) Let m LA represent the treatment mean number of words remembered with the list of long, abstract
words. Similarly for m LC, m SA, and m SC.
H0 : m LA = m SA = m LC = m SC (the underlying treatment means are the same for the 4 treatments)
Ha: at least one treatment results in a different average number of words remembered
The standard deviations are reasonably similar (1.553/1.235 < 2) and this was a randomized experiment.
It is fairly questionable whether the treatment distributions follow a normal distribution. Since the sample
sizes are moderate, we should proceed with caution.
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Minitab output:
An F statistic of 3.49 is moderately surprising if the null hypothesis is true based on randomization alone.
With a p-value of .023, we would reject the null hypothesis at the 5% level but not the 1% level. There is
some evidence that the treatment mean number of words memorized differ depending on the type of list.
We can draw a cause and effect conclusion since this was an experiment, which applies for students on this
campus (we have to be cautious about generalizing to this larger population since the sample was chosen by
convenience). But this p-value may be inaccurate since the treatment distributions do not appear to be
normal.
(f) We are justified in drawing a cause and effect relationship between word type and number of words
memorized since this was a randomized experiment.
(g) Minitab output:
Whether the words are abstract or concrete appears to have a much stronger evidence of an effect (p-value
= .005) than the length of words (p-value = .224). The differences in the group means for the abstract and
concrete words were larger than we would expect by chance alone.
Plots of just the means for these two variables support this observation.
25. Situation B would have higher power. It is more likely that we will reject H0 : m 1 =m 2 =m 3 in situation B
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because the population means are more different (assuming the sample size and standard deviations are
similar in the two situations).
26. (a) Comparing the genders:
The distribution of the ages between the males and females appear quite similar in these samples.
Let m f represent the population mean age of all new female members to this club and m m the mean age of
all new male members to this club.
H0 : m f = m m (the population mean ages for new male and female members are the same)
Ha: m f ≠ m m (the population mean ages differ)
The sample standard deviations are similar (16.58/14.58<2). It appears that the population distributions are
skewed to the right but the sample sizes are large so we will be less concerned about this condition. The
data are systematic samples for each gender but we will consider them independent random samples.
Minitab output:
With such a large p-value (.404 > .1) we fail to reject the null hypothesis. We do not have convincing
evidence that the population means differ. If the population means were the same, we would observe sample
means at least this different in 40.4% of random samples. In conclusion, we do not have evidence that the
mean age of all new female members and all new male members differ.
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mean age of all new female members and all new male members differ.
(b) Comparing the months:
There appears to be very slight tendency for these August new members to be a bit older than these
September new members. The variability is quite similar in the two distributions and the shape is also
similarly (bimodal and a bit skewed to the right).
Let m A represent the mean age of all new August members to this club and m S the mean age of all new
September members to this club.
H0 : m A = m S (the population mean ages for new members in the two months are the same)
Ha: m A ≠ m S (the population mean ages differ between August and September)
The sample standard deviations are similar (15.91/15.32<2). It appears that the population distributions are
not normal but the sample sizes are large so we will be less concerned about this condition. The data are
systematic samples for each month but we will consider them independent random samples.
Minitab output:
With such a large p-value (.339 > .1) we fail to reject the null hypothesis. We do not have convincing
evidence that the population means differ. If the population means were the same, we would observe sample
means at least this different in 33.9% of random samples. In conclusion, we do not have evidence that the
mean age of new August members and September members differ in the overall process (assuming the data
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mean age of new August members and September members differ in the overall process (assuming the data
in this year are representative).
27. (a) ANOVA is more appropriate when we have more than 2 populations or treatment groups to compare.
(b) A two-sample t-test would be more appropriate if the population standard deviations are different or if
we want to determine a one-sided p-value.
28. (a) Descriptive statistics:
diet
Atkins
Ornish
Weight Watchers
Zone
N
40
40
40
40
Mean
2.057
3.28
2.985
3.175
StDev
4.761
7.29
4.854
6.019
Median
0.000
0.000
2.000
0.600
IQR
4.500
5.58
5.350
5.325
The means, medians, standard deviations, and IQR’s are all smaller than they were when only subjects who
completed the diet were considered. The means are also closer together than they used to be.
(b) Boxplots:
These boxplots reflect the smaller medians and IQR’s, as mentioned in (a). They also reveal more outliers
than did the boxplots based only on subjects who completed the diet plan.
(c) Boxplots:
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The “completion 0” group consists of subjects who completed the program. The distribution of their weight
loss appears to be slightly skewed to the right, with a few large outliers. About 3/4 of these subjects
experienced a positive weight loss. The “completion 1” group consists of subjects who did not complete the
study, so their weight losses were all assumed to be zero. Thus, their boxplot consists of nothing more than
the value zero.
(d) ANOVA table:
Source
diet
Error
Total
DF
3
156
159
SS
37.4
5286.3
5323.7
MS
12.5
33.9
F
0.37
P
0.776
The F-statistic of 0.37 is very small, and the p-value of .776 is very large. The data provide no evidence of
a treatment difference in mean weight loss among the four diet plans.
(e) The conclusion does not change at all, but the F-statistic is even smaller and the p-value even larger than
before.
29. Analyzing the weight losses after 2 months produces boxplots and descriptive statistics that do not
reveal much of a difference in weight loss distributions among the four diet plans:
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diet
Atkins
Ornish
Weight Watchers
Zone
N
40
40
40
40
Mean
3.627
3.625
3.465
3.795
StDev
3.255
3.422
3.830
3.590
Median
3.250
3.300
2.900
3.350
IQR
6.950
5.650
5.100
5.725
The ANOVA table (see below) gives an extremely high p-value of .981, suggesting that the experimental
data provide essentially no evidence of a difference in average weight loss among these four diet plans after
two months of the study.
One-way ANOVA: weight loss (2 mos) versus diet
Source
diet
Error
Total
DF
3
156
159
SS
2.2
1944.7
1946.9
MS
0.7
12.5
F
0.06
P
0.981
Repeating this analysis using weight loss after 6 months as the response variable produces very similar
results, although some outliers who have lost a good bit of weight emerge with each plan). Again with this
variable the experimental data provide no evidence of a significant difference among any of the four diet
plans.
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diet
Atkins
Ornish
Weight Watchers
Zone
N
40
40
40
40
Mean
3.213
3.56
3.535
3.353
StDev
4.858
6.67
5.645
5.668
Median
0.300
0.000
1.050
1.200
IQR
5.425
5.28
5.675
5.300
One-way ANOVA: weight loss (6 mos) versus diet
Source
diet
Error
Total
DF
3
156
159
SS
3.2
5150.6
5153.9
MS
1.1
33.0
F
0.03
P
0.992
32. (a) This conclusion would not be valid. A correlation close to zero implies very little linear association,
not a negative association.
(b) We would expect the sample result to be more statistically significant when arising from a larger sample
size.
(c) It would not make sense to calculate the correlation coefficient between class size and gender because
gender is categorical and not quantitative (the correlation coefficient is only calculated between two
quantitative variables).
33. (a) The only change will be that the intercept b0 is larger (the graph shifts up vertically).
(b) The only change will be that the intercept b0 is lower (the graph shifts to the left and so the least squares
line crosses the y-axis at a lower value).
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line crosses the y-axis at a lower value).
(c) The slope b1 becomes larger (line becomes steeper).
(d) The slope b1 becomes smaller (line becomes flatter).
(e) The slope b1 becomes closer to zero (line becomes flatter).
34. (a)
(b) No, in that case we would get
. These equations differ by
whether r is in the numerator or the denominator.
(c) They will be the same equation when r = ± 1.
(d) The calculation of r is in not effected since the roles of x and y are separable and related through
products (where order of operation doesn’t matter).
35. (a) b1 = (.711)(3.445/5.004) = .49
b0 = 28.50 - .49(67.75) = -4.70
= -4.70 + .49 height
(b) For each additional inch in height, we predict a person’s foot length to be .49 cm longer.
(c) -4.70 + .49(66) = 27.6 cm.
(d) slope: cm/inch
intercept: cm
correlation coefficient: no units
(e) foot = (height-38.3)/1.03 = .97height – 37.2
(f) The lines are not the same.
39. (a) There appears to be a weak negative linear relationship with 2 outliers (1940, 1964), indicating that
the date in of the ice break is occurring progressively earlier in the year as time goes on.
(b) Fitted Line Plot
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r = -.312, r 2 = 9.7%
There does appear to be a weak (r = .312) linear relationship between date of ice breaking and year, but that
only explains 9.7% of the variability in the ice break data. Each year that passes predicts that the date will
occur .07 days closer to April 1.
(c) The residuals vs. year plot does not reveal any unusual patterns apart from the 2 outliers. The normality
of the histogram of the residuals also supports the assumption of normality in the distribution of the
response for each year. The technical conditions appear to be met. We are assuming the observations from
different years are independent.
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Let b1 represent the true population slope (where the observed slope differs from this value by measurement
error)
H0 : b1 = 0 (there is no relationship between ice break date and time
Ha: b1 ≠ 0 (there is a relationship)
Minitab output:
Note: You could also consider testing for a negative relationship since the researchers suspected the
date was coming earlier over time; in which case, the p-value would be .0015.
With a test statistic of t 0 = -3.04 and a two-sided p-value of .003, we have strong evidence against the null
hypothesis. If there was no underlying relationship between the ice break date and time, we would observe
a relationship at least this strong in only .3% of random samples from such a process. We reject the null
hypothesis and conclude that there is a linear association between ice break date and year.
(d) The p-value reveals strong evidence of an association. To measure the strength of the association,
examine the correlation coefficient. Because the correlation is so small, we do not see evidence of a strong
association here.
(e) Slightly, the r 2 value (9.7%) is pretty low suggesting that 9.7% of the variability in the dates in explained
by the regression on year. We have reduced the unexplained variation by just 9.7% over using to make
our predictions. But the p-value is small enough that we have strong evidence that there really is a
relationship, albeit a weak one, between year and date of ice breaking.
(f) The predicted date for 2005 is: 178-.073(2005) = 31.63 days starting on April 1, which is about May 2.
The predicted date for 2020 is: 178-.073(2020) = 30.54 days starting on April 1, which is about May 1.
However the data used to create the regression model only extend to 2004 and we should be cautious in
extrapolating this relationship too far out in time. Also, the relationship is quite weak, so there is a lot of
uncertainty in these predictions.
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40. (a) Temperature is being used as the response variable and frequency of chirping as the explanatory
variable.
(b)
The scatterplot shows a strong positive linear relationship between temperature and frequency of chirping.
(c) The correlation coefficient of r =.978 confirms our observation of a strong positive linear relationship.
(d) The regression equation is temp = 35.8 + 0.251 chirps/min
For each additional chirp per minute by the crickets, we predict the temperature to be .251 degrees warmer.
(e) r 2 = 98.7%. The regression on chirps/min explains 98.7% of the observed variation in temperatures.
(f) 35.8 + .251(120) = 65.9o F
41. (a) The graph of the residuals vs. chirps/min does not show any marked evidence of curvature or
unequal variance. This normal probability plot of the residuals also supports assuming normality of the
response at each x value.
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(b) Let b1 represent the true slope in the relationship between temperature and chirps per minute.
H0 : b1 = 0 (there is no relationship between the chirping frequency and the temperature)
Ha; b1 > 0 (there is a positive relationship)
The one-sided p-value would be .000/2 » 0.
With a test statistic of t 0 = 24.89 and a p-value < .001, we reject the null hypothesis. If there was no real
relationship between the chirping frequency and temperature, then we would almost never observe a sample
relationship this extreme by chance alone. In conclusion, there is extremely strong evidence of a positive
association between chirps/min and temperature.
(c) With n = 30-2 = 28 and 95% confidence, t* = 2.048
b1 ± t* SE(b1 ) = .2512 ± 2.048(.01009) = (.2305, .2708)
We are 95% confident that the true regression slope between temperature and chirps/min is between .2305
and .2708.
(d) Using Minitab:
We are 95% confident that the average temperature when the cricket is chirping at 120 chirps per minute is
between 65.46 and 66.38 degrees Fahrenheit.
(e) We are 95% confident that the temperature on a particular night when the cricket is chirping at 120
chirps per minute is between 63.39 and 68.45 degrees Fahrenheit.
(f) The midpoints are the same (65.92) but the prediction interval is wider than the confidence interval since
it is more difficult to predict an individual value than to predict an average value.
(g) The prediction interval would be narrowest at =117.07, because the standard error for the fitted value
is smallest when the prediction is made at .
50. (a) Graphical displays:
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The residual plots suggest a slight increase in the variability with larger body weights and the normal
probability plots provides evidence that the residuals do not follow a normal distribution. Both of these
observations suggest trying a transformation.
(b)
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The transformed variables appear to better follow the basic regression model.
(c) Let b1 represent the true population regression slope between log backpack weight and log body weight
H0 : b1 = 0 (no relationship between log backpack weight and log body weight)
Ha: b1 ≠ 0 (there is a relationship)
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With a p-value of .024, the relationship would be considered statistically significant at the 5% level.
However, we might consider analyzing the data without the three lowest values for log backpack weight (the
three lightest students) as they appear to be exerting a strong influence on the model.
(d) predicted log backpack weight = -.445 + .669log(150) = 1.01
predicted backpack weight = 101.01 = 10.25 lbs.
(e) Using log(150) = 2.176 as the predictor value, Minitab calculates
So that the 90% prediction interval for the backpack weight is (10 .6268 , 101.3926) = (4.23, 24.69).
We are 90% confident that a 150 pound individual will carry a backpack that weighs between 4.23 pounds
and 24.69 pounds.
51. (a) Graphical displays:
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The regression equation is
ln(backpack wgt) = - 1.03 + 0.669 ln(body wgt)
Predictor
Constant
ln(body wgt)
S = 0.528298
Coef
-1.025
0.6686
SE Coef
1.463
0.2916
R-Sq = 5.1%
T
-0.70
2.29
P
0.485
0.024
R-Sq(adj) = 4.1%
(b) The regression coefficients have changed, but the equation will be the same if we back-transform
appropriately.
(c) r 2 (and p-value) is the same.
(d) Predicting the ln backpack weight for ln(150) = 5.01 gives -1.03 + .669(5.01) = 2.32, so the predicted
backpack weight is e 2.32 » 10.18 pounds. Except for rounding errors, this is the same prediction as with log
base 10.
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(e) Minitab output:
The 90% prediction interval for backpack weight is e 1.4429, e 3.2062 = (4.23, 24.69), also the same.
52. (a) The percentage applies to “of variation in the response variable” not “of points”
(b) It is also not the percentage “of correct predictions.” If “correct prediction” means that the point falls
exactly on the line, it is likely that no predictions will be correct.
(c) Because r 2 arises from a squared term it cannot be negative
(d) We really need to check the residual plots to decide if the linear model is appropriate; we cannot rely on
r 2 . In fact, the data values can follow a curved pattern and still produce a large value of r 2 .
(e) A large r 2 value can indicate a statistically significant relationship but does this does not necessary
imply causation. We would have to consider the type of study conducted. If the study is observational, we
can not draw a cause-and-effect conclusion no matter how large r 2 might be.
53. (a) The sum of residuals is:
=
. Now, we know that the intercept
coefficient for a least-squares line is
, so this becomes
=0
(b) It follows that the mean will also be equal to zero because the mean equals the sum divided by n.
(c) This does not follow as the median and the sum of the residuals are not directly related. It is quite
possible for the sum to equal zero but the median not to (e.g., residual values of -3, 1, and 2).
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