Review problems
Identify the error in the following analyses:
1) A confidence interval comparing the rate of developing a flu-like illness from a vaccinated and an unvaccinated group p1-p2 is determined to be (-.056, .016).
(a) This indicates that 90% of the sample proportions are contained in this interval.
The confidence interval aims to
specify p1-p2, the unique difference in the population proportions. It is not a confidence interval for a
difference in sample proportions or for a single proportion.
(b) This indicates that we are 90% confident that between 1.6% and 5.6% of the population developed a flu-like illness.
The confidence interval here is for
the difference in the population proportions of success, not for the overall
proportion of success of for just one of the population. That is why this interval may contain both positive
and negative values. Negative values
correspond to p2 > p1 and positive values
correspond to p2 < p1.
2) A confidence interval comparing the average number of words remembered for two different ways of presenting the words (familiar chunks and unfamiliar chunks) is determined to be (1.65, 8.60) with a p-value of .003.
(a) Since the interval does not contain 0, we are 95% confident that the scores for the first group are larger than the scores for the second group.
We are 95% confident that the difference in true treatment means is
greater than 0. It is important that
your interpretation does not sound like all values in the first population are
larger than all values in the second population. Of course, it is also best to spell out what
you mean by “first group” and “second group.”
In particular, it should be clear that the confidence interval estimates
the difference in population means or
the true treatment effect; it does
not merely pertain to the samples in this study.
(b) Those receiving the letters in familiar chunks will perform better 95% of the time than those receiving the letters in unfamiliar chunks.
The “95%” of the confidence
interval does tell you that something will be true “95% of the time.” The “of the time” refers to repeating the
random process many times and observing the resulting differences in sample
means. Your conclusion here is either
that the average recall score with the familiar packets is higher or it is not,
not that sometimes it is higher and sometimes it is not higher.
(c) The p-value of .003 indicates that there is a .003 probability that those receiving the letters in familiar chunks will perform better than those receiving the letters in unfamiliar chunks.
The interpretation reads too much
like the p-value is a probability statement about whether the null hypothesis
is true. The p-value tells you the
probability of observing a difference in sample means at least this large assuming the null hypothesis is true. There is a .3% chance that the difference in
mean recall scores between these sample groups would be larger than 5.12 if
there was actually no treatment effect.
3) Another study of housing prices (in
thousands of dollars) found the equation (for
predicted price = 30.15 + .0695 sq ft. r2 = 56.4%, p-value < .001
(a) The sample slope coefficient reveals that a house’s price goes up by $69.50 for each additional square foot of size.
the slope is .0695, and the
response variable is “thousands of dollars,” so this is close to a correct
interpretation, but it does not take the variability of the prices about the
regression line into account. A better
interpretation would be to say that the price of a house is increases by an average of $69.50 for each additional square foot of
size or at least that it’s the predicted
price that is associated is such an increase.
(b) If the technical conditions are met, the very small p-value suggests that there is no linear association between a house’s size and its price.
The very small p-value would
suggest strong evidence against the null hypothesis. But the null hypothesis says that the
population slope equals zero, which means that there is no linear association between house price and size in the
population. So this conclusion is backwards:
The small p-value actually provides strong evidence that there is a linear association between house
price and size in the population.
(c) If the technical conditions are met and if the p-value had been larger than .10, you could have concluded that the sample data provide strong evidence that there is no association between a house’s size and its price.
This conclusion commits the error
of “accepting H0.” Remember that a test of significance assesses the
strength of evidence against the null
hypothesis, not in favor of it. A large p-value would suggest no evidence
against the null hypothesis of no association, but it would not suggest strong
evidence of no association.
(d) Adding square footage to a house causes the price to increase by $69.50 on average.
The conclusion states that the
small p-value implies a causal relationship.
Since this was an observational study, such a conclusion is not valid.
4) Suppose we want to predict hiking time from hiking distance for Day Hikes in San Luis Obispo County and find predicted time = -1.27 + 31.5distance with r2 = .838. Identify at least one problem with each of the following interpretations.
(a) The slope shows that for each additional minute, we predict the hike is 31.5 miles longer.
The interpretation of the slope is
incorrect. The student has reversed the
role of the x and y variables. When interpreting the slope, you discuss how
a unit increase in the explanatory variable
is associated with a change in the predicted value of the response variable.
(b) The slope shows that a hiker’s time increases by 31.5 minutes for each additional mile.
This interpretation of the slope is
too definitive; it makes it sound as if
there is no variability present. A
better interpretation would be that the predicted time of a hike increases by
31.5 for each additional mile, or that the hike time increases on average by
31.5 minutes for each additional mile.
Although the y-value of the
corresponding position on the line changes by the exact same amount each time
we move 1 unit in x, this is not
exactly true for the data values themselves.
The interpretation of the slope applies to the line and not to
individual hikers.
(c) The predicted time for a 5-mile hike is about 156.
The calculation is correct here,
but what’s missing are the units: the predicted time f or a 5-mile hike is 156 minutes.
(d) About 84% of hikes have times that are correctly predicted by the line.
This is a completely incorrect
interpretation of r2. It
may well be that none of these hikes have a time that is perfectly predicted by
the line. That is not what r2 means.
(e) About 84% of the variability in hikes is explained by time.
This interpretation of r2 comes closer but is still
lacking because it does not indicate what variable about the hikes is being
explained. A good interpretation is that
about 84% of the variability in the times
of the hikes is explained by the least squares line based on the hikes’
distances.
Sample Problems from
a previous Stat 512 Final Exam (different instructor)
1. The
MINITAB output that follows resulted from taking observations on the percentage
of body fat taken from teenagers defined as clinically obese.
T Confidence Intervals
Variable N
Mean StDev SE Mean
99.0 % CI
Body Fat 18 28.95
4.53 1.07 (
25.86, 32.05)
a) Define the parameter(s) of interest.
m = mean percentage of body fat for
an appropriate population of teenagers defined as clinically obese (from what
population was this sample selected, all?)
b) Is the value of 28.95 a parameter or a statistic? Explain.
A
statistic since it was calculated from the sample of 18 teenagers
c) True or false:
i) For this interval to be valid, it is necessary that the population of percentages of body fat be normally distributed.
Probably
since the sample size is fairly small, we need to have the population
distribution of values to be well behaved or else the t-procedure will not be valid.
ii) A larger sample would make the population distribution more normal.
NO! The size
of the sample has no affect of the shape of the population (only on the shape
of the sampling distribution of sample means).
iii) For this interval to be valid, it is necessary that the population standard deviation of the percentages of body fat be known.
No, the
point of a t-interval is that we can
use s for the sample standard
deviation and we don’t need to know s, the population standard
deviation.
iv) 99% of the time, the mean percentage of body fat will fall between 25.86 and 32.05.
False (bad
interpretation of confidence, m is either in the interval or it’s
not, see above)
v) You can be absolutely sure that the mean percentage of body fat is between 25.86 and 32.05.
False, we
are only 99% confidence it is
vi) It is possible that the mean percentage of body fat of the population is not between 25.86 and 32.05.
True
vii) 99% of all intervals created in this fashion will contain the mean percentage of body fat.
True
viii) A 95% confidence interval obtained from the same data would be wider.
False, if
the confidence level is smaller, the width of the interval will be less (all
else staying equal)
ix) If a two-tailed hypothesis test of H0: m = 30 were performed using these data, H0 would be rejected.
False, since
30 is inside the CI, the two-sided p-value > .01.
2. “Photo-volume
and weight tables for
The regression equation is
Field = 34.4 + 1.14 Photo
Predictor Coef SE Coef T P
Constant 34.37 72.64 0.47
0.642
Photo 1.13710 0.08953 12.70
0.000
S = 209.8 R-Sq = 90.0% R-Sq(adj) = 89.4%
Analysis of Variance
Source DF SS MS F P
Regression 1 7099245
7099245 161.31 0.000
Residual Error 18 792170 44009
Total 19 7891416
Unusual Observations
Obs Photo
Field Fit SE Fit
Residual St Resid
10 1944
2000.0 2244.9 127.5 -244.9 -1.47 X
17 957
1876.0 1122.6 55.8 753.4 3.73R
R denotes an observation with a large standardized residual
X denotes an observation whose X value gives it large influence.
a) From the output, read or calculate the values of the following.
i) The y-intercept. 34.37
ii) The estimate of the average change in field volume for an increase of one in photo volume. 1.137
iii) The quantity that the least squares line minimizes. SSE = 792170
iv) The standard deviation in the estimate of the slope. .08953
v) The quantity that measures the proportion of error removed from the estimation of field volume by using a linear regression model with photo volume rather than using the sample mean field volume as the estimate. 90.0%
vi) The sample correlation coefficient. sqrt(.900) = +.949 (positive since slope is positive)
b) Test to see if there is a linear relationship between photo volume and field volume.
i) Define the parameter of interest. b = population slope
ii) What are the hypotheses? H0: b = 0, Ha: b ≠ 0
iii) Give the values from the MINITAB output of the two test statistics that may be used to perform the test. t = 12.70, p-value =. 000 or F = 161.31, p-value=.000
iv) Reach and justify a decision at a = .05. Provide an interpretation of the decision. Since p-value < .05, we reject the null hypothesis. We have convincing evidence that there is a linear relationship between photo volume and field volume in the population.
3. For a class project, a student took a sample of students and determined their age, gender, whether they belonged to a fraternity or sorority, how many years they had been attending college, and the number of alcoholic drinks they had in a week. A regression analysis by MINITAB resulted in the following output.
The
regression equation is
num.drinks/week
= 2.6 + 1.01 age - 0.01 sex - 9.67 frat/sor
- 0.10 yr. in school
Predictor Coef
SE Coef T P
Constant 2.59 12.48 0.21
0.838
age 1.0091 0.7192 1.40
0.181
sex -0.013 1.671 -0.01
0.994
frat/sor -9.666 1.811 -5.34
0.000
yr.
in s -0.102 1.164 -0.09
0.931
S
= 3.571 R-Sq = 66.0% R-Sq(adj) = 56.9%
Analysis
of Variance
Source DF SS MS F P
Regression 4
371.26 92.81 7.28
0.002
Residual
Error 15 191.29 12.75
Total 19 562.55
a)
Perform a test to
see if at least one of the predictors aids in the prediction of the response.
i)
What are the
hypotheses? H0: b1=b2=b3=b4 = 0 vs. Ha: at
least one bI ≠ 0
ii)
Make and justify
your decision at a = .05 and provide an interpretation. F =
7.28, p-value = .002, since p-value < .05 we reject the null hypothesis and
conclude that at least one of the predictors (age, sex, fat/sor, year in
school) aids in the prediction of the response (number of drinks per week)
b)
Perform a test to
decide if sex is a significant part
of the current model.
i)
Give the
hypotheses H0: b2 = 0 vs. Ha: b2 ≠ 0
ii)
Make and justify
your decision at a = .05 and provide an interpretation. t = -.01
and p-value = .994, with such a large p-value, we fail to reject the null
hypothesis. We conclude that sex is not
a significant part of the current model (it does not aid in the prediction of
num of drinks / week once we know the other explanatory variables).
c)
The student
believed that whether a student belonged to a fraternity or sorority was they
only variable that would predict alcoholic ingestion, and that the other
variables would not be necessary components of the model. Based on the t-tests and their associated
p-values (use a = .05), does the student have sufficient justification for her
claim? Explain. Frat/Sort does
appear to have the strongest effect but we would need to remove the other
variables one at a time, e.g., sex, year in school as perhaps another variable
like age will also become significant.
Additional Review Problems
1) Suppose that instructors A, B, and C are each
teaching three large sections of a course, and each instructor wants to study
whether the mean exam scores differ significantly across the three
sections. Suppose that each takes a
random sample of ten students, and calculates the following descriptive
statistics:
|
|
A1 |
A2 |
A3 |
B1 |
B2 |
B3 |
C1 |
C2 |
C3 |
|
Sample size |
10 |
10 |
10 |
10 |
10 |
10 |
10 |
10 |
10 |
|
Sample mean |
50 |
60 |
70 |
50 |
60 |
70 |
57 |
60 |
63 |
|
Sample std dev |
24 |
24 |
24 |
5 |
5 |
5 |
5 |
5 |
5 |
(a) Based on these statistics, which instructor has
the strongest evidence that the mean scores differ significantly across his/her
three sections? Which has the least
evidence? Explain your answers.
Since the sample
sizes are the same, the question becomes – which has the largest different in
means compared to the amount of variability in the data. Between A and B, the differences between the
three sample means are larger, but for instructor B these differs appear much
more significant since there is so much more consistency in the exam scores (as
shown by the smaller standard deviations).
Instructor B will have strong evidence than instructor C since the
differences in the group means are larger.
So instructor B has the strongest evidence overall. It’s hard to tell between A and C since both
the differences in means and the sample standard deviations differ – that’s why
we need ANOVA!
2) Consider the following
four data sets, each consisting of four (x,
y) data points:
A: (1,3) (2,5) (3,6) (4,8) B: (1,4) (2,7) (3,2) (4,4)
C: (1,8) (2,6) (3,2) (4,3) D: (1,5) (2,3) (3,5) (4,2)
Based on the changes in the x and y values, arrange these data sets in order from the most negative
correlation to the most positive.
Explain your reasoning.
In Data Set A, as x increases from 1 to 4, y also increases by a similar amount
each time (1 or 2), so r will be
close to 1.
In Data Set B, as x increases, y increases and decreases.
The changes in y are 3, -5,
and 2. The overall tendency is negative
but the association should be pretty weak.
In Data Set C, as x increases, y tends to decrease (-2, -4, 1).
This should result in a fairly strong negative correlation.
In Data Set D, as x increases, y increases and decreases (-2, 2, -3). This should result in a
moderate negative correlation.
Ordering, from
most negative to most positive: C, D, B, A.
Note: A: r = .998; B: r = -.313; C: r = -.891; D: r = -.602
3) An article in the May 24,
2004 issue of Sports Illustrated
raised two separate questions about seven-game series in professional team
sports. One question concerns the proportion of seven-game series that
have gone to the full length of seven games.
The article reported that through the year 2003, 44 of 131 (34%) series
went to the full length in baseball, compared to 111 of 471 (24%) in hockey and
85 of 303 (28%) in basketball.
(a) Conduct a chi-square
analysis of whether these percentages differ more than would be expected by
random variation. Begin with graphical
displays and numerical summaries, and then proceed to a chi-square test. Which type of chi-square test did you do?
Summarize your conclusions.
Two-way table:
|
|
Baseball |
Hockey |
Basketball |
|
Full length |
44 |
111 |
85 |
|
Less than full
length |
87 |
361 |
218 |
|
Total |
131 |
471 |
303 |
Segmented bar
graph:
Baseball has the
largest proportion of series that lasted for the full seven games (.336),
following by basketball (.281) and then hockey (.236).
Let’s pretend
these results are independent random samples from the populations of games in
each sport and focus on the proportion of each series in the population that
lasts the full seven games (homogeneity of proportions).
Let pbaseball represent the probability that a baseball series
lasts for the full seven games, and similarly define phockey and pbasketball.
We can conduct a chi-square
test of H0: pbaseball = phockey = pbasketball
vs. Ha:
at least one p differs from the rest.
The expected
counts are all at least five (smallest = 35.74). If we regard the observed
series as independent random samples from each sport’s process, we can perform
the Chi-square test. Minitab output:
Expected counts are printed below
observed counts
Chi-Square contributions are printed
below expected counts
baseball hockey basketball
Total
1 44 111 85
240
34.74 124.91
80.35
2.468 1.548 0.269
2 87 360
218 665
96.26 346.09 222.65
0.891 0.559 0.097
Total 131
471 303 905
Chi-Sq = 5.831, DF = 2, P-Value = 0.054
The p-value of .054 is not terribly
small. It says that if the three sports
all had the same probability that a series would last for the full seven games,
then there’s about a 5.4% chance that randomness alone would produce sample
proportions as different as the three reported here. This p-value is somewhat small but not terribly
small (.05 
p-value .10), so the data
provide some, but not strong, evidence that the sports differ with regard to probability
that a series lasts for the full seven games.
(b) Comment on whether these
data come from random samples or from randomization to groups, or whether the
randomness is hypothetical here.
We analyzed all
playoff series in these sports, so there is no real randomness here. We have to assume that these series can be
regarded as a random sample from each sport’s process.
(c) The other question posed
by the article compares the proportion of “game sevens” that are won by the
home team across these sports. The article reported that 23 of 44 (52%)
were won by the home team in baseball, compared to 70 of 111 (63%) in hockey
and 70 of 85 (82%) in basketball. Analyze these data to assess whether
they provide evidence that the three proportions differ significantly, and
write a paragraph or two summarizing your conclusions.
Two-way table:
|
|
Baseball |
Hockey |
Basketball |
|
Won by home team |
23 |
70 |
70 |
|
Won by visiting
team |
21 |
41 |
15 |
|
Total |
44 |
111 |
85 |
Segmented bar
graph:
These proportions appear to differ fairly considerably. Basketball has the highest proportion of
seven-game series won by the home team (.824), with hockey in the middle (.631)
and baseball much lower (.523). All
sports do see the home team win more than half of the game sevens. Conducting a chi-square test of the null
hypothesis that all three sports have the same probability that a game seven is
won by the home team produces the following Minitab output:
Expected counts are printed below
observed counts
Chi-Square contributions are printed
below expected counts
baseball hockey basketball
Total
1 23 70 70 163
29.88 75.39 57.73
1.586 0.385 2.608
2 21
41 15 77
14.12 35.61 27.27
3.356 0.815 5.521
Total 44 111 85 240
Chi-Sq = 14.272, DF = 2, P-Value = 0.001
The p-value is quite small (.001), so the
observed data would be very unlikely to occur by chance alone if the three
probabilities were equal. Thus, we have
strong evidence that the three sports do not have the same probability that a
game seven would be won by the home team.
This conclusion is valid as long as the data observed are representative
of the overall process for each sport.
Note that the biggest discrepancy comes in the basketball, won by
visiting team category where we observed far fewer games (15) than we would
have expected if the sports were performing the same as each other (27.27).
4) A student conducted a
student to examine the ages of people who joined a local health club. The participants were chosen by systematically
sampling the men and the women who joined the health club in August and
September 2004. The data are in the
Minitab worksheet GymMembership.mtw.
Analyze the data to compare
the mean ages of the men and women and also the mean ages of those who joined
the club in August and in September, conditional on the other variable. Produce and comment on numerical and
graphical summaries, state hypotheses, check the technical conditions for each
procedure. Also comment on whether there
appears to be a statistically significant interaction between gender and month
joined and which factor (gender or month) appears to be more strongly related
to the ages of new members.
Response variable
= age
Explanatory
variable 1 = gender
Explanatory
variable 2 = month joined
The sample male
and female age distributions look very similar.
The sample means (35.17 years vs. 36.82 years) are very close and the
variability is similar (s1 = 14.58 years and s2=16.58
years). Both distributions have a
skewness to the right and perhaps a bimodal shape with peaks around 24 years
and 40 years.
The sample August
and September age distributions look very similar. The sample means (36.90 years vs. 35.02
years) are very close and the variability is similar (s1 = 15.32
years and s2=15.91 years).
Both distributions have a skewness to the right and a bimodal shape with
peaks around 24 years and 50 years.
Treating the data
as a random sample of new members, we can use two-way ANOVA to analyze the
“effect” of each EV on the response, conditional on the other variable.
Since the sample
sizes are unequal in this observational study, we will use the “General Linear
Model” command in Minitab. We saw above
that the variability was similar in each group and the sample sizes are large enough
(over 100 in each group) that we will not be concerned with the normal
population condition (good thing since the samples provide strong evidence that
these population distributions are not normal).
General
Linear Model: age versus gender, month
Factor Type
Levels Values
gender fixed
2 female, male
month fixed
2 Aug, Sep
Analysis
of Variance for age, using Adjusted SS for Tests
Source DF
Seq SS Adj SS Adj MS
F P
gender 1
170.0 155.2 155.2
0.64 0.426
month 1
208.4 208.4 208.4
0.85 0.356
Error 249
60720.5 60720.5 243.9
Total 251
61099.0
S
= 15.6159 R-Sq = 0.62% R-Sq(adj) = 0.00%
This output
indicates that we would fail to reject H0: mf = mm due to a large p-value of .426. There is not significant evidence of a
difference in the mean age of male and female new members after adjusting for
the month joined. This output also
indicates that we would fail to reject H0: mS = mA due to a large p-value of .356. There is
not significant evidence of a difference in the mean age of new members that
join in August or September, after adjusting for the age of the member.
Considering an
interaction between gender and month:
We do see some
graphical evidence of an interaction. It
appears that the male and female ages are similar in August but much different
in September. However:
Analysis of Variance for
age, using Adjusted SS for Tests
Source DF
Seq SS Adj SS Adj MS
F P
gender 1
170.0 169.9 169.9
0.70 0.405
month 1
208.4 207.1 207.1
0.85 0.358
gender*month 1
218.8 218.8 218.8
0.90 0.345
Error 248
60501.7 60501.7 244.0
Total 251
61099.0
We see
that this interaction is not statistically significant (p-value = .345) at any
reasonable level of significance, so we will fail to reject the null hypothesis
of no interaction. None of these factors
seem related to age, but if we had to pick one, we would focus on the
interaction since the p-value is smallest (and when we have an interaction it
becomes more problematic to interpret the “main effects.”)
5) In a “matched pairs”
experiment, each subject receives both treatments, in random order. This allows us to see if the treatment is
consistently effective, comparing each person to themselves, instead of across
individuals, allowing a more direct comparison and “controlling for” the person
to person variability. To analyze the
data, we just take the differences in the results for each person and see if
the average difference is significantly different from zero. A “repeated measures” design is just this
idea extended to 2 or more treatments.
“Blocking” is the same logic, but we group experimental units that are
very similar to each other instead of using the same unit more than once. We have still minimized the variability for
trying to detect the treatment effect itself.
In both these cases, we include “subjects” or “blocks” as one of the
variables in an ANOVA analysis (assuming a quantitative response).
Researchers who are studying a new shampoo
formula plan to compare the condition of hair for people who use the new
formula with the condition of hair for people who use the current formula. Twelve volunteers are available to
participate in this study. Information
on these volunteers (numbered 1-12) is shown in the table below.
|
Volunteer |
Gender |
Age |
|
1 |
Male |
21 |
|
2 |
Female |
20 |
|
3 |
Male |
47 |
|
4 |
Female |
60 |
|
5 |
Female |
62 |
|
6 |
Male |
61 |
|
7 |
Male |
58 |
|
8 |
Female |
44 |
|
9 |
Male |
44 |
|
10 |
Female |
24 |
|
11 |
Male |
23 |
|
12 |
Female |
46 |
(a) The researchers want to
conduct an experiment involving the two formulas (new and current) of
shampoo. They believe that the condition
of hair changes with age but not gender.
Because researchers want the size of the bocks in an experiment to be
equal to the number of treatments, they will use blocks of size 2 in their
experiment. Identify the volunteers (by
number) that would be included in each of the six blocks and give the criteria
you used to form the blocks.
|
Block |
Volunteers |
Ages |
|
1 |
1,2 |
20,21 |
|
2 |
10,11 |
23,24 |
|
3 |
8,9 |
44,44 |
|
4 |
3,12 |
46,47 |
|
5 |
4,7 |
58,60 |
|
6 |
5,6 |
61,62 |
Since these researchers believe that the condition of hair changes
with age but not gender, the volunteers are sorted from youngest to oldest. The
volunteers in the sorted list are paired to form six blocks of size two. More
specifically, the youngest two volunteers are placed in the first block. The
next two volunteers in the sorted list are placed in the second block. This
pairing continues until all six blocks of two are formed, with the oldest two
volunteers in the sixth block.
(b) Other researchers believe
that hair condition differs with both age and gender. These researchers will also use blocks of
size 2 in their experiment. Identify the
volunteers (by number) that would be included in each of the six blocks and
give the criteria you used to form the blocks.
|
Block |
Volunteers |
Ages |
|
Female 1 Female 2 Female 3 Male 1 Male 2 Male 3 |
2, 10 8, 12 4, 5 1, 11 3, 9 6, 7 |
20, 24 44, 46 60, 62 21, 23 47, 44 61, 58 |
Since these researchers believe that the condition of hair changes with
both age and gender, the women are sorted from youngest to oldest and then the
men are sorted from youngest to oldest. The women (men) in the sorted list are
paired to form the blocks of size two. More specifically, the youngest two
women (men) are placed in a block. The next two youngest women (men) are placed
in another block. Finally, the oldest two women (men) are placed in another
block.
(c) The researchers in (b)
decide to select three of the six blocks to receive the new formula and to give
the other three blocks the current formula.
Is this an appropriate way to assign treatments? If so, describe a method for selecting the
three blocks to receive the new formula.
If not, describe an appropriate method for assigning treatments.
No, the researchers in part (b) should not randomly select three
blocks to receive the new formula and then give the current formula to the
other three blocks. They blocked on both age and gender to form homogeneous
groups because they believe hair condition differs with both age and gender.
Giving the youngest or oldest women (men) the same formula defeats the purpose
of blocking. In a block design, randomization should be carried out separately
within each block. That is, for each block, two random numbers are generated
(via a random number generator or a table of random digits) and assigned to the
two volunteers. The volunteer with the smallest random number is given the new
formula and the other volunteer is given the current formula.
Review problems from text
1) p. 236, problem 7 see back of book
p. 244, problem 43
The 95% confidence
interval for the difference in the two population proportions (p82-p74) that would report watching no television:
.031 - .038 ± 1.96sqrt(.031(.969)/350 + .038(.962)/1965) =
-.007 ± .02 => (-.027, .013).
We are 95%
confident that the proportion in 1982 is up to .013 higher but could also be
.027 smaller than the proportion in 1974. Since zero is included in this confidence
interval, the difference is not statistically significant at the 5% level of
significance.
p. 244, problem 47 see back of book (don’t worry about knowing what McNemar’s test
is, we would just say we don’t know a method for comparing dependent samples
like this)
2) p. 288, problem 5 see back of book, there is definitely an association here –
the probability of having young children differs greatly depending on whether
the woman has grey hair. However, this is an observational study so no causal
conclusions can be drawn, no matter how much you might believe it! Perhaps the women with grey hair are older
and have reach child bearing age but the other women living on her block
without gray hair are much younger.
3) p. 351, problem 33 see back of book
(a) The units of
slope are dollars/x unit so if we convert the y variable units, the slope will
go through the same conversion.
(b) The
correlation coefficient does not change with changes in scale.
p. 351, problem 39 see back of book
(a) no constant
linear trend
(b) not constant
variance in the contributions with income level (megaphone effect)
(c) not constant
linear trend
(d) not constant
linear trend
4) p. 434, problem 45
5) p. 481, problem 17 see back of book