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98-2 P (X < c) = .95?
105-2 the probability of obtaining an F-ratio exceeding
1278-1 The appropriate statistical test for this comparison is:
1430-1 chicken soup consumption between sexes statistically significant?
1536-1 Set 95% confidence limits for the difference between means
1612-1 F > 9.48773 and the computed value of F from the data is .86.
1614-2 and your P-value satisfies .01 < P < .05.
1650-1 What useful information can you supply future investigators when
1687-4 "In ANOVA, if we wish to investigate the difference among five"
1872-3 Suggest some factors that might influence variability in weights
1873-2 Data containing less variability than would be expected by chance
1895-1 You are to develop plans for a follow-up study.
1917-1 s**2 = (Sum of Squares)/(n-1)
1953-2 The degrees of freedom associated with a error (or within)
1953-3 the degrees of freedom associated with within sum of squares depend
2102-3 which of the following is an estimate of the variance of
2118-1 MEAN SQUARE
2548-1 "If known variation is large compared to unknown variation,"
2550-1 sees that his MS within groups is larger than his MS among groups.
2551-1 What decision would be made regarding H(0):
2552-1 Then we know that the experimenter used
2553-1 Items i. thru iii. are based on a teaching experiment involving
2567-1 what is the number of degrees of freedom for
2570-1 How many degrees of freedom do we have for the within groups sum of
2577-2 F = 3.5 so hypothesis is
2582-1 A fisheries researcher wishes to conclude that there is a difference in
2584-2 "If the sample means for each of the treatment groups were identical,"
2586-1 exceeding that reported in the F table
2588-1 -0.10 to 1.5. Which one of the following statements is true?
2591-1 Samples from 3 classes were given an identical math test.
2593-1 "If you reject, which shelving policies are different? "
2598-1 "the hypothesis that ""mothers are more likely to be success"
2601-1 """The wealthier a person, the more likely he will be relatively"
2603-1 hypothesis that male students are more assertive than female students
2609-1 Complete the ANOVA Table and test if all routes are equally
2612-1 What hypothesis is tested using this F ratio?
2702-1 Complete the following analysis of variance table:
2705-1 what is the value of (corrected) total sums of squares?
2706-1 what is the sum of squares for between groups?
2709-1 Consider each of the pairs of methods.
2719-4 indicates how a particular sum-of-squares value is calculated.
2723-4 "In a one-way ANOVA, the between mean square plus the within mean square"
2771-2 Are you satisfied with this conclusion? Why or why not?
2779-1 but use of an LSD is much less exposed to Type II errors.
2781-2 Which of the following pairs of sample means should NOT be judged
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Q: If X has an F distribution with df=4,5, what is the value of c where
P (X < c) = .95?
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Q: True or False? If False, correct it.
In a one-way classification ANOVA, when the null hypothesis is false,
the probability of obtaining an F-ratio exceeding that reported in the
F table at the .05 level of significance is greater than .05.
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Q: In a study, subjects are randomly assigned to one of three groups:
control, experimental A, or experimental B. After treatment, achieve-
ment test scores for the three groups are compared. The appropriate
statistical test for this comparison is:
a. the correlation coefficient
b. chi square
c. the t-test
d. the analysis of variance
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Q: A survey of 114 men and 126 women produced the result that the mean
amount of chicken soup consumed by the men in a month's time was .67
liters, compared with a mean of .54 liters for the women. The var-
iance of the chicken soup consumption for the men was 25% greater
than that for the women. A 95% confidence interval for the
difference between the means (men's mean - women's) was found to be
-.07 to +.33 liters. At the .05 level, is the difference in mean
chicken soup consumption between sexes statistically significant?
a) no
b) yes
c) Can't tell from the data given
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Q: The report for an investigation includes:
Means for weight loss by dieters
Cream puff diet 0
Frogurt diet 5
Beer and raw eggs diet 7
Saltines and kind words diet 8
HSD(.05) 2
Set 95% confidence limits for the difference between means for:
a. Cream puff diet vs. beer and raw eggs
b. Beer and raw eggs vs. saltines and kind words
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Q: Suppose the critical region for a certain test of hypothesis is of the
form F > 9.48773 and the computed value of F from the data is .86.
(F refers to an F statistic.) Then:
a) H(O) should be rejected.
b) H(A) is two-tailed.
c) The significance level is given by the area to the right of 9.48773
under the appropriate F distribution.
d) None of these.
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Q: Suppose H(O): MU(1) = MU(2) and your P-value satisfies .01 < P < .05.
Which of the following conclusions can be drawn?
(a) Fail to Reject H(O) because P is small.
(b) Reject H(O) because P is small.
(c) Fail to Reject H(O) because P is large.
(d) Reject H(O) because P is large.
(e) Fail to Reject H(O) at ALPHA = .01.
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Q: A carefully designed experiment has just been concluded. Execution of
the experiment was flawless. Unfortunately use of ALPHA (.05) indic-
ated no significant differences among treatments.
What useful information can you supply future investigators when you
report on this experiement?
Indicate agreement with yes or disagreement with no for each of the fol-
lowing items.
a. No useful information. These are negative results and there is
nothing useful to report.
b. The estimated variance (and its df) can be useful to future invest-
igators.
c. A careful description of experimental conditions and treatments may
be useful as an indication of circumstances where responses are
about the same.
d. Significant differences can be reported by changing the Type I
error rate from .05 to .10, .20 or whatever is needed to declare
significance.
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Q: True or False? If False, correct it.
In ANOVA, if we wish to investigate the difference among five
means, it is good statistical procedure to perform a t-test on
each pair of means.
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Q: Suggest some factors that might influence variability in weights
measured on a group of men.
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Q: True or False? If False, correct it.
Data containing less variability than would be expected by chance
are always preferable.
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Q: A report on an investigation includes the following information
related to the influence of several mouth washes to length of
time that breath remains "great".
Analysis of Variance
Source df SS Mean Square
Treatments 3 500 166.67
Error 196 19,600 100
Total 199 20,100
Means (hours that breath remained "great")
Whiskey 12
Brand X 11
Water 9
Brand L 8
F(critical) (.05) = 3.92 with 196 df
You are to develop plans for a follow-up study. In particular,
you are to re-examine the difference between Brand X and Brand L.
In looking at methods for estimating number of replicates needed
you find that you need values for -
1. the size of the difference to be detected
2. the anticipated standard deviation
3. the anticipated variance
On the basis of the above report what values will you use for
each of the above 3 items? Why?
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Q: Given the following observed number of pigs for 8 litters, the numerator
of the formula for s**2 is called the corrected sum of squares as illus-
trated.
s**2 = (Sum of Squares)/(n-1)
Find the sum of squares.
X(1) = 9 X(5) = 10
X(2) = 6 X(6) = 7
X(3) = 14 X(7) = 8
X(4) = 9 X(8) = 9
a) 5 b) SQRT(5) c) 40 d) 80
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Q: True or False? If False, correct it.
The degrees of freedom associated with a error (or within) term
in ANOVA depends only on sample size.
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Q: True or false? IF FALSE, CORRECT IT.
In ANOVA the degrees of freedom associated with within sum of squares
depend on how many independent sample values were used and how many
other independent parameters have been estimated.
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Q: In a simple analysis of variance problem, which of the following is an
estimate of the variance of individual measurements (after the
various effects have been accounted for)? (MS means SS/df so each of
answers is a Mean Square.)
a. MS(between)
b. MS(within)
c. MS(total)
d. none of the above
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Q: Define the following term and give an example of its use.
Your example should not be one given in class or in a handout.
MEAN SQUARE
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Q: The total variation in response, assuming no bias, is due to error
(unexplained variation) plus differences due to treatments (known
variation). If known variation is large compared to unknown variation,
which of the following conclusions is the best?
a) There is no difference in response due to treatments.
b) There is a difference in response due to treatments.
c) The treatments are not comparable.
d) The cause of the response is due to something other than treatments.
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Q: An investigator randomly assigns 30 college students into three equal
size study groups (early morning, afternoon, late night) to determine
if the period of the day at which people study has an effect on their
retention. The students live in a controlled environment for one
week, on the third day of which the experimental treatment (study of
predetermined material) is administered. The seventh day the
investigator tests for retention, and in computing his analysis he
sees that his MS within groups is larger than his MS among groups.
What is the indication of this result?
a. An error in calculation was made.
b. There was more than the expected variability between groups.
c. There was more variability between subjects within the same
group than there was between groups.
d. That there should have been additional controls in the experi-
ment.
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Q: Source SS df
------ -- --
Between 30.5 4
Within
Total 165.0 99
What decision would be made regarding H(0): population means are
equal?
(a) Reject H(0) at the .05 level
(b) Fail to reject H(0) at the .01 level
(c) Insufficient information is given to answer
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Q: In reading a scientific article you encounter the following table:
Analysis of Variance
------------------------------------------------------------------------
Source SS df MS F
------------------------------------------------------------------------
Between samples 722.7 4 180.68 15.3**
Within samples 473.3 40 11.83
------------------------------------------------------------------------
Total 1196.0 44
Further reading indicates that all sample sizes are equal. Then we know
that the experimenter used
(a) 4 samples of size 10.
(b) 5 samples of size 10.
(c) 4 samples of size 9.
(d) 5 samples of size 9.
(e) None of these
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Q: NOTE: Items i. thru iii. are based on a teaching experiment involving
four elementary statistics classes. Below are scores for 24
students who took the same final examination.
Statistics Statistics Statistics Statistics
Cookbook with Humor Made Useful in Story Form
------------------------------------------------------------------------
78 51 64 54
78 57 54 61
79 64 61 79
70 75 66 69
83 42 57 69
74 83 71 65
------------------------------------------------------------------------
462 372 373 397
Suppose further that calculations yield SS(total) = 800, and
SS(between) = 300.
i. What is the observed value of the statistic one computes to test
H(0): MU(1) = MU(2) = MU(3) = MU(4) against H(1): not all 4 means
are equal?
(a) 2.5 (b) 5.2 (c) 4.0 (d) 11.1
(e) None of these
ii. If ALPHA = .01, then the critical value of the statistic is
(a) 3.10 (b) 4.94 (c) 8.66 (d) 26.7
(e) None of these
iii. Suppose that the observed value of the statistic in Item i. is
5.6 while the critical value in Item ii. is 7.21. With only this
information, which of the following conclusions is most logical?
(a) The four populations do not all have the same mean.
(b) The four populations have the same mean.
(c) The four populations do not all have the same mean.
Statistics Cookbook and Statistics in Story Form produce
higher means than the other two books.
(d) Statistics Cookbook has the highest population mean.
(e) The four population means may be different but these samples
fail to demonstrate any difference.
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Q: An experiment was conducted as a oneway random ANOVA design yielding K
sample means, each based on n scores. If the between and within mean
squares are represented by S(m)**2 and S(p)**2, respectively, what is
the number of degrees of freedom for S(m)**2?
a. n - 1
b. K - 1
c. n - K
d. (n - 1)(K - 1)
e. none of the above
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Q: Suppose we have five independent groups (or samples), each of size 9.
How many degrees of freedom do we have for the within groups sum of
squares in the ANOVA assuming a single factor experiment?
1. 4
2. 41
3. 44
4. 45
5. Not given.
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Q: Samples of size 5 are taken from 3 populations and the following
analysis of variance table found. Test the hypothesis that the
three populations have the same means.
Source d.f. M.S. F
------ ---- ---- -
Between means 350
Within samples 100
Total
a. F = 3.5 so hypothesis is rejected at 5% level
b. F = 3.5 so hypothesis is not rejected at 5% level
c. F = 3.5 so hypothesis is not rejected at 10% level
d. F = 3.5 so hypothesis is rejected at 1% level
e. Do not have enough information to perform test.
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Q: A fisheries researcher wishes to conclude that there is a difference in
mean weights of three species of fish caught in a large lake near Lin-
coln, Nebraska. The data are as follows: (Use ALPHA = .05.)
SPECIES
-------
X Y Z
1.5 1.5 6.0
4.0 SUM(X) = 13 1.0 SUM(Y) = 9 4.5 SUM(Z) = 20.5
4.5 4.5 4.5
3.0 2.0 5.5
ANOVA Table (incomplete) :
Source of Variation SS df MS F
Between Groups 17.04 2 8.52
Within Groups 14.19 9 1.58
TOTAL 31.23 11
i) The null hypothesis is:
a) H(O): BETA = 0
b) H(O): MU = 0
c) H(O): MU(X) = MU(Y) = MU(Z)
d) H(O): BETA(X) = BETA(Y)= BETA(Z)
ii) The test statistic is:
a) t(calc) = 2.52
b) t(calc) = 3.09
c) F(calc) = 1.20
d) F(calc) = 5.40
iii) The critical value is:
a) t(.05,9) = 2.262
b) t(.10,9) = 1.833
c) F(.05,2,9) = 4.26
d) F(2.5,2.9) = 5.71
iv) What is your conclusion?
a) Reject H(O) because F(calc) > F(crit), (at least 1 pair
has different means).
b) Reject H(O) because ]t(calc)] > t(crit), (all means are
different).
c) Fail to reject H(O) because F(calc) < F(crit), (insuffi-
cient evidence that means are different).
d) Fail to reject H(O) because ]t(calc)] < t(crit), (means
are equal).
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Q: If the sample means for each of the treatment groups were identical,
what would be the observed value of the ANOVA F-ratio?
a. 1.0
b. 0.0
c. A value between 0.0 and 1.0
d. A negative value
e. Infinite
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Q: Assuming a single factor experiment is being analyzed and that the
null hypothesis being tested by the F-test in the ANOVA is false, the
probability of obtaining a mean-square ratio exceeding that reported
in the F table as the 95th percentile is
a. greater than .05.
b. less than .05.
c. equal to .05.
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Q: A one-way classification analysis of variance is performed on experi-
mental data for which there were 10 subjects in each of two groups.
The .95 confidence interval around the difference YBAR(1) - YBAR(2) is
-0.10 to 1.5. Which one of the following statements is true?
(a) The F ratio obtained in the analysis of variance was less
than 4.41
(b) The F ratio obtained in the analysis of variance was greater
than 8.28
(c) The true difference MU(1) - MU(2) must lie between -0.10 and
1.50.
(d) The best estimate of MU(1) - MU(2) possible from the results
is 1.50.
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Q: Samples from 3 classes were given an identical math test. The scores
are given as follows. Test the hypothesis that the performances of
the 3 classes are equal. (ALPHA = .05)
Class A Class B Class C
------- ------- -------
5 10 7
4 2 7
6 1 6
10 5 5
4 2 7
7 7 8
3 8 10
9 2 1
--- --- ---
Total 48 37 51
Source of variation SS df MS F
Between classes 13.583 2 6.7915
Within classes (error) 220.75 21 10.5
Total 234.333 23
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Q: To test the hypothesis that shelf placement influences sales, a
marketing researcher has collected data on sales in a random sample
of 15 comparable supermarkets with 3 different shelving policies
for an identical brand of soup. The data is weekly sales figures
(in tens of cans). Perform the appropriate test at the 5% level.
If you reject, which shelving policies are different?
bottom shelf middle shelf top shelf
sales sales sales
------------ ------------ ---------
10 25 10
5 20 10
10 25 20
10 30 20
15 50 40
MEANS FOR SHELF
TREATMENT MEAN(SALES)
MIDDLE 30
TOP 20
BOTTOM 10
ANALYSIS OF VARIANCE (incomplete) :
---------------------
SOURCE OF VARIATION DF SS MEAN SQUARE F(CALC.)
SHELF 2 1000.0000 500.00000
EXPERIMENTAL ERROR 12 1200.0000 100.00000
TOTAL 14 2200.0000
HSD FOR ABOVE MEANS IS 16.86 at CONF.LEVEL 95
(Note: HSD means Tukey's Honest Significant Difference.)
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Q: A reseaecher assigns each of his interviewers a list of 7 families,
drawn randomly from a region, to be interviewed. Each interviewer
is instructed to administer a successful parenting scale (SPS) to
each parent in his sample. The SPS scores, Y(i), are defined as
ranging from 0 (no parenting skills deemed successful) to 100
(successful parenting skills consistently and skillfully applied).
An interviewer returns with data for both parents. Use this data to
test, using classical analysis of variance, at the 90% level of con-
fidence, the hypothesis that "mothers are more likely to be success-
ful parents".
Mothers Fathers
Y(i) Y(j)
68 63
72 48
48 30
54 52
83 55
92 41
87 57
MBAR = 72.00
FBAR = 49.43
ANOVA Table (incomplete) :
Source of Variation SS df MS F(calc)
Between Groups 1783.143 1
Within Groups 2391.714 12 199.310
Total 4174.857 13
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Q: Use one-way analysis of variance, with an F test, to test the hypothesis
that "The wealthier a person, the more likely he will be relatively
politically conservative," at the 90% level of confidence. Note that,
for purposes of research, the researcher operationally defined "wealthy"
as those with an annual income of $50,000, while "poor" subjects re-
ceived less than $5,000/year. Note, too, that the "political conser-
vatism" scale used produced scores of 0 for "extremely liberal", and 100
for "extremely conservative". Sample data:
X, Income Category
Wealthy Poor
+---------+--------+
| Y Y | Y |
| 90 | 50 |
| 80 | 60 |
| 70 | 40 |
| 60 | 50 |
| 90 | 30 |
+---------+--------+
ANOVA Table (incomplete) :
Source of variation | SS | df | MS | F(calc.)
--------------------+--------+--------+--------+-----------
Between groups | 2560 | 1 | |
Within groups | 1200 | 8 | |
Corrected total | 3760 | 9 | |
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Q: A sociologist conducted a study of assertion by having one of her top
students, after appropriate training, note the number of assertive acts
performed in a day by each of 10 randomly selected coeds, producing the
following sample of data, in acts per day: [5,3,10,6,4,9,5,5,7,5].
Another sociologist wonders whether the male and female students at
Bedrock indeed differ in assertiveness and, by a similar procedure,
gathers the following data for male students, in acts per day: [8,3,5,
8,12,10,7,7,9,7]. Use a one-way analysis of variance to test the
hypothesis that male students are more assertive than female students
at Bedrock College, at the 90% level of confidence.
ANOVA table (incomplete) :
Source of variation | SS | df | MS | F(calc.)
---------------------+--------+--------+--------+----------
Between Groups | 14.45| 1 | |
Within Groups | 99.30| 18 | |
Total | 113.75| 19 | |
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Q: Mr. Martin can drive to work along four different routes, and
the following are the number of minutes in which he timed him-
self on five different occasions for each route:
Route 1 Route 2 Route 3 Route 4
________________________________________
22 25 26 26
26 27 29 28
25 28 33 27
25 26 30 30
31 29 33 30
________________________________________
T.(j) = 129 135 151 141
ANOVA Table (incomplete)
Source df SS M.Sq. F ratio
Routes .3 52.8
Error 16 100.4
Total 19 153.2
Complete the ANOVA Table and test if all routes are equally
fast (ALPHA = 5%).
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Q: An imaginary study has been conducted on the effects of three brands of
laxatives on regularity of TV actresses where each brand was tested by
one actress belonging to each of 10 age groups. Results obtained
included: F= (brand M.Sq.)/(Error M.Sq.) = 2.1 with 2 and 18 df.
a. What hypothesis is tested using this F ratio?
b. Interpret these results using a significance level of 5%.
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Q: Samples of size 11 are taken from each of 5 populations. Complete
the following analysis of variance table:
Source S.S. d.f. M.S. F
------ ---- ---- ---- -
Between means 1000 a c e
Within samples 5000 b d
Total 6000
a. a = 4 b = 44 c = 250 d = 113.6 e = 2.2
b. a = 4 b = 44 c = 250 d = 113.6 e = 0.2
c. a = 5 b = 55 c = 200 d = 90.9 e = 0.2
d. a = 5 b = 50 c = 200 d = 100 e = 2.0
e. a = 4 b = 50 c = 250 d = 100 e = 2.5
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Q: In a single factor experiment with four levels if the mean square
(between)=25, mean square(within)=10, n(1)=n(2)=n(3)=8 and n(4)=10,
what is the value of (corrected) total sums of squares?
a. 435
b. 786
c. 1221
d. Insufficient information
e. Sufficient information but correct value is not given
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Q: In the ANOVA for a single factor experiment with four levels all n's
equal 5 and YBAR(1)=22, YBAR(2)=24, YBAR(3)=20, and YBAR(4)=18. What
is the sum of squares for between groups?
a. 25.00
b. 33.33
c. 100.00
d. Cannot be determined without more data
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Q: Nine children were randomly split into three groups of three each. In
a spelling unit, individuals in one group were criticized each time they
misspelled a word. The individuals in another group were praised each
time they correctly spelled a word, while the individuals in the third
group were neither praised or criticized. At the end of the unit, each
child was given ten words to spell with the following results (number
correct is given for each child):
Praised Neutral Criticized
8 3 9
9 2 10
7 5 7
MEANS FOR RESPONSE
TREATMENT MEAN(WORDS CORRECT)
CRITICIZED 8.66667
PRAISED 8
NEUTRAL 3.33333
ANALYSIS OF VARIANCE:
--------------------
SOURCE OF VARIATION DF SS MEAN SQUARE
RESPONSE 2 50.6667 25.33333
EXPERIMENTAL ERROR 6 11.3333 1.88889
TOTAL 8 62.0000
PROBABILITY LEVEL FOR COMPARING MEANS = 0.05
HSD FOR ABOVE MEANS IS 3.444 AT PROB.LEVEL 0.05
Is there any evidence for significant differences among the methods?
Consider each of the pairs of methods. Are there significant differ-
ences within any of the pairs?
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Q: NOTE: Items i. thru iv. are based on the following situation. Each
item indicates how a particular sum-of-squares value is
calculated. Use the following code in responding to each item.
(a) Between
(b) Within
(c) Total (corrected)
(d) Either (a) or (b)
(e) None of these
i. Squaring the deviations of each score from its sample (or group)
mean.
ii. Squaring the deviations of each score from the overall mean.
iii. Squaring the deviations of each sample mean from the overall mean.
iv. Squaring the deviations of one sample mean from all other sample
means.
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Q: True or False? If False, correct it.
In a one-way ANOVA, the between mean square plus the within mean square
must equal the total mean square if computations are correct.
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Q: A physiologist reports an investigation of potential plant hormones. He
reports the following averages for lengths of 20 stem segments treated.
Compound X 1.18
Compound Y 1.17
Compound Z 1.15
Control 1.14
The conclusion is that there are no treatment differences. Are you
satisfied with this conclusion? Why or why not?
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Q: When various methods for comparing means are discussed two methods
that are often cited as extremes are -
a. LSD for which the Type I error rate is ALPHA per comparison
b. Tukey's HSD, for which the Type I error is ALPHA per experiment
regardless of the number of comparisons made.
If there are many comparisons to be made in an experiment, use of
an LSD is much more exposed to Type I error than use of Tukey's
method, but use of an LSD is much less exposed to Type II errors.
In each of the items below indicate whether or not you would use
an LSD to compare means. Explain your choice.
a. An experiment involves 10 treatments. If you commit a Type I
error, there will be little immediate loss. If you commit a
Type II error there will be somewhat greater losses.
b. An experiment involves 2 treatments. If you commit a Type I
error, there will be appreciable loss. If you commit a Type
II error, there will be little loss.
c. An experiment involves 20 treatments. If you commit a Type I
error, there could be widespread injury (e.g. loss of life).
If you commit a Type II error, there will be bothersome
losses (loss of investment in developing treatments), but
less loss than from Type I error.
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Q: A F-ratio signficant at the .05 level was found in a one-factor ANOVA
in which there are 3 levels and n=25. A set of simultaneous confidence
intervals with confidence coefficient .95 is established around
differences between pairs of means using Tukey's HSD method. The
following intervals are obtained:
YBAR(1) - YBAR(2): (-7.50, -7.00)
YBAR(1) - YBAR(3): (8.50, 8.75)
YBAR(2) - YBAR(3): (-.80, 1.50)
Which of the following pairs of sample means should NOT be judged to
be significantly different?
(a) YBAR(1) - YBAR(2)
(b) YBAR(1) - YBAR(3)
(c) YBAR(2) - YBAR(3)
(d) None of the above
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A: F(95th percentile, df=4,5) = 5.19
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Look at this question's identification
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A: True
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Look at this question's identification
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A: d. the analysis of variance
Of the four tests given, the analysis of variance is the only
exact test that can be performed. The t-test does not compare
more than two means and chi square is not exact.
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Look at this question's identification
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A: a) no
The confidence interval includes zero.
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A: a. (7-0) +/- HSD
7 +/- 2
Interval is from 5 to 9 (significant difference)
b. (8-7) +/- 2
Interval is from -1 to +3 (not a significant difference)
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A: c) The significance level is given by the area to the right of 9.48773
under the appropriate F distribution.
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A: (e) Fail to Reject H(O) at ALPHA = .01.
Since P > .01 you cannot reject H(O) at ALPHA = .01.
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A: a. no
b. yes, a variance estimate is needed in order to estimate sample size
required when someone else investigates this phenomenon in the
future.
c. yes, if circumstances that produce no response aren't reported, the
reports on record sometimes will give the misleading impression that
there generally is a response.
d. no, Type I error rate is to be chosen before the investigation
and is not to be changed in order to declare significance.
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A: False, good statistical procedure would use an F-test and Tukey's
test to avoid the problem of increased type I error in the
multiple comparisons.
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A: Factors that might influence variability in weights include:
variability in height, age, level of activity,
diet, and random error.
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A: False - When data contains less variability than would be expected by
chance, it is time to be suspicious of the techniques or procedures
used in collecting the data. But business managers are typically con-
servative and prefer data with less variability.
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A: 1. The estimated difference between Brands X and L in this trial was 3.
Since a follow up study has been commissioned, it would seem that a
difference at least this small is to be detected.
2. The mean square for experimental error gives us an estimate of
the variance and an estimate of the standard deviation would be
the square root of that, so SQRT(100) = 10.
3. As explained above, it is 100.
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A: c) 40
SUM(X) = 72
XBAR = 72/8 = 9
Corrected Sum of Squares = (0**2) + (-3**2) + (5**2) + (0**2) +
(1**2) + (-2**2) + (-1**2) + (0**2)
= 9 + 25 + 1 + 4 + 1
= 40
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A: False. It also depends on how many independent parameters (or treatment
level means) are being estimated.
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A: True.
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A: b. MS(within)
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A: Definition: A sum of squares divided by its degrees of freedom.
Example: If an ANOVA table includes:
Source df SS
Experimental Error 12 3600,
the mean square for experimental error is
300 = 3600/12.
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A: b) There is a difference in response due to treatments.
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A: c. There was more variability between subjects within the same
group than there was between groups.
This is precisely the null hypothesis of the F-test in this analysis.
Since the MS(within) is larger than MS(among), this F-test will have
a calculated value less than one. Based on this result we will
undoubtedly not reject the null hypothesis.
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A: (a) Reject H(0) at the .05 level
Completed ANOVA table:
Source SS df MS
------ -- -- --
Between 30.5 4 7.625
Within 134.5 95 1.416
Total 165.0 99
F(calc.) = 7.625 / 1.416
= 5.385
F(crit., ALPHA=.05, df=4,95, one-tail) = 2.47
F(crit., ALPHA=.05, df=4,95, one-tail) = 3.59
H(0) would be rejected at both ALPHA levels
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A: (d) 5 samples of size 9.
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A: i. (c) 4.0
ANOVA Table:
Source SS df MS F
------ -- -- -- -
Between samples 300 3 100 4.00
Within samples 500 20 25
Total (corrected)800 23
ii. (b) 4.94
F(crit., ALPHA=.01, df=3,20, one-tail) = 4.94
iii. (e) The four population means may be different but these samples
fail to demonstrate any difference since F(obs.) < F(crit.).
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A: b. K - 1
Total Categories = K
Degrees of Freedom = K - 1
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A: 5. Not given.
df(within) = (total # of observations) - 1 - df(between)
= (5*9) - 1 - 4
= 45-5
= 40
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A: b. F = 3.5 so hypothesis is continued at 5% level.
Completed ANOVA table:
Source d.f. M.S. F
------ ---- ---- -
Between samples 2 350 3.5
Within samples 12 100
Total 14
F(critical, df = 2, 12, ALPHA = .05) = 3.89
Therefore the hypothesis should not be rejected.
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A: i) c) H(O): MU(X) = MU(Y) = MU(Z)
ii) d) F(calc) = 5.40
iii) c) F(.05,2,9) = 4.26
iv) a) Reject H(O) because F(calc) > F(crit), (at least 1 pair
has different means).
ANOVA Table:
Source of Variation SS df MS F
Between Groups 17.04 2 8.52 5.40
Within Groups 14.19 9 1.58
TOTAL 31.23 11
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A: b. 0.0 because Sum of Squares between would be zero and the F-ratio
is:
F(calc.) = [SS(between)/df(between)]/[SS(within)/df(within)]
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A: a. greater than .05.
The null hypothesis being tested is:
H(O): MU(1) = M(2) = ... = MU(K)
where K is the number of experimental groups or samples. But if
this is false, then the K population means are not equal and the
between groups mean square will be greater than the within groups
mean square. So the sampling distribution of this ratio will be
centered further to the right than the F-distribution associated
with the null hypothesis.
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A: (a) The F ratio obtained in the analysis of variance was less than 4.41
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A: H(O): MU(1) = MU(2) = MU(3)
H(A): MU(1) =/= MU(2) =/= MU(3)
F(calculated) = MS for classes/MS for error
= 6.7915/10.5
= .646
F(critical, df=2, 21, ALPHA =.05) = 3.47
Since F(calculated) < F(critical) do not reject H(O). Therefore
performances of the 3 classes are not found to be different.
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A: PROBABILITY LEVEL FOR COMPARING MEANS = .05
F(critical, df=2,12, ALPHA=.05) = 3.88
Therefore, reject the null hypothesis that shelving policy does not
influence sales.
Based on the HSD given above, it appears that there is a significant
difference between the middle shelf and the bottom shelf.
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A: n(M) = n(F) = 7
H(O): MU(females) = MU(males)
H(A): MU(females) =/= MU(males)
ANOVA Table:
Source of Variation SS df MS F(calc)
Between Groups 1783.143 1 1783.143 8.947
Within Groups 2391.714 12 199.310
Total 4174.857 13
F(crit., ALPHA=.10, one-tail, df=1,12) = 3.18
Since F(calc) > F(crit), H(O) should be rejected. Thus, based on this
sample evidence, it appears that the means for mothers and fathers are
different and the indication from the sample means is that mothers are
more likely to be successful parents.
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A: H(0): MU(Wealthy) = MU(Poor)
H(A): MU(Wealthy) =/= MU(Poor)
ANOVA Table:
Source of variation | SS | df | MS | F(calc.)
--------------------+--------+--------+--------+-----------
Between groups | 2560 | 1 | 2560 | 17.1
Within groups | 1200 | 8 | 150 |
Total | 3760 | 9 | |
F(crit., df=1,8, ALPHA=.10) = 3.46
Since F(calc.) > F(crit.) reject H(0) and based on the fact that wealthy
people have a higher mean income assume that the wealthier a person, the
more likely he will be relatively politically conservative.
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A: H(0): MU(Males) = MU(Females)
H(A): MU(Males) =/= MU(Females)
ANOVA table:
Source of variation | SS | df | MS | F(calc.)
---------------------+--------+--------+--------+----------
Between Groups | 14.45| 1 | 14.45 | 2.62
Within Groups | 99.30| 18 | 5.52 |
Total | 113.75| 19 | |
F(crit., df=1,18, ALPHA=.10) = 3.01
Since F(calc.) < F(crit.), do not reject H(0). Therefore, conclude that
there is no difference between men and women at Bedrock College.
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A: H(0): The means for the routes are all equal.
H(A): The means for the routes are not all equal.
ANOVA Table
Source df SS M.Sq. F ratio
Routes .3 52.8 17.6 2.805
Error 16 100.4 6.275
Total 19 153.2
F(ALPHA = .05, df = 3,16) = 3.24
Since the F ratio is less than 3.24, we continue the null
hypothesis that all routes are equally fast.
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A: a. The null hypothesis tested is that all brand effects equal zero,
H(O): MU(1) = MU(2) = MU(3).
b. F(calculated) = 2.1
F(critical, df = 2.18, ALPHA = .05) = 3.55
Since the calculated F value is smaller than the critical F value, we
do not reject the null hypothesis that all brand effects are equal.
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A: e. a = 4 b = 50 c = 250 d = 100 e = 2.5
a = number of treatments - 1 = 5 - 1 = 4
b = (a+1)*(number of replications - 1) = (4+1)*(11-1) = 50
c = 1000/a = 250
d = 5000/b = 100
e = c/d = 250/100 = 2.5
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A: e. Sufficient information but correct value is not given
Known parts of ANOVA table are:
Source of Variation df SS MS
------------------- -- -- --
Between (Treatment) 3 25
Within 30 10
(Corrected) Total 33
Therefore between Sum of Squares is:
SS(B) = mean square(between) * df(between)
= (25*3)
= 75
Within Sum of Squares is:
SS(W) = mean square(within) * df(within)
= (10*30)
= 300
Corrected Total SS is:
SS(T) = SS(B) + SS(W)
= 75+300
= 375
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A: c. 100.00
Overall Mean = (22+24+20+18)/4
= 84/4
= 21
SS(between) = [5*((22-21)**2)]+[5*((24-21)**2)]+[5*((20-21)**2)]
+[5*((18-21)**2)]
= [5*1]+[5*9]+[5*1]+[5*9]
= 5 + 45 + 5 + 45
= 100.00
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A: F(calculated) = (25.3/1.9)
= 13.40
--------------------------------------------------------------
H(O): No difference between treatments
H(A): Not all treatment means are the same
Critical Region: F(df=2,6) > 5.14
Reject H(O), and conclude that amount and/or kind of praise
makes a difference in spelling ability.
Using the HSD and the treatment means given above, we see that
"criticizing vs. neutral" are significantly different and "prais-
ing vs. neutral" are significantly different.
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A: i. (b) Within
ii. (c) Total (corrected)
iii. (a) Between
iv. (e) None of these
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A: False, in a one-way ANOVA, the between SS plus the within SS must equal
the total SS if computations are correct.
SS(TOTAL) = SS(Between) + SS(Within)
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A: On the basis of the information given I am not satisfied with this con-
clusion. No indication of an estimate for the background variation has
been given, nor any indication of a basis for comparing these means,
such as an HSD with a specified significance level.
Due to the lack of information reported, there is no basis for making
a conclusion one way or another.
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A: a. I would use an LSD because I want to protect against Type II
error, not Type I errors.
b. I could use either method in this case because when there
are only two treatments, the two methods are identical.
c. I would not use an LSD in this case, I would use Tukey's test.
Tukey's method protects us from Type I errors when more than
one comparison is to be made.
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A: (c) YBAR(2) - YBAR(3)
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