# Test Selection Selecting a statistical test for the analysis of research data is a key step in the investigative process In fact experienced researchers are known t by fpt11744

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```									                              Test Selection

Selecting a statistical test for the analysis of research data is a key step in the investigative
process. In fact, experienced researchers are known to weigh that decision very carefully
beforecollecting their data. Such prior planning assures that the statistical test under
consideration is both appropriate for the data and capable of providing information that meets the
goals of the study.

Unfortunately, the sequential topic-by-topic coverage in an academic course provides limited
opportunities to practice test selection. When confronted with the practice problems at the end of
the t-test chapter, you knew to use a t test, and so forth for other problems in other chapters.
The present workbook provides opportunities to practice test selection outside of such structure.

As you encounter the research scenarios on the following worksheets, your task is to select and
run the analysis that best meets the needs of the researcher. The solutions appear off to the right
of each worksheet in an area initially out of view (column O and beyond). Do your analyses on
the "my work" worksheets, then go back to the problem worksheets to compare your results to the
solutions.
he investigative
very carefully
est under
mation that meets the

provides limited
roblems at the end of
in other chapters.
de of such structure.

appear off to the right
Data for Part I
These are the data for the first group of problems. Excel will instantly forward any changes you
enter on this worksheet to the data and solutions that follow. But for your first run through the
workbook, it's fine to use the original data. Select the statistical analysis that is appropriate for the
data and the research question, click a "my work" worksheet tab, and do the analysis. When done,
go back to the problem worksheet and scroll right to see the solution. No peeking! Try the
problems first on your own before looking at the solutions.

8             2
4             1
7             3          Any changes to the Part I
9             6          data must be made here,
13            10          NOT on the problem
14            16          worksheets.
19            18
17            15
20             5
12            11
d any changes you
run through the
is appropriate for the
analysis. When done,
eking! Try the

the Part I
de here,
After inspecting complaints filed with the police, 210    dogs of ten different breeds were identified
as excessive barkers. As part of a court-ordered attempt to address the problem, all the dogs were fitte
with a special collar that administers an unpleasant shock each time it detects a vocalization above a
specific decibel level. Was the success rate of the device relatively uniform across the 10 breeds
(retain Ho), or did some breeds respond better to the collar device than others? First do your analysis.
on the "my work 1-1" worksheet, then return here to compare your results to the solution.
Barking
Breed       Stopped Continued
A             8          2
B             4          1                   Solution
C             7          3                  (Columns O to U)
D             9          6
E            13         10
F            14         16
G            19         18
H            17         15
I            20          5
J            12         11
reeds were identified         Solution for the chi-square test of i
m, all the dogs were fitted
degrees of
vocalization above a
freedom
oss the 10 breeds                            c2(9)=

Barking
Breed    Stopped
A         8
B         4
C         7
D         9
E         13
F         14
G         19
H         17
I        20
J         12
Totals   123
r the chi-square test of independence. See workbook 10-2.

11.486         p= 0.244

Barking
Continued   Totals           Expected Frequencies
2        10                5.86      4.14
1         5                2.93      2.07
3        10                5.86      4.14
6        15                8.79      6.21
10        23               13.47      9.53
16        30               17.57     12.43
18        37               21.67     15.33
15        32               18.74     13.26
5        25               14.64     10.36
11        23               13.47      9.53
87       210
This worksheet is blank except for data set 1. Use it to carry out the statistical analyses for problem 1-1.
Then compare your answers to the solution on the problem 1-1 worksheet.

Data:
8           2
4           1
7           3
9           6
13          10
14          16
19          18
17          15
20           5
12          11
analyses for problem 1-1.
The data below represent homes sold in 2005 by 20 agents of Blank & Best Real Estate, Inc.
For the past five years, mean home sales for these 20 agents has been a fairly steady 6.75.
Are the data for 2005 consistent with the historical mean sales performance of 6.75 (retain
Ho), or was there a significant change in sales performance in 2005? Click the "my work 1
tab.

Home Sales
for 20 Agents
8          2
4          1                 Solution
7          3                (columns O to S)
9          6
13         10
14         16
19         18
17         15
20          5
12         11
Solution for the single-sample t tes
st Real Estate, Inc.
t=
e of 6.75 (retain
df =
k the "my work 1-2"
p=

Home Sales
for 20 Agents
8         2
4         1
7         3
9         6
13         10
14         16
19         18
17         15
20         5
12         11
r the single-sample t test. See workbook 6-2.

2.835
19
0.011

mean = 10.5
Use this sheet to carry out the statistical analyses for problem 1-2.
Then compare your answers to the solution on the problem 1-2 worksheet.

Data:
8           2
4           1
7           3
9           6
13          10
14          16
19          18
17          15
20           5
12          11
A researcher randomly divided 20 seedlings into two groups. One group was maintained
using fertilizer A, and the other was maintained using fertilizer B. At maturity the researcher
counted the number of blossoms on each plant. Was blossom count basically the same for
the two fertilizers (retain Ho), or was there a difference in mean blossom count between the
two groups of mature plants? Click the "my work 1-3" tab.

Fertilizer A Fertilizer B
8            2
4            1
7            3                 Solution
9            6                (Columns O to Y)
13            10
14            16
19            18
17            15
20             5
12            11
s maintained      Solution for the independent-group
y the researcher
ly the same for
nt between the

Fertilizer A
8
4
7
9
13
14
19
17
20
12
Means:       12.3
r the independent-groups t test.   Solution for the Mann-Whitney U.            Note: The ranks of the
original data contain no ties.
t = 1.394                 U = 32                   z = 1.361  But, because of the way
df = 18            With n 1 = n 2 = 10, U crit (a=.05)         Excel resolves tied ranks,
p = 0.180         equals  23 or 77.                         changes you make to the
data set 1 worksheet that
result in tied ranks will
Fertilizer B              Fertilizer A Ranks Fertilizer B Ranks cause the U value reported
2                         8            8           2           2 here to be slightly different
1                         4            4           1           1 from the output of other
3                         7            7           3           3 statistical software. For the
6                         9            9           6           6 more precise answer, type
10                        13           13          10          10 over the "Ranks" values as
16                        14           14          16          16 necessary to resolve the
18                        19           19          18          18 ties. Review the "rank and
percentile" worksheet in
15                        17           17          15          15
workbook 1-2 for the
5                        20           20           5           5 instructions on resolving
11                        12           12          11          11 ties.
8.7                           R1=     123             R2= 87
nks of the         Not        combined   ranks
contain no ties.   intended       8         8
of the way       for            4         4
s tied ranks,     viewing        7         7
make to the                      9         9
rksheet that
13        13
ranks will
value reported                   14        14
ghtly different                  19        19
ut of other                     17        17
tware. For the                   20        20
" values as                  2         2
resolve the                      1         1
the "rank and                    3         3
orksheet in
6         6
n resolving                      10        10
16        16
18        18
15        15
5         5
11        11
Use this worksheet to carry out the statistical analyses for problem 1-3.
Then compare your answers to the solution on the problem 1-3 worksheet.

Data:
8          2
4          1
7          3
9          6
13         10
14         16
19         18
17         15
20          5
12         11
Students took a test that assessed their knowledge of current events followed by another test that
assessed reading comprehension. The two tests had the same number of questions. The numbers
below represent the number of correct answers on each test. Do the data show a significant
relationship between reading comprehension scores and knowledge of current events? If so,
assemble the information you need to predict the variable "reading comprehension" from the variable
"knowledge of current events." Click the "my work 1-4" tab.

Students:    Events    Comprehension
Jim          8          2
Steve         4          1                 Solution
Matt          7          3                (Columns P to AA)
Paul          9          6
Dave         13         10
Josh        14         16
Larry        19         18
Jerry        17         15
Ethan        20          5
Jacob        12         11
Solutions for the correlation and re
ollowed by another test that
r of questions. The numbers
r XY =
ata show a significant                  p=
current events? If so,          Here is the
prehension" from the variable   regression
equation for
Y on X.

Students:
Jim
Steve
Matt
Paul
Dave
Josh
Larry
Jerry
Ethan
Jacob
Solutions for the correlation and regression analyses (See workbooks 3-3, 3-6, and 6-4.)
0.700                               Testing the r value for significance yields:
0.024                                        t = 2.775     p = 0.024
Here is the
regression      Y' =      ay      + by X            To predict reading comprehension score
equation for       Y'=   -1.29    + 0.812 X         (Y) from knowledge of current events
(X), substitute a current events score for
Current Reading                         X in the regression equation -- or use
Events (X) Comp. (Y)
the FORECAST paste function.
8          2
4          1
7          3
9          6                            r 2= 0.49
13         10           The r2 statistic reports the proportion of
14         16           the variability in the "reading
19         18           comprehension" variable Y that is
17         15           predictable from the "current events"
20          5           variable X. See workbook 3-6.
12         11
s 3-3, 3-6, and 6-4.)

prehension score

t events score for
Use this worksheet to carry out the statistical analyses for problem 1-4.
Then compare your answers to the solution on the problem 1-4 worksheet.
Data: Events Comprehension
8           2
4           1
7           3
9           6
13          10
14          16
19          18
17          15
20           5
12          11
Ten obese volunteers agreed to take part in a diet study for six months in which their food intake and
exercise would be precisely managed. One expectation was that after a period of rapid weight loss
in the first three months, pounds would come off more slowly during the second three months.
Others disagreed, believing that weight loss need not slow during the second three months if food
intake and exercise are precisely controlled. Which position do the data support? The data below
are pounds lost by the 10 volunteers in each three-month period of the diet program. Click the "my
work 1-5" tab.

Jan.-Mar. April-June
Jim         8          2
Steve         4          1                  Solution
Matt         7          3                 (columns O to AG)
Paul         9          6
Dave         13         10
Josh        14         16
Larry        19         18
Jerry        17         15
Ethan        20          5
Jacob        12         11
Solution of the paired-samples t te
ch their food intake and
(See workbook 7-3.)
od of rapid weight loss
nd three months.
three months if food               t=     2.515
port? The data below               df =    9
rogram. Click the "my              p=     0.033

Month 1
Jim         8
Steve         4
Matt         7
Paul         9
Dave          13
Josh         14
Larry         19
Jerry         17
Ethan         20
Jacob         12
MEANS:        12.3
the paired-samples t test.       Solution of the Wilcoxon and Sign Tests                     (#DIV/0! error messages in cells W3 an
following changes you make to "data set 1" indicate that an "illegal" tie is present in a row of data.)
The Wilcoxon T is the smaller of these two values:                         3      or
To achieve statistical significance (n =10 pairs) T must be 8 (a=.05) or 3 (a=.01).

The split is       1       negative       and           9                      positive values, so
the sign-test output for these data is a p value = 0.021
rank of
absolute value
Month 2 differences                    Month 1       Month 2        differences of differences abs. value
2          6             Jim            8            2                  6            6          9
1          3           Steve            4            1                  3            3          5
3          4            Matt            7            3                  4            4          8
6          3            Paul            9            6                  3            3          5
10          3           Dave            13            10                 3            3          5
16         -2            Josh           14            16                -2            2          3
18          1           Larry           19            18                 1            1          1
15          2           Jerry           17            15                 2            2          3
5         15           Ethan           20            5                15            15         10
11          1           Jacob           12            11                 1            1          1
8.7                                                                                   sums: 50
messages in cells W3 and Y3                                                Illustration based on the original "data set 1" values
present in a row of data.)
Because of the way Excel resolves tied
47                                                                             Excel's
ranks, if some of the signed ranks are
05) or 3 (a=.01).           ties, the value of T reported here may        absolute value    rank of
differ slightly from the solution of other    of differences absolute value
positive values, so         statistical software. For the more precise            6             9
answer, re-rank the "absolute value of
3             5
differences" (col. W) and enter the new
signed        ranks in column X (rank of absolute                   4             8
ranks        value). Try it yourself first, then see the           3             5
9         illustration (using the original data) in             3             5
5         The "Rank and Percentile" worksheet in                2             3
8         workbook 1-2 explains the precise                     1             1
5         method for resolving tied ranks.                      2             3
5                                                              15            10
-3                                                               1             1
1
3                                                       T now equals 3.5 rather than 3.0,
10                                                       and p .05, but not .01.
1
44
n the original "data set 1" values

Precise
rank of     signed
absolute value ranks
9           9
6           5
8           8
6           5
6           5
3.5        -3.5
1.5          1
3.5          3
10         10
1.5          1

ather than 3.0,
Use this worksheet to carry out the statistical analyses for problem 1-5.
Then compare your answers to the solution on the problem 1-5 worksheet.

Data:
8          2
4          1
7          3
9          6
13         10
14         16
19         18
17         15
20          5
12         11
Data for Part II
In Part II, the data array expands from 2 columns to 4. As for Part I, select and run the
correct statistical analysis, then scroll right to column O and beyond for the solution. To
practice computational skills with new data, enter one or more new values below.
(Make all changes to data here, not on the problem worksheets.)

53          44         40          52
54          59         61          63
42          47         55          59            Any changes to the Part II
47          41         48          57            data must be made here,
49          50         53          49
NOT on the problem
47          42         56          55
worksheets.
57          50         43          53
40          39         55          54
42          39         45          40
46          38         52          54
ct and run the
e solution. To

changes to the Part II
on the problem
Ten fourth-grade children were evaluated by a team of psychologists with respect to four areas
of development: reading skill, math skill, gross-motor development, and fine-motor development.
Is there any association among these measures? For example, do students who read well also
tend to do well in math? Determine the pattern of association among these behavioral
measures, noting any relationships that are significant.

Reading        Math      Gross Motor Fine Motor
Jim        53            44           40          52
Steve       54            59           61          63          Solution
Matt        42            47           55          59         (Columns O to S)
Paul        47            41           48          57
Dave        49            50           53          49
Josh        47            42           56          55
Larry       57            50           43          53
Jerry         40           39            55           54
Ethan         42           39            45           40
Jacob         46           38            52           54
respect to four areas
motor development.
se behavioral

Solution
(Columns O to S)
Solution for the intercorrelation matrix of the four variables.

An example of using the correlation tool to generate an inter-correlation
matrix is in workbook 7-1.

Reading      Math     Gross Motor Fine Motor
Math                           1.000       0.332       0.480
Gross Motor                                1.000       0.573
Fine Motor                                             1.000

With n -2 = 8 degrees of freedom, r crit = 0.6319.

To be significant, the absolute value of an r obt (above, in red)
must equal or exceed the r crit value of .6319.
Use this worksheet to carry out the statistical analyses for problem 2-1.
Then compare your answers to the solution on the problem 2-1 worksheet.

Data:          53          44         40          52
54          59         61          63
42          47         55          59
47          41         48          57
49          50         53          49
47          42         56          55
57          50         43          53
40          39         55          54
42          39         45          40
46          38         52          54
Forty male participants were randomly divided into 4 groups of 10 each. Each group had the
same jigsaw puzzle to assemble, but under four different conditions. Group 1 participants worked
on the puzzle while carrying on a hands-free cell-phone conversation, Group 2 worked while
instant messaging on a hand-held device, Group 3 worked while listening to a talk-radio sports
show, and Group 4 worked in silence. No participants completed the puzzle in the time allowed.
The dependent variable is the number of puzzle pieces that the participant successfully joined to
others. Do the data support the hypothesis that processing verbal information slows completion of
Group 1       Group 2      Group 3      Group 4
53            44           40           52
54            59           61           63        Solution
42            47           55           59        (Columns O to AB)
47            41           48           57
49            50           53           49
47            42           56           55
57            50           43           53
40            39           55           54
42            39           45           40
46            38           52           54
oup 1 participants worked
oup 2 worked while
zle in the time allowed.
nt successfully joined to
ation slows completion of

Solution
(Columns O to AB)
Solution for the one-way ANOVA for independent groups.

Group 1      Group 2   Group 3   Group 4
53           44        40        52      Using ESC file 8-3 as a guide,
54           59        61        63      the Tukey HSD was determined
42           47        55        59      to equal 7.57
47           41        48        57
49           50        53        49      So, to meet the criterion for
47           42        56        55      significance of an unplanned
57           50        43        53      comparison, means must be
40           39        55        54      7.57     units apart.
42           39        45        40
46           38        52        54               See chart
Means:  47.7         44.9      50.8      53.6

Source of Variation       SS       df        MS        F       F crit    p value
Between Groups          426.5     3       142.17     3.62     2.87       0.022
Within Groups (error)    1413      36       39.25
Total                   1839.5     39
56
SC file 8-3 as a guide,
y HSD was determined                             54
Mean Pieces Assembled   52
50
eet the criterion for
nce of an unplanned                               48
son, means must be                                46
units apart.
44

See chart                                         42
40
1   2           3   4
Group
Use this worksheet to carry out the statistical analyses for problem 2-2.
Then compare your answers to the solution on the problem 2-2 worksheet.

Data:          53          44         40          52
54          59         61          63
42          47         55          59
47          41         48          57
49          50         53          49
47          42         56          55
57          50         43          53
40          39         55          54
42          39         45          40
46          38         52          54
A professor gave his 10 graduate students four examinations over the course of the semester, one
on each quarter of the course content. Was overall student performance about the same on the
four examinations, or did grades differ among the four tests?

Students     Exam 1       Exam 2       Exam 3       Exam 4
Jim         53           44           40           52
Steve         54           59           61           63
Matt         42           47           55           59        Solution
Paul         47           41           48           57        (Columns O to AD)
Dave         49           50           53           49
Josh         47           42           56           55
Larry        57           50           43           53
Jerry        40           39           55           54
Ethan         42           39           45           40
Jacob         46           38           52           54
urse of the semester, one

Solution
Solution for one-way repeated-measures ANOVA. (See workbook 8-2.)
Students Exam 1     Exam 2   Exam 3    Exam 4 Row Means
Jim       53        44      40        52     47.25
Steve      54        59      61        63     59.25
Matt      42        47      55        59     50.75
Paul      47        41      48        57     48.25
Dave       49        50      53        49     50.25
Josh       47        42      56        55     50.00
Larry      57        50      43        53     50.75
Jerry      40        39      55        54     47.00
Ethan       42        39      45        40     41.50
Jacob       46        38      52        54     47.50
Col. Means:  47.7      44.9    50.8      53.6

Source          SS     df       MS       F       p value
Between Subjects    717.00    9      79.67    3.09      0.011
Examinations        426.50    3     142.17    5.52      0.004
Error               696.00   27      25.78
Total              1839.50   39
Using ESC file 8-3 as a guide,                56
the Tukey HSD was determined                  54
to equal 3.78 .                               52

Exam Score
50
So, to meet the criterion for
48
significance of an unplanned
comparison, means must be                     46
must be 3.78 units apart.                     44
42
See chart                            40
1   2          3   4
Exam
Use this worksheet to carry out the statistical analyses for problem 2-3.
Then compare your answers to the solution on the problem 2-3 worksheet.

Data:          53          44         40          52
54          59         61          63
42          47         55          59
47          41         48          57
49          50         53          49
47          42         56          55
57          50         43          53
40          39         55          54
42          39         45          40
46          38         52          54
SONY Corp. wanted to determine the age demographic of consumers who purchased plasma TV
monitors in 2004. An inspection of 40 randomly selected post-purchase questionnaires (a form
that consumers complete and return to the company to activate their warranty) revealed the 40
ages below. Based on these data, estimate the average age of consumers who purchased
plasma TV monitors.
Ages of 40 Buyers of Plasma TV Monitors
53          44            40         52
54          59            61         63         Solution
42          47            55         59         (Columns O to T)
47          41            48         57
49          50            53         49
47          42            56         55
57          50            43         53
40          39            55         54
42          39            45         40
46          38            52         54
o purchased plasma TV
questionnaires (a form
anty) revealed the 40
rs who purchased

Solution
(Columns O to T)
Solution for an interval estimate of the mean
age of the population of Plasma TV purchasers.

Lower limit =   47.06

Upper limit =   51.44

Therefore, we are 95% confident that the mean age of
plasma TV buyers is between 47.06 and 51.44

The topic of confidence intervals is covered in workbook 6-3.
Use this worksheet to carry out the statistical analyses for problem 2-4.
Then compare your answers to the solution on the problem 2-4 worksheet.

Data:          53          44         40          52
54          59         61          63
42          47         55          59
47          41         48          57
49          50         53          49
47          42         56          55
57          50         43          53
40          39         55          54
42          39         45          40
46          38         52          54
Participants suffering from osteoarthritis of the hands were randomly assigned to one of four
treatment conditions with an experimental drug (no drug, placebo, low dose, or high dose). The drug
therapy was paired either with a cold compress or a hot compress. The eight treatment
combinations (5 per cell) are evident from inspection of the table below. The data represent minutes
(post treatment) needed to successfully master a finger flexibility task. What do the data tell you
about the effectiveness of the two therapies working alone or in combination?

no drug      placebo      low dose     high dose
cold compress       53            44           40            52
cold compress       54            59           61            63       Solution
cold compress       42            47           55            59       (Columns O to Z)
cold compress       47            41           48            57
cold compress       49            50           53            49
hot compress        47            42           56            55
hot compress        57            50           43            53
hot compress        40            39           55            54
hot compress        42            39           45            40
hot compress        46            38           52            54
Solution for the two-way
d to one of four
or high dose). The drug
treatment
data represent minutes
o the data tell you
A1
A2
Col. Means:

Solution
(Columns O to Z)        Source
Compress
Drug Dose
Interaction
Error

Total
olution for the two-way ANOVA. (See workbooks 9-2 and 9-3.)
60

Cell Means               Row                             55
B1        B2         B3        B4       Means:                          50

Flexibility Scores
49.00     48.20      51.40     56.00       51.15
46.40     41.60      50.20     51.20       47.35                         45
47.70     44.90      50.80     53.60                                     40

Analysis of Variance Summary Table                                   35
SS          df        MS        F       p value                        30
Cold Compress
144.40        1       144.40    3.77       0.061
25
426.50        3       142.17    3.71       0.021                                   Hot Compress

42.60        3        14.20    0.37       0.775                         20
1226.00       32        38.31                                                  No Drug        Plscebo

1839.5       39
See chart
Cold Compress

Hot Compress

Plscebo       Low Dose   High Dose
Drug Treatment
Use this worksheet to carry out the statistical analyses for problem 2-5.

no drug     placebo    low dose high dose
cold compress       53           44         40        52
cold compress       54           59         61        63
cold compress       42           47         55        59
cold compress       47           41         48        57
cold compress       49           50         53        49
hot compress        47           42         56        55
hot compress        57           50         43        53
hot compress        40           39         55        54
hot compress        42           39         45        40
hot compress        46           38         52        54
end of workbook

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