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Pagano-Understanding Statistics in Behavioral Sci (Homework)

James Finch

Statistics, Fall 2010

Instructor: Dr. Friendly

Current Score : 0 / 49

Due : Tuesday, November 30, 2010 23:00 EST

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Question
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Total
0/49

Description

Here is a sample assignment with questions to accompany Understanding Statistics in Behavioral Sciences 9/e by Robert R. Pagano published by Wadsworth. Click here for a list of all of the questions coded in WebAssign.

Questions 1, 2, 3, and 4 are Tutorial questions. They provide step-by-step walk-throughs that demonstrate common problem types.

Questions 5, 6, 7 and 8 are Additional Practice Problems. These problems are similar to the practice problems in the textbook. Each problem presents an ideal solution which is especially useful to students for learning the solution and for correcting their own attempts at solving the problem.

Questions 9, 10, 11, and 12 are End-of-Chapter problems from your text. These questions are exactly the same as the questions in your textbook except for minor wording changes to make them work in a web environment. As often as possible, variables, and numbers are generated so that each student will receive a unique version and so students can attempt the problem multiple times.


Instructions

This demo assignment allows many submissions and allows you to try another version of the same question for practice.



1. –/1 points Notes Question: PagUnderStat9 2.Tut.01.
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2. –/1 points Notes Question: PagUnderStat9 2.Tut.02.
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3. –/1 points Notes Question: PagUnderStat9 4.Tut.01.
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4. –/1 points Notes Question: PagUnderStat9 5.Tut.01.
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5. –/2 points Notes Question: PagUnderStat9 4.APP.03.
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Calculate the standard deviation of the following set of sample scores.
(a)    12, 14, 16, 19, 22, 24, 27
Enter a number.


(b)    0.1, 0.4, 0.8, 1.2, 1.7, 2.8
Enter a number.


Solution or Explanation
(a)
X X2 Calculation of SS Calculation of s
12
14
16
19
22
24
27
ΣX = 134
N = 7     		144
196
256
361
484
576
729
ΣX2 = 2746
SS = ΣX2 − 
(ΣX)2
N
 
 = 2746 − 
(134)2
7
 
 = 180.8571
s = 
SS
N − 1
 = 
180.8571
6
 
 = 5.49

(b)
X X2 Calculation of SS Calculation of s
0.1
0.4
0.8
1.2
1.7
2.8
ΣX = 7.0
N = 6     		0.01
0.16
0.64
1.44
2.89
7.84
ΣX2 = 12.98
SS = ΣX2 − 
(ΣX)2
N
 
 = 12.98 − 
(7.0)2
6
 
 = 4.8133
s = 
SS
N − 1
 = 
4.8133
5
 
 = 0.98

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6. –/1 points Notes Question: PagUnderStat9 5.APP.02.
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Scores obtained on a particular aptitude test by psychology students are normally distributed, with a mean of 150 and a standard deviation of 25. Assume the distribution is a population set of scores. What is the percentile rank of a score of 180?
Enter a number.


Solution or Explanation
In solving problems involving areas under the normal curve, it is wise, at the outset, to draw a picture of the curve and locate the relevant areas on it. Such a picture is shown below.

Maple Generated Plot

zA = 
X − μ
σ
 = 
180 − 150
25
 = 
30
25
 = 1.20


To find the area between the mean and a z score of 1.20, we enter Table A, locate the z value in column A, and read off the corresponding entry in column B. This value is 0.3849. Thus, the proportion of the total area between the mean and a z score of 1.20 is 0.3849. From the picture above, we can see that the remaining scores below the mean, labeled B in the picture, occupy 0.5000 proportion of the total area. If we add these two areas together, we shall have the proportion of scores lower than 180. Thus, the proportion of scores lower than 120 is 0.3849 + 0.5000 = 0.8849. The percentile rank of 180 is then 0.8849 × 100 = 88.49.

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7. –/3 points Notes Question: PagUnderStat9 14.APP.01.
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The correlated t test is very much like the t test for single samples. The main difference is that with the correlated t test, we analyze a single set of difference scores, while with the t test for single samples we analyze a single set of raw scores. Let's first do a bare bones problem for practice.

Given the data shown in the table below, determine whether there is a significant difference between conditions. In the experiment, each subject was run in both conditions, resulting in paired scores for each subject. Assume that Condition 1 and Condition 2 were made as similar as possible, except that in Condition 1 the IV was present and in Condition 2 it was absent. Use α = 0.052 tail in making your decision.

Subject Condition 1 Condition 2
1
2
3
4
5
6
7
8 		25
22
18
36
24
11
31
27 		28
20
22
39
23
15
36
29

tobt =
Enter a number.
tcrit =
Enter an exact number.

What is your conclusion?





Solution or Explanation
Subject Condition 1 Condition 2 Difference
Score
D 		D2
1
2
3
4
5
6
7
8
N = 8 		25
22
18
36
24
11
31
27 		28
20
22
39
23
15
36
29 		3
−2
4
3
−1
4
5
  2
ΣD = 18 		9
4
16
9
1
16
25
  4
ΣD2 = 84
Dobt = 
ΣD
N
 = 
18
8
 = 2.25

Conclusion, using α = 0.052 tail
Step 1: Calculate the appropriate statistic. For this design, of the inference tests covered so far, both the sign test and the t test are possible choices. Because we want to use the most sensitive test, the t test has been chosen. From the above table, N = 8, and
Dobt
= 2.25, ΣD = 18, and ΣD2 = 84. The calculation of tobt is as follows:
tobt  = 
Dobt − μD
SSD
N(N − 1)
 
 
 = 
2.25 − 0
43.5
8(7)
 
 
 =  2.55
SSD  =  ΣD2 − 
(ΣD)2
N
 
 =  84 − 
(18)2
8
 
 =  43.5

Step 2: Evaluate the statistic. Degrees of freedom = N − 1 = 8 − 1 = 7. From Table D, with α = 0.052 tail and 7 df,

tcrit = ±2.365

Because |tobt| > 2.365, it falls within the critical region for rejection of H0. Therefore, you reject the null hypothesis and conclude that there is a significant difference between conditions. It appears that the IV decreases the value of the DV.

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8. –/5 points Notes Question: PagUnderStat9 15.APP.02.
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A clinical psychologist conducts an experiment to evaluate two methods for promoting smoking cessation, nicotine patch and nicotine gum. Eighteen volunteers from a heavy smoking population are assigned six each to one of three groups. Conditions are similar for all groups, except for the treatment. Subjects in Group 1 get the nicotine patch. Subjects in Group 2 get the nicotine gum and subjects in Group 3 chew a placebo gum. All subjects are asked to do their best to stop smoking. The treatment for each group lasts for three months. On the last treatment day the number of cigarettes smoked by each subject is recorded. The results are shown in the following table.

Group 1
Nicotine Patch 		Group 2
Nicotine Gum 		Group 3
Placebo Gum
10
8
6
7
10
9 		11
6
9
7
8
5 		15
12
14
16
15
12
(a) What is the alternative hypothesis?





(b) What is the null hypothesis?





(c) What is the conclusion? Use α = 0.05.
Fobt =
Enter a number.
Fcrit =
Enter a number.





Solution or Explanation
Group 1
Nicotine Patch 		Group 2
Nicotine Gum 		Group 3
Placebo Gum
X1 X12 X2 X22 X3 X32
10
 7
 6
 9
10
 9
51 		100
 49
 36
 81
100
 81
447 		11
 6
 9
 7
 8
 7
48 		121
 36
 81
 49
 64
 49
400 		15
12
14
16
15
12
84 		225
144
196
256
225
144
1190
n1 = 6
X1 = 
51
6
 = 8.5
n2 = 6
X2 = 
48
6
 = 8.0
n3 = 6
X3 = 
84
6
 = 14.0
N = 18
 
all scores
 
X = 183
 
 
all scores
 
X2 = 2037
 
XG = 
all scores
 
X
N
 
 = 
183
18
 = 10.17

(a) Alternative hypothesis: The alternative hypothesis states that at least one of the treatments affects smoking cessation differently from the other two. Therefore, at least one of the means (μ1, μ2, or μ3) differs from at least one of the others.

(b) Null hypothesis: The null hypothesis states that the treatments have the same effect on smoking cessation. Therefore, the three sets of sample scores are random samples from populations where μ1 = μ2 = μ3.

(c) Conclusion, using α = 0.05
A. Calculate Fobt.

Step 1: Calculate the between-groups sum of squares, SSB.

SSB = 
X1
2
 
n1
 
 + 
X2
2
 
n2
 
 + 
X3
2
 
n3
 
 − 
all scores
 
X
2
 
N
 = 
(51)2
6
 + 
(48)2
6
 + 
(84)2
6
 − 
(183)2
18
 = 133.0


Step 2: Calculate the within-groups sum of squares, SSW.

SSW = 
all scores
 
X2
 − 
X1
2
 
n1
 
 + 
X2
2
 
n2
 
 + 
X3
2
 
n3
 
 = 2037 − 
(51)2
6
 + 
(48)2
6
 + 
(84)2
6
 = 43.5


Step 3: Calculate the total sum of squares, SST.

SST = 
all scores
 
X2
 − 
all scoresX
2
 
N
 
 = 2037 − 
(183)2
18
 = 176.5

Note, this step is a check on the calculations for SSB and SSW.
SST = SSB + SSW
176.5 = 133.0 + 43.5
176.5 = 176.5
The calculations are correct.


Step 4: Calculate the degrees of freedom for each estimate.
dfB = k − 1 = 3 − 1 = 2
dfW = N − k = 18 − 3 = 15
dfT = N − 1 = 18 − 1 = 17


Step 5: Calculate the between-groups variance estimate, sB2.
sB2 = 
SSB
2
 = 
133.0
2
 = 66.5

Step 6: Calculate the within-groups variance estimate, sW2.
sW2 = 
SSW
dfW
 = 
43.5
15
 = 2.9

Step 7: Calculate Fobt.
Fobt = 
sB2
sW2
 = 
66.5
2.9
 = 22.93


B. Evaluate Fobt. From Table F, with α = 0.05, dfnumerator = 2, and dfdenominator = 15,
Fcrit = 3.68
Since Fobt > Fcrit, we reject H0. The independent variable has had an effect on the dependent variable. Increasing levels of the independent variable appear to increase the value of the dependent variable. A summary of the solution is shown in the following table.

Source SS df s2 Fobt
Between groups
Within groups
Total 		133.0
43.5
176.5 		2
15
17 		66.5
2.9 		22.93*
*With α = 0.05, Fcrit = 3.68. Therefore, H0 is rejected.

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9. –/5 points Notes Question: PagUnderStat9 2.EOC.11.
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Round the following numbers to one decimal place.
(a) 1.923
Enter a number.


(b) 22.250
Enter a number.


(c) 101.550
Enter a number.


(d) 42.777
Enter a number.


(e) 33.473
Enter a number.



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10. –/6 points Notes Question: PagUnderStat9 4.EOC.35.
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A social psychologist interested in the dating habits of college undergraduates samples 10 students and determines the number of dates they have had in the last month. Given the scores 2, 6, 12, 3, 6, 13, 5, 4, 6, 17 compute the following:
(a) Mean
Enter an exact number.


(b) Median
Enter an exact number.


(c) Mode
Enter an exact number.


(d) Range
Enter an exact number.


(e) Standard deviation
Enter a number.


(f) Variance
Enter a number.

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11. –/4 points Notes Question: PagUnderStat9 6.EOC.15.
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In a large introductory sociology course, a professor gives two exams. The professor wants to determine whether the scores students receive on the second exam are correlated with their scores on the first exam. To make the calculations easier, a sample of eight students is selected. Their scores are shown in the accompanying table.

Student Exam 1 Exam 2
1 58 62
2 74 101
3 70 78
4 74 70
5 56 74
6 81 99
7 81 84
8 63 90

(a) Construct a scatter plot of the data, using exam 1 score as the X variable.

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Does the relationship look linear?
    


(b) Assuming a linear relationship exists between scores on the two exams, compute the value for Pearson r.
Enter a number.


(c) How well does the relationship account for the scores on exam 2?
Enter a number.
%

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12. –/6 points Notes Question: PagUnderStat9 10.EOC.09.
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A primatologist believes that rhesus monkeys possess curiosity. She reasons that, if this is true, then they should prefer novel stimulation to repetitive stimulation. An experiment is conducted in which 12 rhesus monkeys are randomly selected from the university colony and taught to press two bars. Pressing bar 1 always produces the same sound, whereas bar 2 produces a novel sound each time it is pressed. After learning to press the bars, the monkeys are tested for 15 minutes, during which they have free access to both bars. The number of presses on each bar during the 15 minutes is recorded. The resulting data are as follows.

Subject Bar 1 Bar 2
1 23 51
2 25 36
3 22 36
4 12 26
5 1 23
6 24 29
7 13 26
8 27 33
9 14 29
10 5 19
11 29 38
12 29 30

(a) What is the alternative hypothesis? In this case, assume a nondirectional hypothesis is appropriate because there is insufficient empirical basis to warrant a directional hypothesis.





(b) What is the null hypothesis?





(c) Using α = 0.052-tail, what is your conclusion?
p(obtained probability) =
Enter a number.






(d) What error may you be making by your conclusion in part (c)?





(e) To what population does your conclusion apply?




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13. –/13 points Notes Question: PagUnderStat9 15.EOC.25.
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A university researcher knowledgeable in Chinese medicine conducted a study to determine whether acupuncture can help reduce cocaine addiction. In this experiment, 18 cocaine addicts were randomly assigned to one of three groups of 6 addicts per group. Once group received 10 weeks of acupuncture treatment in which the acupuncture needles were inserted into points on the outer ear where stimulation is believed to be effective. Another group, a placebo group, had acupuncture needles inserted into points on the ear believed not to be effective. The third group received no acupuncture treatment; instead, addicts in this group received relaxation therapy. All groups also received counseling over the 10-week treatment period. The dependent variable was craving for cocaine as measured by the number of cocaine urges experienced by each addict in the last week of treatment. The following are the results.

Acupuncture +
Counseling 		Placebo +
Counseling 		Relaxation Therapy +
Counseling
3 8 11
8 11 6
8 12 11
5 10 8
1 9 10
5 9 7

(a) Using α = 0.05, what do you conclude?
Fill in the missing values.
Source SS df s2 Fobt
Between
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Within
Enter a number.
Enter a number.
Enter a number.
Total
Enter a number.
Enter a number.

What is the critical F value?
Fcrit =
Enter a number.


What do you conclude?





(b) If there is a significant effect, estimate the size of the effect, using omega hat2. (If there is not a significant effect, enter NO EFFECT.)
omega hat2 =
Enter a number.


(c) This time estimate the size of the effect, using η2.
η2 =
Enter a number.

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