Download File - Eric Chavez MD MMI

Chavez, Eric
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ClinRes 490
Midterm Exam
Instructions for Midterm Exam:
1. Please enter your last name and first name in the header above.
2. Save your document using the following naming convention:
lastname_firstname_clinres409_midtermexam.doc (or .docx)
3. For SPSS output, please paste the appropriate tables in the Word document and make reference
to each table and why you are including it. If a table is associated with one of the hypothesis
testing steps, it must be clear which step and why you are including it there. Please also discuss
and interpret as appropriate.
4. Please submit exam through Blackboard by the date and time listed on your syllabus. Thank
you – I can only grade exams I can read and interpret clearly!
1. You have just received an excel dataset called “CardioStudy.xls”
This dataset contains data from a cardiovascular study, which looked at the relationship between
cardiovascular disease and a set of risk factors. The data set contains one record per subject and has the
following variables:
Variable
ID
Data Type
Numeric
Range of Values
101-96017
CVD
Description
5-Digit Unique Subject
Identifier
Cardiovascular Disease
Numeric
TRT
Treatment Group
Numeric
AGE
HDL
LDL
SMOKER
Subject’s Age
HDL Cholesterol
LDL Cholesterol
Smoking Status
Numeric
Numeric
Numeric
Numeric
No = 0
Yes = 1
0 = Placebo
1 = Drug
34-60
18-98
155-428
0= No
1 = Yes
Using the dataset CardioStudy Run the appropriate analysis to answer each
of the following questions. Paste the appropriate tables or graphs into your response as
appropriate.
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Chavez, Eric
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a. What percent of the subjects are in the PLACEBO treatment group? (1 pt)
50% of subjects are in the placebo group
TRT
Frequency
Valid
Percent
Valid Percent
Cumulative
Percent
placebo
10
50.0
50.0
50.0
drug
10
50.0
50.0
100.0
Total
20
100.0
100.0
b. How many subjects have cardiovascular disease? (1 pt)
12 subjects have cardiovascular disease
CVD
Frequency
Valid
2
Percent
Valid Percent
Cumulative
Percent
no CVD
8
40.0
40.0
40.0
yes CVD
12
60.0
60.0
100.0
Total
20
100.0
100.0
Chavez, Eric
c. Select the best way to describe HDL cholesterol by smoking status graphically. Paste
the graphic descriptions here. Interpret the graphs (4 pts).
This box and whisker plot describes the HDL values of non-smokers and of smokers. The bottom of each
box represents the first quartile of the HDL values: for the non-smokers this is approximately 20 and for
the smoker is 56. The band inside the box represents the second quartile or the median: for the nonsmokers this is approximately 38 and for the smokers is 73. The top of each box represents the third
quartile: for the non-smokers this is 60 and for the smokers it is 87. The whiskers which extend above
and below the boxes represent the range of HDL values with minimum on the bottom and maximum on
the top: for non-smokers the range is 62 with minimum of 18 and maximum of 80, for the smokers the
range is 62 with a minimum of 33 and maximum of 95.With these descriptive statistics it appears that
non-smokers have lower HDL values than smokers.
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Chavez, Eric
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d. Select the best way to describe LDL by smoking status numerically. Paste the SPSS
tables with the numeric descriptions here. Interpret the output (3 pts).
Descriptives
SMOKING_STATUS
LDL
NON-SMOKER
Statistic
Mean
95% Confidence Interval
for Mean
226.70
Lower Bound
165.31
Upper Bound
288.09
5% Trimmed Mean
222.44
Median
168.00
Variance
27.140
7365.789
Std. Deviation
SMOKER
Std. Error
85.824
Minimum
155
Maximum
375
Range
220
Interquartile Range
150
Skewness
.693
.687
Kurtosis
-1.455
1.334
Mean
284.30
30.368
95% Confidence Interval
for Mean
Lower Bound
215.60
Upper Bound
353.00
5% Trimmed Mean
283.28
Median
291.50
Variance
Std. Deviation
9222.456
96.034
Minimum
159
Maximum
428
Range
269
Interquartile Range
202
Skewness
.061
.687
-1.216
1.334
Kurtosis
This table shows descriptive statistics for LDL values for the subjects divided by smoking status. In the
table we can see that the mean LDL for non-smokers is 226.7 with a standard deviation of 85.8. For the
non-smokers the minimum LDL is 155 and maximum is 375 which gives a range of 220. The 95%
confidence interval for LDL values for non-smokers is (165.3, 288.1).
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Chavez, Eric
The mean LDL for smokers is 284.3 with a standard deviation of 96.0. For the smokers, the minimum LDL
is 159 and maximum is 428 which gives a range of 269. The 95% confidence interval for LDL values for
smokers is (215.6, 353.0).
This table shows the quartiles for LDL values divided by smoking status. For the non-smokers the first
quartile is 158, the second quartile is 168, and the third quartile is 305. For the smokers the first quartile
is 178, the second quartile is 291, and the third quartile is 370.
With these descriptive statistics it appears that non-smokers have lower LDL values than do smokers.
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d.
What is the mean age in the data set for subjects who are
smokers? What is the mean age for non-smokers (2 pts)?
The mean age for smokers is 49.1.
The mean age for non-smokers is 48.6.
Descriptives
SMOKING_STATUS
AGE
NON-SMOKER
Statistic
Mean
95% Confidence Interval
for Mean
48.60
Lower Bound
41.70
Upper Bound
55.50
5% Trimmed Mean
48.78
Median
51.50
Variance
93.156
Std. Deviation
Minimum
35
Maximum
59
Range
24
Interquartile Range
19
Kurtosis
Mean
95% Confidence Interval
for Mean
.687
-1.716
1.334
49.10
3.096
Lower Bound
42.10
Upper Bound
56.10
49.33
Median
50.00
Variance
95.878
9.792
Minimum
34
Maximum
60
Range
26
Interquartile Range
21
Skewness
Kurtosis
6
-.378
5% Trimmed Mean
Std. Deviation
3.052
9.652
Skewness
SMOKER
Std. Error
-.296
.687
-1.537
1.334
Chavez, Eric
e. Test the hypothesis that there is a difference in HDL cholesterol between the
treatment and control groups at the 5% confidence level (5 pts).
1. Set up the hypothesis.
H0: μ1 = μ2 (mean HDL in the treatment group = mean HDL of the control group)
H1: μ1 ≠ μ2 (mean HDL in the treatment group is not equal to the mean HDL of the control group)
Set the level of significance, α=0.05
2. Select the appropriate test statistic. Will use independent samples t test since the HDL is measured on
two independent samples, the population σ is unknown and the n<30. Will use SPSS to calculate the test
statistic t since the population variances are unknown and may not be equal.
3. Generate the decision rule. Assuming degrees of freedom is 18 (n-2 = 20-2 = 18) then for a two-sided
test at α=0.05 the critical value of t is 2.101. Reject H0 if t ≤ -2.101 or if t ≥ 2.101.
Do not reject H0 if -2.101 < t < 2.101.
4. Compute the value of the test statistic. The output from SPSS pasted below.
Group Statistics
TRT
HDL
N
Mean
Std. Deviation
Std. Error
Mean
placebo
10
58.60
29.530
9.338
drug
10
55.20
23.470
7.422
In the Levene’s test, the F value is not significant. Therefore population variances are assumed equal (we
do not reject the null hypothesis that states that the population variances are equal). This means that
we use the top line of the output in the table above.
The test statistic is computed as t=0.285.
5. Draw a conclusion about H0 by comparing the test statistic to the decision rule.
Because the test statistic lies within the critical t values, -2.101 ≤ 0.285 ≤ 2.101 we do not reject the null
hypothesis. We do not have significant evidence, α=0.05, to show that there is a difference in HDL
values between the treatment (drug) and control (placebo) groups. The mean HDL for the treatment and
control groups are equal. The treatment does not affect HDL levels compared to placebo.
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2.
A healthy eating program was held in a college dorm. When freshmen students moved into
the dorm, they completed a survey that included the number of servings of vegetables
typically eaten. Following the program, they repeated the survey. Test the hypothesis that
the program improved vegetable consumption at a 5% level of significance (5 pts).
Subject ID
# servings
# servings
before
after
program
program
1
5
6
2
4
6
3
7
4
4
3
4
5
9
5
6
6
7
7
3
6
8
8
6
9
5
7
10
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1. Set up the hypothesis.
H0: μbefore = μafter or the mean difference = 0 (mean #servings_before = mean #servings_after)
H1: μbefore < μafter or mean difference > 0 (mean #servings_before < mean #servings_after)
Set the level of significance, α=0.05
2. Select the appropriate test statistic.
We will use the dependent samples t test since we are comparing two paired measures taken on the
same subject. Will use SPSS to calculate the test statistic t.
3. Generate the decision rule.
Degrees of freedom is 9 (n-1 = 10-1 = 9), then for a one-sided test at α=0.05 the critical value of t is
1.833. Reject H0 if t ≥ 1.833. Do not reject H0 if t < 1.833.
4. Compute the value of the test statistic. The output from SPSS is pasted below.
Paired Samples Statistics
Mean
Pair 1
N
Std. Error
Mean
Std. Deviation
servings_before
5.60
10
2.011
.636
servings_after
5.90
10
1.287
.407
Paired Samples Correlations
N
Pair 1
8
servings_before &
servings_after
Correlation
10
-.017
Sig.
.962
Chavez, Eric
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The test statistic t = -0.394.
5. Draw a conclusion about H0 by comparing the test statistic to the decision rule.
Since the calculated t of -0.394 < 1.833 the critical value of t, we do not reject the null hypothesis. The
mean difference is zero. We do not have significant evidence, α=0.05, that the mean difference of
number of servings before and after is greater than zero. The healthy eating program did not improve
vegetable consumption.
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