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2011Fall MED_INF_409-DL_SEC55 Introduction to Biostatistics 2011FA_MED_INF_409-DL_SEC55Grade Center Grade Test: Midterm

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A __________ is a quantity computed from the scores in a population; a ______ is a quantity computed from thescores in a sample. (2 pts)

Given Answer: a. parameter; statistic

Correct Answer: a. parameter; statistic

Question 1: Multiple Choice 2 out of 2 points

Which of the following is an example of ordinal scale data? (2 pts)

Given Answer: b. Health rating of excellent, very good, good or fair

Correct Answer: b. Health rating of excellent, very good, good or fair


You are designing an operations manual for a research project which includes the statistical plan. Thestatistician indicated the study should include randomization without replacement. What steps would youinclude and why? (2 pts)GivenAnswer:

There are three classic methods for selecting a simple random sample without replacement:1. Lottery Method: The members of the population are assigned numbers from 1 to N, and thenumbers are written on slips of paper, round balls, or some other medium of identical size andshape. The slips or balls are placed in a container and mixed thoroughly. Then a slip or ball is pickedout, the slips/balls are mixed again, another slip is picked, and so on. The procedure repeats untilthe desired number of slips (sample size=n) has been picked. Since the slips are not replacedbetween rounds of choosing, the sampling is "without replacement." The method works well for small

Question 3: Short Answer 2 out of 2 points

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populations, but is cumbersome for large populations.2. Using a Random Number table: Again, the members of the population are assigned numbers from1 to N (or from 0 to N-1). The researcher then consults a random number table (many available bothin print or on line) to choose the random sample. For example, suppose we want to select a randomsample of 10 from a population of 100. We number the population units from 00 to 99 and select tentwo-digit numbers from a random number table. If any number occurs twice, we ignore the secondoccurrence because the sampling is without replacement.Computer Algorithm: There are a large number of standardized computer methods in print and online for selecting a simple random sample without replacement. Most of them depend on selectingrandom (actualy pseudorandom) numbers using one of many available algorithms. Many that I havefound on line reference Donald Knuth's classic The Art of Computer Programming volume 2(Seminumerical Algorithms). One example written in Python would be:## {{{ http://code.activestate.com/recipes/272884/ (r1)def sample(n, r): "Generate r randomly chosen, sorted integers from [0,n)" rand = random.random pop = n for samp in xrange(r, 0, -1): cumprob = 1.0 x = rand() while x < cumprob: cumprob -= cumprob * samp / pop pop -= 1 yield n-pop-1

# Example call to select three samples in range(0,10)>>> list(sample(10, 3))[2, 7, 8]## end of http://code.activestate.com/recipes/272884/ }}}Retrieved from http://code.activestate.com/recipes/272884-random-samples-without-replacement/

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For a normal distribution with a mean of 80 and a standard deviation of 10, find the probability value for p(X <95). (1 pts)Given Answer: Z= X-mu/sigma=95-80/10=1.5. P(X < 95)=P(Z < Z1.5)=0.9332.




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For a normal distribution with a mean of 80 and a standard deviation of 10, find the probability value for p(85 <X < 95). (1 pts)Given Answer: Z=X-mu/sigma=85-80/10=0.5. P(X < 85)= P(Z < Z0.5)=0.6915.

P(85 < X < 95)= P(X < 95) - P(X < 85)= 0.9332 - 0.6915=0.2417

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1 out of 1 points

For a normal distribution with a mean of 140 and a standard deviation of 55 find each value requested. (2 pts)a. What is the score needed to be in the bottom 10% of the distribution?b. What scores form the boundaries for the middle 75% of the distribution.

GivenAnswer:

a. X=mu+(Z x sigma) so X=140+ (Z0.1 x 55). Z0.1 = -1.285. Therefore X=140+(-1.285 x 55)=140-




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70.675=69.325. Any score less than 69.325 will be in the bottom 10% of the distribution.b. Scores in the middle 75 % of the distribution will be between Z0.125 and Z0.875.Z0.125 = -1.15 and Z0.875 = 1.15. X=mu+(Z x sigma).X=140+(-1.15 x 55)=140-63.25=76.75 and X=140+(1.15 x 55)=140+63.25=203.25.Therefore the middle 75% of the distribution lies between 76.75 and 203.25.

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We are testing the effects of a new flu medicine in a randomly drawn sample of 10 year old children with the fluand compare the length of their illness to the mean length of illness in the population of untreated 10 year oldchildren with the flu. The t statistic for our sample falls within the critical region. We should ________. (2 pts)

Given Answer: d. Both a and b

Correct Answer: a. Reject the Null hypothesis


We are testing the effects of a new therapy for depression in a randomly drawn sample of depressed patientsand compare their depression scores at the end of therapy to the mean depression score in the population ofuntreated depressed patients. Our single sample t statistic does not fall within the critical region. We should____________.(2 pts)

Given Answer: b. Fail to reject the H0

Correct Answer: b. Fail to reject the H0


When we know the value of the population standard deviation, we use which of the following tests? (2 pts)

Given Answer: b. z test

Correct Answer: b. z test


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Class discussion later showed that answer d was correct.

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In general, the interval width of a confidence interval _______ as sample size (n) gets larger and ________ as theconfidence level increases. (2 pts)

Given Answer: gets smaller, gets larger

Correct Answer: gets smaller, gets larger


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You are designing a study to determine whether music has an effect on learning. Previous research indicates thatthe standard deviation to learn a task is 3.4 minutes. How many subjects would you need to ensure with 90%confidence that the estimate is within 1 minute of the true mean time required? (2 pts)Given Answer: n=((Z1-alpha/2 x sigma)/E)2 = ((Z0.95 x 3.4)/1)2 = ((1.645 x 3.4)/1)2 = (5.593)2 = 31.28 so

we need 32 subjects.CorrectAnswer:

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A Gallup Poll shows that 60% of voters support the new health care reform bill. If nine voters are randomly selected,what is the probability that at least three voters favor the new health care reform bill? (2 pts)GivenAnswer:

The result is calculated using the binomial distribution with n=9 and p=0.6. The probability that at least 3voters favor the bill is:P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)=0.0743+0.1672+0.2508+0.2508+0.1612+0.0605+0.0101=0.9749

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Use the attached Excel dataset titled “CardioStudy.xls” for this problem. 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 Description Data Type Range of Values

ID 5-Digit Unique Subject

Identifier

Numeric 101-96017

CVD Cardiovascular Disease Numeric No = 0

Yes = 1

TRT Treatment Group Numeric 0 = Placebo

1 = Drug

AGE Subject’s Age Numeric 34-60

HDL HDL Cholesterol Numeric 18-98

LDL LDL Cholesterol Numeric 155-428

SMOKER Smoking Status Numeric 0= No

1 = Yes

1. 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.

a. What percent of the subjects are in the PLACEBO treatment group? (1 pt)

TREATMENT FREQUENCIES

Frequency Percent Valid Percent

Cumulative

Percent

Valid 0=Placebo 10 50.0 50.0 50.0

1=Drug 10 50.0 50.0 100.0

Total 20 100.0 100.0

Fifty percent of the subjects are in the Placebo treatment group.

Midterm Exam Part II

MED_INF 409: Biostatistics

MED_INF 409: Biostatistics Midterm Exam Part II

© 2011 Northwestern University School of Continuing Studies 2

b. How many subjects have cardiovascular disease? (1 pt)

Twelve subjects have cardiovascular disease.

c. Select the best way to describe HDL cholesterol and smoking status graphically. Paste the graphic

descriptions here. Interpret the graphs (2 pts).

STATISTICS: HDL LEVELS IN

NONSMOKERS

HDL

N Valid 10

Missing 0

Mean 43.10

Std. Error of Mean 7.856

Median 38.50

Std. Deviation 24.844

Skewness .441

Std. Error of Skewness .687

Kurtosis -1.535

Std. Error of Kurtosis 1.334

Range 62

Minimum 18

Maximum 80

Cardiovascular Disease

Frequency Percent Valid Percent

Cumulative

Percent

Valid 0=No 8 40.0 40.0 40.0

1=Yes 12 60.0 60.0 100.0

Total 20 100.0 100.0



STATISTICS: HDL LEVELS IN SMOKERS

HDL

N Valid 10

Missing 0

Mean 70.70


Median 73.00


Skewness -.698


Kurtosis -.293


Range 62

Minimum 33

Maximum 95



T-Test

Group Statistics

SMOKING STATUS N Mean Std. Deviation Std. Error Mean

HDL 0=Nonsmoker 10 43.10 24.844 7.856

1=Smoker 10 70.70 19.721 6.236

Independent Samples Test

Equal variances assumed

Levene's Test for

Equality of Variances t-test for Equality of Means

F Sig. t df

Sig. (2-

tailed)

Mean

Difference

Std. Error

Difference

95% Confidence Interval of

the Difference

Lower Upper

HDL 1.665 .213 -2.752 18 .013 -27.600 10.031 -48.673 -6.527

Interpretation: Nonsmokers have a lower mean HDL than smokers and the difference is statistically significant

(p=0.013).



d. Select the best way to describe LDL and smoking status numerically. Paste the SPSS tables with

the numeric descriptions here. Interpret the output (2 pts).

STATISTICS: LDL LEVELS IN

NONSMOKERS

LDL

N Valid 10

Missing 0

Mean 226.70


Median 168.00


Skewness .693


Kurtosis -1.455


Range 220

Minimum 155

Maximum 375

STATISTICS: LDL LEVELS IN SMOKERS

LDL

N Valid 10

Missing 0

Mean 284.30


Median 291.50


Skewness .061


Kurtosis -1.216


Range 269

Minimum 159

Maximum 428



T-Test



Levene's Test for


F Sig. t df

Sig. (2-

tailed)

Mean

Difference

Std. Error

Difference


the Difference

Lower Upper

LDL .018 .894 -1.414 18 .174 -57.600 40.729 -143.168 27.968

Interpretation: There is no significant difference between the mean LDL values of smokers and nonsmokers.

e. What is the mean age in the data set for subjects who are

smokers? What is the mean age for non-smokers (2 pts)?

Group Statistics

Smoking Status N Mean Std. Deviation Std. Error Mean

LDL 0=Nonsmoker 10 226.70 85.824 27.140

1=Smoker 10 284.30 96.034 30.368

STATISTICS: AGES OF NONSMOKERS

AGE

N Valid 10

Missing 0

Mean 48.60


Median 51.50




Interpretation: The mean age for smokers is 49.1 years, and the mean age for nonsmokers is 48.6 years.

f. Test the hypothesis that there is a difference in HDL cholesterol between the treatment and

control groups at the 5% confidence level (5 pts).

T-Test

Group Statistics

Treatment Group N Mean Std. Deviation Std. Error Mean

HDL 0=Placebo 10 58.60 29.530 9.338

1=Drug 10 55.20 23.470 7.422

Interpretation: There is not a significant difference in HDL cholesterol between the treatment and placebo groups at

the 5% confidence level.

STATISTICS: AGES OF SMOKERS

AGE

N Valid 10

Missing 0

Mean 49.10


Median 50.00




Levene's Test for


F Sig. t df

Sig. (2-

tailed)

Mean

Difference

Std. Error

Difference


the Difference

Lower Upper

HDL 1.031 .323 .285 18 .779 3.400 11.928 -21.661 28.461



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

before

program

# servings

after 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 6 8

T-Test

Paired Samples Statistics

Mean N Std. Deviation Std. Error Mean

Pair 1 Svg_After 5.90 10 1.287 .407

Svg_Before 5.60 10 2.011 .636

Paired Samples Correlations

N Correlation Sig.

Pair 1 Svg_After & Svg_Before 10 -.017 .962



Interpretation: The program did not significantly improve vegetable consumption at the 5% level of significance.

Paired Samples Test

Paired Differences t df Sig. (2-tailed)

Mean

Std.

Deviation

Std. Error

Mean

95% Confidence

Interval of the

Difference

Lower Upper

Pair

1

Svg_After - Svg_Before .300 2.406 .761 -1.421 2.021 .394 9 .703

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Grade Test: Midterm - medinfodoc.net Test: Midterm 11/9/11 9:03 PM €¦%3Fcourse_id%3D_291438_1%26cmd%3Dnext%26sequenceId%3D_291438_1