Transcript Midterm Review

Midterm Review
PU515 – Applied Biostatistics
Dana Colbert-Wheeler, MHA, MCHES
• How are you all feeling about the course thus
far?
• What specific things are you struggling with, and
what have you mastered at this point?
• Reminder: please make sure you are reviewing
the examples in your textbook/workbook!
First things first…
• The midterm consisted of 8 questions
• Calculations
• True/False
• The total points for the exam was 100 points.
• The midterm covered Chapters 1-5, but focused
mostly on probability
PU515 Midterm Review
• As you will see from this review and from your
Units 6-9 assignments, the key to solving these
problems is thorough review of the
textbook/workbook.
• For the midterm, in particular, Table 5-10
provides a Summary of Key Formulas. This is
what you need to address every problem!
• Every chapter from has this summary – find it!
PU515 Midterm Review
• Question #1: Glucose levels in patients free of
diabetes are assumed to follow a normal
distribution with a mean of 120 and a standard
deviation of 16.
• In other words, patients who do not have diabetes
generally have an average glucose level of 120, but this
level can vary. Since the standard deviation is 16, it can
be as low as 104, and as high as 136.
PU 515 Midterm Review
•
A.) What proportion of patients has glucose levels exceeding
115?
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In other words, of the total # of patients, how many of them
have glucose levels higher than 115?
So what should be our first step here? How do we know where
to begin?
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We need to find the appropriate formula for this problem.
Remember the scenario. It stated that glucose levels followed a
normal distribution, so we need to use that formula (see section
5.6.2)
PU515 Midterm Review
• Since 120 is the mean, and 115 is the level we have been
given (which is lower than the mean), we subtract by 115
from 120 and then divide that value by the standard
deviation, which is 16.
• 115 – 120 / 16 = -0.3125.
• We then find the z value of -0.3125 using the table in the
back of your textbook. The z value is 0.3783.
• Since Table 1 gives us the probability that z < 0.378, we need to
subtract it from 1  1-.03783 = 0.6217
• Here’s the summary calculation:
P(X > 115) = P(Z > 115-120/16) = P(Z > -0.31) =
1-0.3783 = 0.6217
PU515 Midterm Review
• B.) If a patient has a glucose level of 140, what
percentile is this?
• In other words, if the patient’s glucose exceeds the mean of
120, what percentile do they fall in?
• Note: since the mean represents the 50% percentile, you know that this
answer must be greater than 50% given that 140 is higher than the mean of
120.
• Since the mean is 120, and we have been given a value or
140 (which is higher than the mean) we take 140 and
subtract 120, and then divide that value by the standard
deviation, which is 16.
• 140 – 120 / 16 = 1.25.
PU515 Midterm Review
• We then find the z value of 1.25 using the table in
the back of your textbook.
• The z value is 0.8944.
• Here’s the summary calculation:
P(X < 140) = P(Z < 140-120/16) =
P(Z < 1.25) = 0.8944, 89th percentile
PU515 Midterm Review
• C.) What is the probability that the mean glucose level exceeds 115
in a sample of 12 patients?
• In other words, what is the possibility of having an average glucose level
of 115 in a group of 12 patients?
• Note: this question is different from a and b. This is referencing a sample
patient population so we need to use a different formula
• See Table 5-10, p. 88
• In this case, we need to use the formula for the Central Limit
Theorem – this is the last formula in Table 5-10.
• So how do we complete this problem?
• Here’s the summary calculation:
P(X > 115) = P(Z > 115-120/16 divided by sq12) =
P(Z > -1.08) = 1-0.1401 = 0.8599
PU515 Midterm Review
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Question #2: The following are body mass
index (BMI) scores measured in 12 patients
who are free of diabetes and participating in a
study of risk factors for obesity. Body mass
index is measured as the ratio of weight in
kilograms to the height in meters squared.
25 27 31 33 26 28 38 41 24 32 35 40
PU515 Midterm Review
• A.) Compute the mean BMI.
• So what do we do?
• You need to take all 12 BMIs and add them up. This
will give you a value of 380. Then, divide that value by
12 since we have 12 patients.
• 25+27+31+33+26+28+38+41+24+32+35+40/12
• This will give you a final mean of 31.66 or 31.7.
PU515 Midterm Review
• B.) Compute the standard deviation of BMI.
• In other words, calculate the value of patients who fall
above or below the mean.
• To perform this calculation, you need to know/use the
formula for standard deviation. Once you plug in the
values, you should end up with a final answer of 5.9
• See p. 47 (Chapter 4)
PU515 Midterm Review
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C.) Compute the median BMI using the
following data:
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24, 25, 26, 27, 28, 31, 32, 33, 35, 38, 40, 41
To calculate the median, you need to locate the
middle value. In this case, we don’t have a single
middle value, we have two: 31 and 32. As a result,
we need to add the two values together and divide
them by 2.
31 + 31 / 2 = 31.5
PU515 Midterm Review
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D.) Compute Q1 and Q3.
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Remember that Q1 and Q3 represent the values
with 25% of the data above/below them. The
calculations are simple once you identify the
values. This is very similar to the median
calculation above.
24, 25, 26, 27, 28, 31 | 32, 33, 35, 38, 40, 41
Q1 = (26+27)/2 = 26.5 and Q3 = (35+38)/2 = 36.5
PU515 Midterm Review
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E.) Are there outliers in the distribution of
BMI? Justify your answer.
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To determine this, you need to use the values for Q1
and Q3 and plug them into the formula (see p.48)
Check:
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Q1-1.5(Q3-Q1) = 26.5-1.5(36.5-26.5) = 11.5
Q3+1.5(Q3-Q1) = 36.5+1.5(36.5-26.5) = 51.5
There are no outliers
PU515 Midterm Review
PU515 Midterm Review
•
Question #3: The following table shows the
numbers of patients classified as underweight,
normal weight, overweight, and obese according to
their diabetes status.
Underweight
Diabetes
No Diabetes
Normal Weight
Overweight
Obese
8
34
65
43
12
85
93
40
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A) If a patient is selected at random, What is
the probability that they are overweight?
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To calculate this, you take the total # of overweight
diabetic patients (65) and add the total # of
overweight non diabetic (93) and divide that value
by the total # of patients.
65 + 93 / 380 = 0.416 or .42
PU515 Midterm Review
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B.) If a patient is selected at random, What is
the probability that they are obese and
diabetic?
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To calculate this, you take the total # of obese
diabetic patients (43) and divide that by 380.
43/380 = 0.11
PU515 Midterm Review
• C.) If a patient is selected at random, What
proportion of the diabetics are obese?
• Note: this is not asking for probability, this is asking of
the 380 total patients, what proportion of the diabetics
is obese?
• To calculate this, we take the total # of obese diabetics
(43) and divide that by the total # of diabetics (150)
• 43/150 = 0.29
PU515 Midterm Review
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D.) If a patient is selected at random, What
proportion of normal weight patients are not
diabetic?
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To calculate this, we take the number of non
diabetic patients (85) and divide that by # of normal
weight patients (119).
85/119 = 0.71
PU515 Midterm Review
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E.) If a patient is selected at random, What
proportion of patients is normal weight or
underweight?
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Note: this is not two separate answers! Many of you did
this on the exam. There should be one final answer.
To calculate this, we take the # of normal weight
patients (119) and add that to the # of underweight
patients (20), then we divide that by the # of patients
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119 + 20 / 380 = 0.37
PU515 Midterm Review
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Question #4: Approximately 30% of obese patients
develop diabetes. If a physician sees 10 patients who
are obese,
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A.) What is the probability that half of them will develop
diabetes?
• To calculate this, you need to plug the values into the
binomial distribution formula to find the probability
that x = 5 (see Table 5-10, p.88).
• We are given the following values: 30% (0.3), 10, and
5. Now we just plug them in. This will give us a final
answer of 0.1029.
PU515 Midterm Review
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B.) What is the probability that none will
develop diabetes?
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Similar concept to Part a; the only difference is that our
values are now 0.3, 10, and 0. Final answer should be
0.0282.
C.) How many would you expect to develop
diabetes?
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Given the percentage of 30 that we were given, as well
as they total # of patients (10), we just multiply 10 by
0.3. Final answer is 3.
PU515 Midterm Review
•
Question #5: A new non-invasive screening
test is proposed that is claimed to be able to
identify patients with impaired glucose
tolerance based on a batter of questions related
to health behaviors. The new test is given to 75
patients. Based on each patient’s responses to
the questions they are classified as positive or
negative for impaired glucose tolerance. Each
patient also submits a blood sample and their
glucose tolerance status is determined.
PU515 Midterm Review
PU515 Midterm Review
Screening Test
•
Not Impaired
Positive
17
13
Negative
8
37
A.) What is the sensitivity of the screening test?
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See sensitivity formula in Table 5-10.
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Impaired Glucose
To calculate this, we take the positive impaired glucose tolerance (17) and
divide it by the total impaired glucose tolerance  17/25 = 0.68
B.) What is the false positive fraction of the screening test?
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See false positive formula in Table 5-10.
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To calculate this, we take non impaired positive value (13) and divide it
by the total not impaired (50)  13/50 = 0.26
• Question #6 - BMI in children is approx. normally distributed
with a mean of 24.5 and a standard deviation of 6.2.
• Note: the calculations for this question are very similar to those in
Question 1 so follow the instructions for that question.
• A.) A BMI between 25 and 30 is considered overweight. What
proportion of children is overweight?
• Here is the summary calculation:
P(25 < X < 30) =
P(25-24.5/6.2< Z < 30-24.5/6.2) =
P(0.08 < Z < 0.89) = 0.8133-0.5319 = 0.2814
PU515 Midterm Review
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B.) A BMI of 30 or more is considered obese. What
proportion of children is obese?
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Here is the summary calculation:
P(X > 30) = P(Z > 30-24.5/6.2) = P(Z > 0.89) = 10.8133 = 0.1867
C.) In a random sample of 10 children, what is the
probability that their mean BMI exceeds 25?
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Here is the summary calculation:
P( > 25) =
P(Z > 25-24.5/6.2 divided by sq 10) =
P(Z > 0.26) = 1-0.6026 = 0.3974
PU515 Midterm Review
PU515 Midterm Review
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Question #7: A national survey is conducted to assess the
association between hypertension and stroke persons over 55
years of age. Development of stroke was monitored over a 5
year follow-up period. The data are summarized below and
the numbers are in millions.
Developed Stroke
Hypertension
No Hypertension
Did Not Develop Stroke
12
37
4
26
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A.) Compute the incidence of stroke in persons
over 55 years of age.
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To calculate this, we need to know the # of patients
who developed stroke, which is 16 (12 + 4). We
then divide that by total # of patients, which is 79.
16/79 = .20. This means that 20 in every 500
persons will develop stroke.
PU515 Midterm Review
• B.) Compute the relative risk of stroke comparing
hypertensive to non-hypertensive persons
• To calculate this, we need the formula for relative risk
(see Summary of Key Formulas on Table 3-7, p. 26).
• RR = PP exposed/PP unexposed
• Now, let’s plug in our data…
PU515 Midterm Review
• We know that 12 hypertensive patients developed
stroke, and there are a total of 49 hypertensive
patients. So, we take 12/49 = 0.24.
• Next, we do the same calculation for nonhypertensive patients. So, we divide 4/30 because
we know that 4 of the 30 non-hypertensive
patients developed stroke. 4/30 = 0.13.
• We then divide these two values: 0.24/1.33 =
1.85.
PU515 Midterm Review
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C.) Compute the odds ratio of stroke
comparing hypertensive to non-hypertensive
persons.
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To calculate this, we need the formula for odds ratio
(see Summary of Key Formulas on Table 3-7, p.
26).
Now, let’s plug in our data…
PU515 Midterm Review
• We know that 12 of the hypertensive patients developed
stroke, while 37 did not
• 12/37 = 0.32
• We know that 4 of the non-hypertensive patients
developed stroke, while 26 did not
• 4/26 = 0.15
• To get the final answer, we divide these two values:
0.32/0.15 = 2.13
PU515 Midterm Review
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Question #8: True/False Questions
If there are outliers, then the mean will be greater than
the median.
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False. See Chap 4, p. 47.
The 90th percentile of the standard normal distribution
is 1.645.
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False. The 95th % of the standard normal dist. is 1.645.
The mean is the 50th percentile of any normal
distribution.
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True, but this is only the case with normal distributions.
The mean is a better measure of location when there are
no outliers.
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True. See Chap 4, p. 46.
PU515 Midterm Review
• Any final questions? Was this helpful to you? I hope
so 
• We will have one more seminar to review for our final
exam. Please note, however, that the purpose of this
seminar is not to present new material; it is to review
the Unit 6 and 7 assignments since your final exam
will be primarily based on this material.
• Final Exam: due at the end of Unit 9
• Tuesday, May 1st @ 11:59pm
Wrapping it up…