Midterm Review

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Midterm Review

• PU515 Applied Biostatistics
• Dana Colbert-Wheeler, MHA, MCHES

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First things first

• 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!

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PU515 Midterm Review

• 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

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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!

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PU 515 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.

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PU515 Midterm Review

• A.) What proportion of patients has glucose
  levels exceeding 115?
• 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?
• 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)

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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 lt
  0.378, we need to subtract it from 1 ? 1-.03783
  0.6217
• Heres the summary calculation
• P(X gt 115) P(Z gt 115-120/16) P(Z gt -0.31)
• 1-0.3783 0.6217

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PU515 Midterm Review

• B.) If a patient has a glucose level of 140, what
  percentile is this?
• In other words, if the patients 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.

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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.
• Heres the summary calculation
• P(X lt 140) P(Z lt 140-120/16)
• P(Z lt 1.25) 0.8944, 89th percentile

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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?
• Heres the summary calculation
• P(X gt 115) P(Z gt 115-120/16 divided by sq12)
• P(Z gt -1.08) 1-0.1401 0.8599

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PU515 Midterm Review

• 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

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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.
• 252731332628384124323540/12
• This will give you a final mean of 31.66 or 31.7.

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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)

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PU515 Midterm Review

• C.) Compute the median BMI using the following
  data
• 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 dont 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

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PU515 Midterm Review

• D.) Compute Q1 and Q3.
• 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 (2627)/2 26.5 and Q3 (3538)/2 36.5

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PU515 Midterm Review

• E.) Are there outliers in the distribution of
  BMI? Justify your answer.
• To determine this, you need to use the values for
  Q1 and Q3 and plug them into the formula (see
  p.48)
• Check
• Q1-1.5(Q3-Q1) 26.5-1.5(36.5-26.5) 11.5
• Q31.5(Q3-Q1) 36.51.5(36.5-26.5) 51.5
• There are no outliers

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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 Normal Weight Overweight Obese
Diabetes 8 34 65 43
No Diabetes 12 85 93 40
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PU515 Midterm Review

• A) If a patient is selected at random, What is
  the probability that they are overweight?
• 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

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PU515 Midterm Review

• B.) If a patient is selected at random, What is
  the probability that they are obese and diabetic?
• To calculate this, you take the total of obese
  diabetic patients (43) and divide that by 380.
• 43/380 0.11

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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

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PU515 Midterm Review

• D.) If a patient is selected at random, What
  proportion of normal weight patients are not
  diabetic?
• 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

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PU515 Midterm Review

• E.) If a patient is selected at random, What
  proportion of patients is normal weight or
  underweight?
• 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
• 119 20 / 380 0.37

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PU515 Midterm Review

• Question 4 Approximately 30 of obese patients
  develop diabetes. If a physician sees 10 patients
  who are obese,
• 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.

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PU515 Midterm Review

• B.) What is the probability that none will
  develop diabetes?
• 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?
• 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.

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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 patients 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.

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PU515 Midterm Review
Screening Test Impaired Glucose Not Impaired
Positive 17 13
Negative 8 37

• A.) What is the sensitivity of the screening
  test?
• See sensitivity formula in Table 5-10.
• 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?
• See false positive formula in Table 5-10.
• To calculate this, we take non impaired positive
  value (13) and divide it by the total not
  impaired (50) ? 13/50 0.26

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PU515 Midterm Review

• 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 lt X lt 30)
• P(25-24.5/6.2lt Z lt 30-24.5/6.2)
• P(0.08 lt Z lt 0.89) 0.8133-0.5319 0.2814

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PU515 Midterm Review

• B.) A BMI of 30 or more is considered obese. What
  proportion of children is obese?
• Here is the summary calculation
• P(X gt 30) P(Z gt 30-24.5/6.2) P(Z gt 0.89)
  1-0.8133 0.1867
• C.) In a random sample of 10 children, what is
  the probability that their mean BMI exceeds 25?
• Here is the summary calculation
• P( gt 25)
• P(Z gt 25-24.5/6.2 divided by sq 10)
• P(Z gt 0.26) 1-0.6026 0.3974

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PU515 Midterm Review

• 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 Did Not Develop Stroke
Hypertension 12 37
No Hypertension 4 26
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PU515 Midterm Review

• A.) Compute the incidence of stroke in persons
  over 55 years of age.
• 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.

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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, lets plug in our data

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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
  non-hypertensive 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.

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PU515 Midterm Review

• C.) Compute the odds ratio of stroke comparing
  hypertensive to non-hypertensive persons.
• To calculate this, we need the formula for odds
  ratio (see Summary of Key Formulas on Table 3-7,
  p. 26).
• Now, lets plug in our data

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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

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PU515 Midterm Review

• Question 8 True/False Questions
• If there are outliers, then the mean will be
  greater than the median.
• False. See Chap 4, p. 47.
• The 90th percentile of the standard normal
  distribution is 1.645.
• False. The 95th of the standard normal dist. is
  1.645.
• The mean is the 50th percentile of any normal
  distribution.
• True, but this is only the case with normal
  distributions.
• The mean is a better measure of location when
  there are no outliers.
• True. See Chap 4, p. 46.

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Wrapping it up

• 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 _at_ 1159pm