PubHlth Introductory Biostatistics Practice Test I (without Unit 3 Questions) SOLUTIONS

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1 1 PubHlth Introductory Biostatistics Practice Test I (without Unit 3 Questions) SOLUTIONS 1. (10 points) a. (2 points). What is the scale of measurement for systolic blood pressure? Continuous, ratio Valid systolic blood pressure values exist on a continuum. Between any two values, it is theoretically possible for there to exist an intermediate value. Division of two systolic blood pressure values produces a meaningful number (not clinically relevant but well defined anyway!) b. (2 points) Construct a figure that displays simultaneously the cumulative relative frequency distribution for nonsmokers and smokers. Cumulative Relative Frequency Smokers v. Nonsmokers Cum Relative Frequency (%) smoker nonsmoker Systolic Blood pressure This plot was produced in excel. A solution BY HAND is fine!.

2 2 c. (2 points) Using your answer to part b, estimate the interquartile range for nonsmokers. [ ] = 29.7 From the table, we estimate that 16 th percentile = 109, 54 th percentile = 129. Thus, 25 th percentile is something between 109 and P = P = Similarly, since we estimate that 54 th percentile = 129 and 83 rd percentile = 149, the 75 th percentile is something between 129 and P = P = Note If your solution used the midpoints of the intervals to obtain guesses of the 25 th and 75 th percentiles, I will give you full credit - cb. d. (2 points) Using Table 2, what is the median systolic blood pressure among non-smokers? From the table, we estimate that 16 th percentile = 109, 54 th percentile = 129. Thus, 50 th percentile is something between 109 and P = P = Note Again, if your solution used the midpoints of the intervals to obtain a guess of the 50th percentiles, I will give you full credit.

3 3 e. (2 points) Construct a box plot for systolic blood pressure among non-smokers. Systolic Blood Pressure A solution by hand is FINE!

4 4 2. (10 points) - omitted 3. (10 points) - omitted 4. (10 points) a. Blood Type Frequency Relative Frequency O = (2672/5400) A = (2041/5400) B = (486/5400) AB = (201/5400) total = b A person cannot be simultaneously type A or type AB blood type; thus these events are mutually exclusive and so the probability of their union is the simple sum. Pr [ Type A or AB ] = Pr [Type A ] = Pr [ Type AB ] c = = Since sampling is without replacement, Pr [first is green and second is yellow ] = Pr [first is green ] Pr [second is yellow first is green ] = [ 8/14 ] [ 6/13 ] =

5 5 d Batch contains 50 defective plus 2450 non-defective (good) tablets and sampling is without replacement. Thus, Pr [ entire batch is rejected ] = Pr [ at least one in sample of 4 is defective ] = 1 - Pr [ all 4 in sample of 4 are good ] = 1 - (2450/2500)(2449/2499)(2448/2498)(2447/2497) = (10 points) a. New mean = 45+10=55 New median = = 31 New standard deviation = 15 (no change) New sample variance = 225 (no change) New interquartile range = 32 (no change) A shift in location will change the values of statistics of location but not variability. b. New mean = (45)(2) = 90 New median = (21)(2) = 42 New standard deviation = (15)(2) = 30 New sample variance = (225)(2 2 ) = 900 New interquartile range = (32)(2) = 64 A rescaling will rescale the location and variability. Because the variance is a measure of variability on a scale of measurement that is the original units squared.

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7 7 6. (10 points) sample mean X = [value of number correct ] [ relative frequency in class ] all values of number correct = [ 10 ] + [ 9 ] + [ 8 ] + [ 7 ] + [ 6 ] = sample variance S 2 2 = [(value of number correct - X) ] all values of number correct frequency in class class size = [ ( ) ] [ ( ) ] [ ( ) ] [ ( ) ] [ ( ) ] = Sample standard deviation = S 2 = =1.0428

8 8 7. a % Sensitivity = Pr[test+ disease+] 11 = 12 = b % Specificity = Pr[test - disease -] 16 = 18 = c % Expected # positive = [#disease] Pr[+ disease] + [#without disease] Pr[+ no disease] = [1](.9167) + [999,999]( ) = [.9167] + [111, ] = 111, Expected proportion postive = [111, ]/[1,000,000] =.1111 = 11.11%

9 9 d. 111,100 Expected # false positive = [#without disease] Pr[+ no disease] = [999,999]( ) = 111, ,100

10 10 8. a. < 1 year The most frequently occurring survival time is the modal class. This is the interval less than one year. b. 4 years The median is required. To estimate the median, notice that because n = 347 and n/2=173.5,, the median survival time is the average of the 173 rd and 174 th ordered survival times. Reading down the table, read that the 173 rd ordered survival time is 4 years. Inasmuch as the next interval is 4-5 years, an answer of 4 years is acceptable. c. No. Not available are the actual survival times of the 47 persons whose survival times are at least 20 years. d. Asymmetric and asymmetry is positive Relative frequencies are largest for smaller survival time intervals and diminish as survival durations increase.

11 11 9. a. 50 As distribution is symmetric, average is center value or 50 b. 25 Again, as distribution is symmetric, average is center value or 25, approximately c. 40 Because of positive skewness, the average is dragged to the right of the center of the peak. d. Statement (i) Undergraduate classes are more likely to include some number of adult learners than precocious teen learners. The result is positive skewness.

Ø Set of mutually exclusive categories. Ø Classify or categorize subject. Ø No meaningful order to categorization.

Ø Set of mutually exclusive categories. Ø Classify or categorize subject. Ø No meaningful order to categorization. Statistical Tools in Evaluation HPS 41 Dr. Joe G. Schmalfeldt Types of Scores Continuous Scores scores with a potentially infinite number of values. Discrete Scores scores limited to a specific number