unit 1 practice test using 2015 test SOLUTIONS - UMasspeople.umass.edu/biep540w/pdf/unit 1 practice test using 2015 test... · BIOSTATS 540 – Fall 2016 Practice Test for Unit 1

BIOSTATS 540 – Fall 2016 Practice Test for Unit 1 – Summarizing Data SOLUTIONS

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1.

1a. The U.S. Census Bureau reports the median family income in its summary of census data. Why do you suppose they use the median instead of the mean? What might be the disadvantages of reporting the mean?

Answer: I will accept a variety of wordings here. In the US and elsewhere , incomes are skewed to the right, as a result of a small percent of the population being extremely wealthy. The median is then a better summary measure of what is “typical”. The medianis the “middle” value , separating the lower half of the distribution from the upper half. The disadvantage of reporting the mean of right -skewed data (here, incomes) is that it is too influenced (“dragged to the right”) by the extremely high incomes . The mean is then misleading for suggesting that incomes are typically higher than they really are.

1b. You have just bought a new car that claims to get a highway fuel efficiency of 31 miles per gallon. Of course, your mileage will “vary”. If you had to guess, would you expect the interquartile range (IQR) of gas mileage attained by all cars like yours to be 30 mpg, 3 mpg, or 0.3 mpg? Why?

Answer: 3 mpg

The IQR is the difference between the 75th percentile and the 25th percentile, thus capturing the middle 50% of the distribution of miles per gallon. A middle 50% spanning just IQR = 0.3 mpg is unrealistically small since, it is unlikely that 50% of all cars have mileages per gallon that are so similar as to be within 3 tenths of a mile per gallon. At the other extreme, a middle 50% spanning IQR = 30 mpg is unrealistically large. This would be saying that only 50% of cars have mileages per gallon in an interval of 30 mpg width. Cars don’t vary that much!

1c. A company selling a new MP3 player advertises that the player has a mean lifetime of 5 years. If you were in charge of quality control at the factory, would you prefer that the standard deviation of life spans of the players you produce be 2 years or 2 months? Why?

Answer: 2 months

The standard deviation is a measure of scatter, addressing “how different is one value from another”. Thus, in settings of production, quality control is about variability. Accordingly, producers of new MP3 players can reasonably be assumed to desire products with life spans that vary as little as possible.



2. A survey is taken by the U.S. Bureau of the Census. It reports that the median incomes in the past 12 months were $52,168 for women and $61,975 for men. It is also reported that the mean earnings in the past 12 months were $59,890 for women and $76,724 for men.

2a. Does this data suggest that the distribution of income for each gender is symmetric, positive (right) skewed or negative (left) skewed? In 1-2 sentences at most, explain your reasoning.

Answer:

Skewness occurs when a distribution includes a small percentage of values that are either very low (left skewness) or very high (right skewness). Comparison of the mean and the median reveals the direction of the skewness. In left-skewed distributions, mean < median. In right-skewed distributions, mean > median. Here, the distributions of incomes are right skewed for both women and men, and more so for men. WOMEN: mean= $59,980 is to the right of (greater than) the median = $52,168 MEN: mean = $76,724 is to the right of (greater than) the median = $61,975

2b. The results of this survey were obtained from 73.8 million women and 83.4 million men. Calculate the overall sample mean income. Show your work.

Answer: $68, 821 Solution:

Overall mean = (73.8x106women)* mean for women[ ] + (83.4x106men)* mean for men[ ](73.8x106women) + (83.4x106men)

= (73.8)* $59,890[ ] + (83.4)* $76,724[ ](73.8) + (83.4)

= $68,821.02



3. The following are 9 observations of cost ($) of compact refrigerators.

1 2 3 4 5 6 7 8 9 $150 $150 $160 $180 $150 $140 $120 $130 $120

3a. By hand (no calculator!) calculate the value of the sample mean. Show your work.

Answer: $144.44

x = x1+x2 +...+xn

n=150+150+...+120

9=1300

9= 144.44



3b. By hand (no calculator!) calculate the value of the sample standard deviation. Show your work.

Answer: $19.44

S2 = (x1-x)2 +(x2 -x)2 +...+(xn -x)2

(n-1)= (150-144.44)2 +(150-144.44)2 +...+(120-144.44)2

8= 3022.2224

8=377.7778

Sample Standard Deviation = S2 = 377.7778 = $19.44



3c. By hand (no calculator!) calculate the value of the sample median. Show your work.

Answer: $150

MedianODD SAMPLE SIZE = n+12

⎡⎣⎢

⎤⎦⎥

th

largest observation = 9+12

⎡⎣⎢

⎤⎦⎥

th

largest observation = 5thobservation = $150



3d. By hand (no calculator!) calculate the values of the 1st (25th percentile) and 3rd (75th percentile) quartiles (Q1 and Q3). Show your work.

Answer: Q1 = 25th percentile = $130 Q3 = 75th percentile = $150

P25 =(n) p100⎡⎣⎢

⎤⎦⎥

th

largest =(9) 25100⎡⎣⎢

⎤⎦⎥

th

largest = 2.25th = 3rd largest = $130

P75 =(n) p100⎡⎣⎢

⎤⎦⎥

th

largest =(9) 75100⎡⎣⎢

⎤⎦⎥

th

largest = 6.75th = 7th largest = $150

cost,&smallest&to&largest

1 $120.002 $120.003 $130.004 $140.005 $150.006 $150.007 $150.008 $160.009 $180.00

P25th&=&3rd&largest&=&$130.00

P75th&=&7thd&largest&=&$150.00

3e. By hand (no calculator!) calculate the value of the interquartile range (IQR). Show your work.

Answer: $20 Solution: IQR = [ Q3 - Q1 ] = [ 75th Percentile – 25th Percentile ] = 150 – 130 = 20



4. A school system employs teachers at salaries between $40,000 and $70,000. The mean and standard deviation are $44,000 and $3,000 respectively. The teachers’ union and the school board are negotiating the form of next year’s increase in the salary schedule. For Questions #4a – #4d ONLY: Suppose that every teacher is given a flat $1000 raise.

4a. By how much will the mean salary increase?

Answer: $1000 Solution:

New mean = (x1+1000) + (x2 +1000) + ... + (xn +1000)n

= x1+x2 +...+xn +1000+1000+...+1000n

= x1+x2 +...+xn

n + 1000+1000+...+1000

n

= old mean + (n)1000n

= old mean + 1000

4b. By how much will the median salary increase?

Answer: $1000 Solution: Because every teacher has been given a $1000 raise, each teacher retains their rank “position” in an ordered listing of salaries from smallest to largest. New median = Old median + $1000



4c. By how much will the interquartile range (IQR) increase?

Answer: $0 Solution: Because every teacher has been given the same raise, $1000, the ordering of teachers from smallest salary to largest salary (“rank” order) is unchanged. Thus, New IQR = Salary of new 75th percentile teacher - Salary of new 25th percentile teacher = (Salary of old 75th percentile teacher + 1000) - (Salary of old 25th percentile teacher + 1000) = (Salary of old 75th percentile teacher – Salary of old 25th percentile teacher) + (1000 – 1000) = Old IQR + 0

4d. By how much will the sample variance increase?

Answer: $0 Solution:

New S2 = [(x1+1000) - (new mean)]2 + ... + [(xn+1000) - (new mean)]2 n-1

= [(x1+1000) - (old mean +1000)]2 + ... + [(xn+1000) - (old mean+1000)]2 n-1

= [x1 - old mean]2 + [xn - old mean]2 n-1

= Old S2



For Question #4e ONLY: Now suppose that, instead of a flat raise, every teacher is given a 5% raise.

4e. By how much will the sample variance increase? Express your answer as a percent.

Answer: 10.25% Each new x = (100%)[old x] + (5%)[old x] = (1.05)*[old x]

Before calculating new S2,obtain new mean:

new mean = (1.05)x1 + ... + (1.05)xn

n = (1.05)*[x1 + ... + xn ]

n = (1.05) * x1+...+xn

n⎡⎣⎢

⎤⎦⎥= (1.05)*[old mean]

New S2 = new x1 - new mean[ ]2 + ... + new xn - new mean[ ]2

(n-1)

= (1.05)x1 - (1.05)old mean[ ]2 + ... + (1.05)xn - (1.05)old mean[ ]2

(n-1)

= (1.05)(x1 - old mean)[ ]2 + ... + (1.05)(xn -old mean)[ ]2

(n-1)

= (1.05)2 (x1 - old mean)[ ]2 + ... + (1.05)2 (xn - old mean)[ ]2

(n-1)

= (1.05)2 (x1 - old mean)2 + ... + (xn - old mean)2

(n-1)⎡

⎣⎢

⎤

⎦⎥

= (1.05)2 Old S2

= (1.1025) Old S2



5. A study examining the health risks of smoking measured the cholesterol (mg/dL) levels of people in two independent groups: 1) SMOKERS: those who had smoked for at least 25 years; and 2) NON-SMOKERS: persons of similar ages who had never smoked. The following are the data.

Smokers 225 211 209 284 258 216 196 288 250 200 209 280 225 256 243 200 213 246 225 237 232 267 232 216 216 243 200 155 216 271 230 309 183 280 217 305 287 217 246 351 200 280 209

NON-Smokers

250 213 300 249 213 310 175 174 328 160 188 321 213 257 292 200 271 227 238 163 263 192 242 249 242 267 243 217 267 218 217 183 228



5a. Complete the following table. Dear BIOSTATS class, There is more than one correct answer here because, unfortunately, there are multiple formulae for Q1 and Q3 (the formulae in the course notes versus taking the median of the lower and upper 50% of the data) and there are multiple formulae for the values of the lower and upper fence (the formulae in the course notes versus using simply Q1 and Q3 + 1.5*IQR). Sigh! - cb

Smokers NON-Smokers Number in group, n = 43 33

P25 = Q1 = Lower Quartile = 212 or 211 213 P50 = Q2 = Median Quartile = 230 238

P75 = Q3 = Upper Quartile = 262.5 or 267 263 Interquartile Range (IQR) = 50.5 50

1.5* IQR= 75.75 75 Value of Lower Fence = 155 or 136.25 160 or 138 Value of Upper Fence = 309 or 338.25 328 or 338

Outliers (if any) below lower fence (LIST) = none none Outliers (if any) above upper fence (LIST) = 351 none

Answer: As a preliminary, I input the data for each group, separately, into Excel. Next, I sorted the data, separately for each group. Finally, using the concatenate function, I created a column of the data in which each cell entry was of the form “label, value” that StatKey requires; eg – smoker, 310. In developing my answers, I then used a mix of StatKey and Excel. From Statkey:



IQR: Smokers: (Q3 - Q1) = (262.5 – 212) = 50.5 Non-smokers: (Q3 - Q1) = (263 – 213) = 50 1.5* IQR: Smokers: 1.5*50.5 = 75.75 Non-smokers: 1.5*50 = 75 Value of lower fence (note – In R lower fence = Q1 -1.5*IQR): Smokers: smallest observation > [ Q1 – 1.5*IQR ] = [ 212 – (1.5)(50.5) ] = 136.25 The smallest observation that is > 136.25 is equal to 155 NON-Smokers: smallest observation > [ Q1 – 1.5*IQR ] = [ 213 – (1.5)(50) ] = 137 The smallest observation that is > 137 is equal to 160 Value of upper fence (note – In R, upper fence = Q3 +1.5*IQR): Smokers: largest observation < [ Q3 + 1.5*IQR ] = [ 262.5 + (1.5)(50.5) ] = 338.25 The largest observation that is < 338.25 is equal to 309 NONSmokers: largest observation < [ Q3 + 1.5*IQR ] = [ 263 + (1.5)(50) ] = 338 The largest observation that is < 338 is equal to 328

5b. In 3-8 sentences at most, write a brief report comparing the cholesterol value distributions of smokers and non-smokers.

Answer:

The distributions of cholesterol values are similar for smokers and non-smokers, with only slight differences. Compared to non-smokers, among smokers, cholesterol values have a slightly lower median, the middle 50% of the distribution is slightly more variable, while the range is slightly smaller.

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unit 1 practice test using 2015 test SOLUTIONS - UMasspeople.umass.edu/biep540w/pdf/unit 1 practice test using 2015 test... · BIOSTATS 540 – Fall 2016 Practice Test for Unit 1