Section A.2 Summarizing Data
Solution 1. Exercise 1
Solution 2. Exercise 3
The graph below shows a ramp up period. There may also be a period of exponential growth at the start before the size of the petri dish becomes a factor in slowing growth. [Figure: A graph is shown with a horizontal axis of "time" and a vertical axis labeled "number of bacteria cells". A curve is shown rising steeply on the left, and as it moves right, it rises more slow until it nearly stops rising as it reaches right side of the graph.]
Solution 3. Exercise 5
Solution 4. Exercise 7
Solution 5. Exercise 9
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Dist 2 has a higher mean since \(20 \gt 13\text{,}\) and a higher standard deviation since 20 is further from the rest of the data than 13.
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Dist 1 has a higher mean since \(-20 \gt -40\text{,}\) and Dist 2 has a higher standard deviation since -40 is farther away from the rest of the data than -20.
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Dist 2 has a higher mean since all values in this distribution are higher than those in Dist 1, but both distribution have the same standard deviation since they are equally variable around their respective means.
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Both distributions have the same mean since they’re both centered at 300, but Dist 2 has a higher standard deviation since the observations are farther from the mean than in Dist 1.
Solution 6. Exercise 11
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About 30.
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Since the distribution is right skewed the mean is higher than the median.
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Q1: between 15 and 20, Q3: between 35 and 40, IQR: about 20.
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Values that are considered to be unusually low or high lie more than 1.5\(\times\)IQR away from the quartiles. Upper fence: Q3 + 1.5 \(\times\) IQR = \(37.5 + 1.5 \times 20 = 67.5\text{;}\) Lower fence: Q1 - 1.5 \(\times\) IQR = \(17.5 - 1.5 \times 20 = -12.5\text{;}\) The lowest AQI recorded is not lower than 5 and the highest AQI recorded is not higher than 65, which are both within the fences. Therefore none of the days in this sample would be considered to have an unusually low or high AQI.
Solution 7. Exercise 13
Solution 8. Exercise 15
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The distribution of number of pets per household is likely right skewed as there is a natural boundary at 0 and only a few people have many pets. Therefore the center would be best described by the median, and variability would be best described by the IQR.
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The distribution of number of distance to work is likely right skewed as there is a natural boundary at 0 and only a few people live a very long distance from work. Therefore the center would be best described by the median, and variability would be best described by the IQR.
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The distribution of heights of males is likely symmetric. Therefore the center would be best described by the mean, and variability would be best described by the standard deviation.
Solution 9. Exercise 17
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The median is a much better measure of the typical amount earned by these 42 people. The mean is much higher than the income of 40 of the 42 people. This is because the mean is an arithmetic average and gets affected by the two extreme observations. The median does not get effected as much since it is robust to outliers.
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The IQR is a much better measure of variability in the amounts earned by nearly all of the 42 people. The standard deviation gets affected greatly by the two high salaries, but the IQR is robust to these extreme observations.
Solution 10. Exercise 19
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The distribution is unimodal and symmetric with a mean of about 25 minutes and a standard deviation of about 5 minutes. There does not appear to be any counties with unusually high or low mean travel times. Since the distribution is already unimodal and symmetric, a log transformation is not necessary.
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Answers will vary. There are pockets of longer travel time around DC, Southeastern NY, Chicago, Minneapolis, Los Angeles, and many other big cities. There is also a large section of shorter average commute times that overlap with farmland in the Midwest. Many farmers’ homes are adjacent to their farmland, so their commute would be brief, which may explain why the average commute time for these counties is relatively low.
Solution 11. Exercise 21
Solution 12. Exercise 23
Solution 13. Exercise 25
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False. Instead of comparing counts, we should compare percentages of people in each group who suffered cardiovascular problems. (ii) True. (iii) False. Association does not imply causation. We cannot infer a causal relationship based on an observational study. The difference from part (ii) is subtle. (iv) True.
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Proportion of all patients who had cardiovascular problems: \(\frac{7,979}{227,571} \approx 0.035\text{.}\)
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The expected number of heart attacks in the rosiglitazone group, if having cardiovascular problems and treatment were independent, can be calculated as the number of patients in that group multiplied by the overall cardiovascular problem rate in the study: \(67,593 * \frac{7,979}{227,571} \approx 2370\text{.}\)
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\(H_0\text{:}\) The treatment and cardiovascular problems are independent. They have no relationship, and the difference in incidence rates between the rosiglitazone and pioglitazone groups is due to chance. \(H_A\text{:}\) The treatment and cardiovascular problems are not independent. The difference in the incidence rates between the rosiglitazone and pioglitazone groups is not due to chance and rosiglitazone is associated with an increased risk of serious cardiovascular problems. (ii) A higher number of patients with cardiovascular problems than expected under the assumption of independence would provide support for the alternative hypothesis as this would suggest that rosiglitazone increases the risk of such problems. (iii) In the actual study, we observed 2,593 cardiovascular events in the rosiglitazone group. In the 1,000 simulations under the independence model, we observed somewhat less than 2,593 in every single simulation, which suggests that the actual results did not come from the independence model. That is, the variables do not appear to be independent, and we reject the independence model in favor of the alternative. The study’s results provide convincing evidence that rosiglitazone is associated with an increased risk of cardiovascular problems.
Solution 14. Exercise 27
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Decrease: the new score is smaller than the mean of the 24 previous scores.
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Calculate a weighted mean. Use a weight of 24 for the old mean and 1 for the new mean: \((24\times 74 + 1\times64)/(24+1) = 73.6\text{.}\) %There are other ways to solve this %exercise that do not use a weighted mean.
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The new score is more than 1 standard deviation away from the previous mean, so increase.
Solution 15. Exercise 29
Solution 16. Exercise 31
The distribution of ages of best actress winners are right skewed with a median around 30 years. The distribution of ages of best actor winners is also right skewed, though less so, with a median around 40 years. The difference between the peaks of these distributions suggest that best actress winners are typically younger than best actor winners. The ages of best actress winners are more variable than the ages of best actor winners. There are potential outliers on the higher end of both of the distributions.
