Continuing, we get (note that unlike the text suggestion, I end the final interval with 95, not 85.

Transcription

1 Chapter 3 -- Review Exercises Statistics Dr. McGahagan Problem 1. Histogram of male heights. Shaded area shows percentage of men between 66 and 72 inches in height; this translates as "66 inches OR more, AND less than 72 inches." Note that since the bars are of equal size and each bar has a base of one inch, height * base = height. Challenge: Guess at the percentage shown (before you look at the answer below: is it closer to 25, 33, 50, 66, or 75 percent? (I would think 66, since the average height of the 6 shaded bars seems greater than put a ruler across the graph at 10 for a quick impression, or try to measure more accurately: my guess for the shaded bars was 9, 12, 14, 13, 11, 8 which sums to 67 percent) Problem 2. Age distribution of US population in Note that the intervals are NOT all of the same size, so we must calculate densities before we draw the histogram, and remember that the densities will be the heights of the histogram bars. Example: (for age 0 to 5): base = 5, height =??, percent = 8 We know percent = height * base, so we have H * 5 = 8 or H = 8 / 5 = 1.6 (for age 5 to 14): base = 9, height =??, percent = 13, so we have H * 9 = 13 or H = 13 / 9 Continuing, we get (note that unlike the text suggestion, I end the final interval with 95, not 85. INTERVAL BASE PERCENT DENSITY The histogram for the data is given on the next page. a. There are more children age 1 (1.6 percent of population) than age 11 (1.44 percent of population) Note that the densities DIRECTLY give the percent for any given year. Only 8 percent of the population are in the 0-5 interval, but for any given age (1 to 2) we would have 8 / 5 = 1.6 percent. b. There are more 31 year olds (1.8 percent of population) than 21 year olds (1.43 percent of population. c. Percent aged = 4 * 1.8 = 7.2 % whereas percent aged = 9 * 1.6 = 14.4 %. There are twice as many aged (an interval covering 9 years) as (an interval of 4 years). d. Percent aged 32 or more: Percent in the interval = 3 * 1.8 = 5.4 percent. Percent above 35: = 46 percent (add up the last 5 percents). Percent above 31: Percent in interval Percent above 35 = = 51.4 percent.

2 Problem 2 (histogram of age distribution of population) Note: to create a histogram of the data in EcLS: > (bind pcts (list )) > (bind bounds (list )) Check that there is one more bound than percentage given: (n pcts) = 11; (n bounds) = 12 > (histogram pcts bounds) and answer "y" when asked if the data is from a table; normally the histogram would expect data on individuals which the program would group into bins. Note that in this case you will get an error message, since the sum of the percentages given in the table is 99 rather than 100. The EcLS program creates a variable called NEWPCTS which forces the data to add up to 100 by dividing the percentages given by the sum of the percentages given (stop a moment and think about why this works). Since the error could have come from a typing error rather than a rounding error, it is better that the program does not make the correction automatically.

3 Problem 3. Housing data, New York City, 1990 In EcLS: > (bind bounds (combine (seq ) 10.5)) Make the dividing lines at 0.5, 1.5, 2.5 and so on, so that the real number of rooms is in the middle of the bin. > (bind oo (list )) distribution of owner-occupied units: only 2 percent of owner-occupied units are one-room, 3.8 percent are two rooms, etc. > (bind rent (list )) distribution of rented rooms. > (hist oo bounds table) since (sum oo) = 99.9 percent, the computer will create a new variable called NEWPCTS and you will have to rerun the command as (hist newpcts bounds table) Since we will want to compare this to the rent variable, replace oo with newpcts: (bind oo (copy-list newpcts)) > (hist rent bounds table) since (sum rent) = percent, redefine rent as well: (bind rent (copy-list newpcts)) re-run the command. Note that in constructing the histograms, there was a printout indicating that the mean of owner-occupied housing was 5.54 rooms and the median was 6 rooms, while the mean of rental housing was only 3.42 rooms and the median 3 rooms.

4 Finally, we can place both histograms back-to-back with the command: > (bihistogram bounds oo rent) which will show the distribution of owner-occupied and rental properties by number of rooms. Note that the low value (1 room) is at the bottom, the high value (10 rooms) is at the top. The questions actually asked in the text: a. Why don't the values add to 100 %? Answer: Rounding error in reporting (say) 2.04 percent as 2.0 percent, and 9.24 percent as 9.2 percent. b. Is the percentage of one room units larger for rented housing only because there is more renter-occupied housing? NO, the purpose of using percentages is to make the comparison INDEPENDENT of the absolute size of the sample. Note that we can compute the number of one room units as: For owner-occupied: * 785,120 = 15, 702 For rented units: * 1,782,459 = 163,986 Note that when we CALCULATE with percentages, we MUST use them in DECIMAL form. Failure to do so will lead to the nonsensical result that there are twice as many owner-occupied one-room units as (all) owner-occupied units. There are 163,986 / 15,702 = times as many rented one-room units as owner-occupied. But when we look at all housing, the ratio or rented to owned units is much less: 1,782,459 / 785,120 = 2.27 times as many rented units as owner occupied units. c. As noted above, the mean for rented units is 3.42 rooms, and for owner-occupied housing How was this calculated? By a weighted average calculation: (example of owner occupied housing):.02 * * * * * * 9.5 = multiplying the percentages in each bar of the histogram and the MIDPOINT of the histogram. This explains why we set up the histogram with bounds 0.5, 1.5 and so on -- except for the last one, which ran from 8.5 to 10.5.

5 ASSIGNED. Problems 4 and 5. Provide a graph for problem 5. Problem 6. Hypothetical histogram of heights: Which of the following lists have this as their histogram? (i). 25 people inches tall, 50 people 68 inches tall, and 25 people 69 inches tall. Yes: since there are 100 people in all, the percentages are 25 / 100 = 0.25 or 25 percent and 50 / 100 = 0.50 or 50 percent; the last two groups have heights EXACTLY at the midpoint, the first group has a height just a bit below the midpoint of the first bar, but still would be in the first bar. (ii) Though there are four groups, the first two groups both fall within the first bar. 10 people inches tall and 15 people inches tall would be 25 people BETWEEN 66.5 and 67.5 inches tall. (iii) Though the three groups are at the midpoints of the bars, the percentages are wrong: there are 100 people in all: 30 / 100 = 0.30 or thirty percent in the first and third groups, 40 / 100 = 0.40 or forty percent in the second group. Problem 7. Natural causes or accident? Which histogram is which? Since one would expect natural causes to be the leading cause of death among the old, and accident or crime the leading cause of death among the young, the identification is easy: The "histogram" sketch (known more exactly as a DENSITY PLOT) with the density highest on the right represents mortality by age from natural causes; and that with the density highest on the left from accident or crime. Note that the histogram with the density highest on the right is called "left-skewed" -- the direction of skew is determined by the tail which extends the furthest away from the main concentration of density.

6 ASSIGNED. Problem 8. Income distribution "histogram"? NO. Provide a clear explanation of why the graph in the book is NOT a historgram. Reading the following may help, but try first to come up with your own answer. Drawing the histogram properly will allow one to see that: (a) is TRUE. Families that earn between $ 10 K and $ 35 K are ARE spread fairly evenly over that range. The bar representing K has a base of 5 and a height which SHOULD be 7.3 / 5 = 1.46 % per $ 1000 The bar representing K has a base of 10 and a height which SHOULD be 15.6 / 10 = 1.56 % per $ 1000 The bar representing K has a base of 10 and a height which SHOULD be 15.0 / 10 = 1.50 % per $ 1000 Since the densities are very close, the percent of families earning $ K is very close to the percentage of families earning K, and so on for any $ 1,000 interval in between. The tallest three bars on the accurately drawn histogram below represent these three families. (b) is FALSE. Although there are similar percentages in total in families making between $ 35 - $ 50 K, and between $ K, the DENSITIES differ greatly: the bar representing $ K has a base of 15 and a height which SHOULD be 19.2 / 15 = 1.28 % per $ 1,000. the bar representing $ K has a base of 25 and a height which SHOULD be 19.6 / 25 = 0.78 % per $ 1,000 Code to draw histogram: > (bind bounds (list )) > (bind pcts (list )) Since (sum pcts) = 100.1, we will have to let the computer adjust the data. > (hist pcts bounds table), followed by (hist newpcts bounds table) Problem 9. There IS a spike at self-reported GPA of 2.0, so (a) and (b) are both true. Probably students reported rounded grades rather than always remembering the decimal points. Problem 10. Birth certificates and great importance placed on exact age rarer in the 19th century, perhaps especially for immigrants. Problem 11. Curious absence of scores between 85 and 90 suggests that we don't have a random distribution. Problem 12. Correlation is not causation. It may be that a riot is hard to start without people already on the street, but then why did the peak of riots occur in the 1960s? Certainly temperature of the 1960s was not hotter than previous or subsequent decades.