Categorical data practice: tables and graphs

Tally each letter in one pass. W appears at positions 1, 4, 7, 10, 11, 15, and 18, so 7 customers bought water. J appears at positions 2, 6, 9, 14, 17, and 20, so 6 bought juice. M appears at positions 3, 8, 13, 16, and 19, so 5 bought milk. S appears at positions 5 and 12, so 2 bought a sports drink.

Check the total before going further: $7 + 6 + 5 + 2 = 20$, which matches the 20 customers, so every purchase is counted exactly once and no category is missing.

(a) The frequency table lists each category with its count.

(b) A relative frequency is the category count divided by the total, so divide each entry by 20. Water: $\frac{7}{20} = 0.35$, or 35%. Juice: $\frac{6}{20} = 0.30$, or 30%. Milk: $\frac{5}{20} = 0.25$, or 25%. Sports drink: $\frac{2}{20} = 0.10$, or 10%.

Check the relative frequency table the way you check any of them: $0.35 + 0.30 + 0.25 + 0.10 = 1.00$, or 100%.

(c) The mode of a categorical variable is the category with the largest frequency. Water has 7, more than any other, so water is the mode. Note that the mode is a category name, not a number.

(d) Water and juice are separate categories and each customer bought exactly one drink, so the two counts can be added without double counting: $7 + 6 = 13$ customers, and $\frac{13}{20} = 0.65$, or 65%.

The four formats are non-overlapping and cover every item, so each item is counted in exactly one column and the four relative frequencies must add to exactly 1.

(a) Add the three printed values: $0.48 + 0.26 + 0.14 = 0.88$. The DVD share is whatever is left, so it is $1 - 0.88 = 0.12$.

(b) A count is the relative frequency times the total, so multiply each share by 250. Print book: $0.48 \times 250 = 120$. Ebook: $0.26 \times 250 = 65$. Audiobook: $0.14 \times 250 = 35$. DVD: $0.12 \times 250 = 30$.

Check the counts against the stated total: $120 + 65 + 35 + 30 = 250$.

(c) Subtract the two counts: $120 - 65 = 55$ more print books than ebooks. You could also work from the shares first, since $0.48 - 0.26 = 0.22$ and $0.22 \times 250 = 55$.

(d) They are right. Part (a) used only the fact that the shares add to 1, and 250 never entered it. The total is needed to turn shares into counts in parts (b) and (c), but not to recover a missing proportion.

Check the heights against the stated turnout: $126 + 84 + 57 + 33 = 300$ votes, so every ballot is on the graph exactly once.

(a) Divide each bar height by 300. Alvarez: $\frac{126}{300} = 0.42$. Brooks: $\frac{84}{300} = 0.28$. Chen: $\frac{57}{300} = 0.19$. Diallo: $\frac{33}{300} = 0.11$. Check: $0.42 + 0.28 + 0.19 + 0.11 = 1.00$.

(b) A majority means more than half of the votes, so more than 150 of the 300. Alvarez received 126, a share of 0.42, which is below 0.50. Alvarez would have needed at least 151 votes, 25 more than the 126 actually received, so the poster is wrong.

(b) What Alvarez does have is a plurality: the largest share of any candidate, 0.42 against the next-highest 0.28. That also makes Alvarez the mode of this categorical variable. Plurality and majority are different claims, and only the plurality one is supported here.

(c) The top two are Alvarez and Brooks: $126 + 84 = 210$ votes, and $\frac{210}{300} = 0.70$, so they took 70% of the vote between them. Equivalently, $0.42 + 0.28 = 0.70$.

(d) Skew describes a quantitative variable, whose values sit on a number line in a fixed order, so that a longer tail on one side means something. Candidate is a categorical variable with no order among its categories, so the horizontal axis is a list of labels, not a scale. These bars happen to be printed tallest to shortest, but they could be printed alphabetically or in any other order without changing a single fact about the election, and the outline would change with them.

(d) Say what the graph does support instead: Alvarez is the mode with 126 votes (0.42), the four shares are 0.42, 0.28, 0.19, and 0.11, and no candidate won a majority.

(a) The slices of a pie chart fill one circle, so their central angles must add to 360 degrees: $126 + 90 + 90 + 54 = 360$. The chart accounts for every senior once.

(b) A slice's angle is its relative frequency times 360 degrees, so divide by 360 to go the other way. Four-year college: $\frac{126}{360} = 0.35$. Two-year college: $\frac{90}{360} = 0.25$. Workforce: $\frac{90}{360} = 0.25$. Gap year: $\frac{54}{360} = 0.15$. Check: $0.35 + 0.25 + 0.25 + 0.15 = 1.00$.

(b) Now multiply each share by the 400 seniors. Four-year college: $0.35 \times 400 = 140$. Two-year college: $0.25 \times 400 = 100$. Workforce: $0.25 \times 400 = 100$. Gap year: $0.15 \times 400 = 60$. Check: $140 + 100 + 100 + 60 = 400$.

(c) They read a degree measure as a count. The number 126 is the slice's angle; the count is $0.35 \times 400 = 140$ seniors. The two would only coincide if the total happened to be 360.

(d) A pie chart encodes each value as an angle, and people judge angles less accurately than lengths, so ranking the 126-degree slice against the 90-degree ones by eye, or reading either as a percent, is harder than reading four bar tops against a labeled axis. The chart also carries no counts on its own: part (b)'s counts came from the stated total of 400, not from the picture.

Check both columns first: $192 + 144 + 96 + 48 = 480$ and $90 + 50 + 40 + 20 = 200$, matching the stated survey sizes.

(a) School A, dividing by its own total of 480: bus $\frac{192}{480} = 0.40$, car $\frac{144}{480} = 0.30$, walk $\frac{96}{480} = 0.20$, bike $\frac{48}{480} = 0.10$. Check: $0.40 + 0.30 + 0.20 + 0.10 = 1.00$.

(a) School B, dividing by its own total of 200: bus $\frac{90}{200} = 0.45$, car $\frac{50}{200} = 0.25$, walk $\frac{40}{200} = 0.20$, bike $\frac{20}{200} = 0.10$. Check: $0.45 + 0.25 + 0.20 + 0.10 = 1.00$.

(b) The count is right and the conclusion is not. School A does report more bus riders, 192 against 90, but School A also surveyed 480 students against School B's 200, which is $\frac{480}{200} = 2.4$ times as many. As a share of its own students, busing is more common at School B: 0.45 against 0.40.

(b) The general point: comparing raw counts across groups of different sizes tells you about the group sizes as much as about the variable. Convert to relative frequencies before comparing.

(c) Biking makes it sharper. School A has 48 bike riders and School B has 20, so A has more than twice as many. Yet $\frac{48}{480} = 0.10$ and $\frac{20}{200} = 0.10$, exactly the same share. Identical shares and very different counts is precisely what the raw numbers hide.

(d) A bar graph of relative frequencies, with School A's and School B's bars side by side for each method. Putting both schools on the same 0 to 1 scale removes the size difference, which is the only thing that made the count comparison mislead. If the question were instead how many buses each school needs, the frequency bar graph would be the right display.

(a) Divide each count by the 150 members. Hiking: $\frac{96}{150} = 0.64$. Climbing: $\frac{84}{150} = 0.56$. Kayaking: $\frac{60}{150} = 0.40$. Their sum is $0.64 + 0.56 + 0.40 = 1.60$, or 160%.

(b) A pie chart divides one circle into slices whose shares add to exactly 1, which requires the categories to be non-overlapping (each unit in at most one) and exhaustive (each unit in at least one). Neither holds here. The three counts total $96 + 84 + 60 = 240$ checks from only 150 members, and if no member had checked more than one box the total could not have exceeded 150, so at least one member is counted in two categories. Shares adding to 1.60 cannot be laid on one circle.

(b) Exhaustiveness fails too: a member who did none of the three activities checked no box and appears in no category, so the slices would not cover the membership even if the overlap were removed.

(c) Change the observational unit from the member to the check. Each of the 240 checks names exactly one activity, so as categories of a check the three are non-overlapping and exhaustive. Shares: $\frac{96}{240} = 0.40$, $\frac{84}{240} = 0.35$, $\frac{60}{240} = 0.25$. Check: $0.40 + 0.35 + 0.25 = 1.00$.

(c) Angles are the share times 360 degrees: hiking $0.40 \times 360 = 144$ degrees, climbing $0.35 \times 360 = 126$ degrees, kayaking $0.25 \times 360 = 90$ degrees. Check: $144 + 126 + 90 = 360$.

(c) What it says: 40% of the activity checks were for hiking. What it does not say: that 40% of members hiked. That figure is 0.64 from part (a). A member who checked all three boxes contributes three times to this chart and once to the membership, so the two questions have different answers.

(d) Ask something every member answers exactly once, with an option that catches everyone. "Which one of these did you do most often this year: hiking, climbing, kayaking, or none of the three?" gives four non-overlapping categories, and the fourth makes them exhaustive, so the four shares add to 1 and the pie is valid. A second workable version reduces the variable to one yes-or-no question, such as "Did you kayak this year?" with categories yes and no.

(a) Add the printed percentages: $35 + 30 + 24 + 15 = 104$, so the table claims 104%. The four sources are non-overlapping and cover every student, so the true percentages add to exactly 100%. A printed total 4 percentage points above that is more than rounding can account for, which part (b) makes precise.

(b) Rounding a value to the nearest whole percent moves it by at most 0.5 percentage points. With four categories, the printed total can therefore differ from the true total by at most $4 \times 0.5 = 2$ percentage points.

(b) The true total is exactly 100%, so the printed total has to land between $100 - 2 = 98$% and $100 + 2 = 102$%, inclusive. The printed 104% is outside that window, so at least one of the four numbers is not a correct rounding of its true value.

(c) Check the counts against the stated sample size first: $70 + 60 + 48 + 22 = 200$.

(c) Divide each count by 200. Social media: $\frac{70}{200} = 0.35$, or 35%. News app: $\frac{60}{200} = 0.30$, or 30%. Print: $\frac{48}{200} = 0.24$, or 24%. Radio: $\frac{22}{200} = 0.11$, or 11%. Check: $0.35 + 0.30 + 0.24 + 0.11 = 1.00$. These four are exact rather than rounded, since all four counts are even and a count out of 200 is half that count as a percent, so each one lands on a whole percent with nothing to round. An odd count would not: $\frac{71}{200} = 35.5\%$.

(c) Line the two tables up. Social media, news app, and print match the printed figures exactly at 35%, 30%, and 24%. Radio was printed as 15% but is 11%, an error of 4 percentage points, which is exactly the amount by which the printed total overshot 100%.

Check each column against its stated total: $160 + 100 + 60 + 80 = 400$, $175 + 125 + 100 + 100 = 500$, and $180 + 150 + 150 + 120 = 600$.

(a) Year 1, dividing by 400: Spanish $\frac{160}{400} = 0.40$, French $\frac{100}{400} = 0.25$, Mandarin $\frac{60}{400} = 0.15$, Latin $\frac{80}{400} = 0.20$. Sum 1.00.

(a) Year 2, dividing by 500: Spanish $\frac{175}{500} = 0.35$, French $\frac{125}{500} = 0.25$, Mandarin $\frac{100}{500} = 0.20$, Latin $\frac{100}{500} = 0.20$. Sum 1.00.

(a) Year 3, dividing by 600: Spanish $\frac{180}{600} = 0.30$, French $\frac{150}{600} = 0.25$, Mandarin $\frac{150}{600} = 0.25$, Latin $\frac{120}{600} = 0.20$. Sum 1.00.

(b) Both are right, because they measure different things. Counted in students, Spanish rose every year: 160, then 175, then 180. As a share of all language students, Spanish fell every year: 0.40, then 0.35, then 0.30.

(b) Why both can happen at once: the total grew from 400 to 600, an increase of $\frac{200}{400} = 0.50$, or 50%, while Spanish grew from 160 to 180, an increase of $\frac{20}{160} = 0.125$, or 12.5%. A category that grows more slowly than the total loses share even while its count rises.

(c) Compare each language's Year 1 share with its Year 3 share. Spanish went 0.40 to 0.30, a loss of 10 percentage points. French held at 0.25 and Latin held at 0.20, no change. Mandarin went 0.15 to 0.25, a gain of 10 percentage points, the largest gain. Its count went from 60 to 150, which is 2.5 times as many students.

(d) Tell them no single bar graph settles it, because growing has not been defined. A frequency bar graph of Spanish across the three years, at heights 160, 175, and 180, answers how many students take Spanish and shows an increase. A relative frequency bar graph, at heights 0.40, 0.35, and 0.30, answers what share of language students take Spanish and shows a decrease. Both are honest. Ask which question they mean, draw that one, and label the axis so the reader can tell which they are looking at.