Free-response practice · Question 2 style · 10 points

Days at the shelter for small and large dogs

A county animal shelter records how many days each dog stays at the shelter before it is adopted. A volunteer selected a random sample of 15 small dogs (dogs that weigh less than 25 pounds) and a random sample of 15 large dogs (dogs that weigh more than 50 pounds) from the records of all dogs adopted from the shelter last year. For each dog, the volunteer recorded the number of days the dog stayed at the shelter before it was adopted.

The data are listed in order in the first table and displayed in the dotplots. Summary statistics are given in the second table.

Days at the shelter before adoption, in order from shortest to longest
Size of dog	Days at the shelter
Small	2, 3, 4, 4, 5, 6, 7, 8, 8, 10, 11, 13, 15, 19, 33
Large	5, 9, 12, 14, 16, 18, 21, 23, 25, 29, 31, 36, 42, 47, 89

Small dogs

Large dogs

Summary statistics for days at the shelter
Size of dog	Number of dogs	Mean (days)	Standard deviation (days)
Small	15	9.9	8.0
Large	15	27.8	20.7

Use the given information to respond to parts A, B, C, D, and E. Label any subparts (e.g., i and ii) that may be present.

A.

Use the data for large dogs to respond to the following. Show the work that leads to your answers.

i. Calculate the median and the interquartile range (IQR) of the number of days at the shelter for large dogs.

ii. Use the $1.5 \times IQR$ rule to determine whether there are any outliers among the stays for large dogs.
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Score part A
- 1 point Calculates the median as 23 days and the IQR as $36 - 14 = 22$ days, with work showing the quartiles. Quartiles from another standard method (for example, $Q_{1} = 15$ and $Q_{3} = 33.5$ ) also earn the point if the work is shown.Earns it: "Median = 23. $Q_{1} = 14$ , $Q_{3} = 36$ , so IQR = 22 days."Does not: "Median 23, IQR 22" with no quartiles or other work shown; "IQR = 89 - 5 = 84" (that is the range, not the IQR); "Median = 25" (the median of 15 ordered values is the 8th value, 23)
- 1 point Calculates the fences, $14 - 1.5 (22) = - 19$ and $36 + 1.5 (22) = 69$ , and concludes that the stay of 89 days is an outlier and that there are no low outliers. Fences that are consistent with the quartiles found in (i) also earn the point. An arithmetic error can be ignored if correct work is shown.Earns it: " $1.5 (22) = 33$ . Fences: $14 - 33 = - 19$ and $36 + 33 = 69$ . 89 is above 69, so 89 days is the only outlier."Does not: "89 is an outlier." (no work is shown); " $36 + 22 = 58$ , so 89 is an outlier." (the IQR was not multiplied by 1.5); Correct fences with no conclusion about which stays are outliers
Model response
i. The median is the 8th of the 15 ordered values, 23 days. The lower half (5, 9, 12, 14, 16, 18, 21) has median $Q_{1} = 14$ days, and the upper half (25, 29, 31, 36, 42, 47, 89) has median $Q_{3} = 36$ days. So IQR $= 36 - 14 = 22$ days.

ii. $1.5 \times IQR = 1.5 (22) = 33$ . Lower fence: $14 - 33 = - 19$ . Upper fence: $36 + 33 = 69$ . No stay is below $- 19$ days, and only the stay of 89 days is above 69 days. So 89 days is an outlier, and it is the only one.
B.

Compare the distributions of the number of days at the shelter for small dogs and large dogs.
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Score part B
- 1 point Center: makes an explicit comparison of center in context, such as the median stay for large dogs (23 days) being greater than the median stay for small dogs (8 days). A correct comparison of the means also earns the point.Earns it: "The median stay for large dogs, 23 days, is greater than the median stay for small dogs, 8 days."; "Large dogs typically stayed longer: mean 27.8 days compared with 9.9 days."Does not: "Small dogs: median 8. Large dogs: median 23." (values are listed but not compared); "Large dogs have the highest value, 89 days." (a single value is not a measure of center)
- 1 point Variability: makes an explicit comparison of variability in context, such as the stays for large dogs being more variable than the stays for small dogs (IQR of 22 days compared with 9 days, a range of 84 days compared with 31 days, or a standard deviation of 20.7 days compared with 8.0 days).Earns it: "Stays for large dogs are more spread out: IQR 22 days versus 9 days for small dogs."Does not: "The range for large dogs is 84 days." (no comparison with small dogs); "Large dogs have more variability because their median is higher." (center and variability are different features)
- 1 point Shape: describes both distributions of stays as skewed to the right. A response may also note that each distribution has a high outlier.Earns it: "Both distributions of days at the shelter are skewed right, each with one very long stay."Does not: "The large-dog stays are skewed right." (the shape of only one distribution is given); "Both distributions are skewed left." (the long tails are toward larger numbers of days); "Both have outliers." (unusual features are not a description of shape)
Model response
Center: the median stay for large dogs (23 days) is much greater than the median stay for small dogs (8 days).

Variability: the stays for large dogs are more variable. Their IQR is 22 days, compared with $13 - 4 = 9$ days for small dogs, and their range is 84 days, compared with 31 days.

Shape: both distributions of stays are skewed to the right, and each has a high outlier (89 days for large dogs, and 33 days for small dogs, since $13 + 1.5 (9) = 26.5$ ).
C.

The shelter's annual report says that "a typical large dog waits about 28 days to be adopted," based on the mean stay for large dogs. Determine whether the mean or the median is the better measure of a typical stay for large dogs. Justify your answer.
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Score part C
- 1 point Determines that the median is the better measure of a typical stay for large dogs.Earns it: "The median is the better measure."Does not: "The mean is better because it uses every value." (using every value is exactly why the 89-day stay pulls the mean up); "Either one works because they are close." (the mean is almost 5 days above the median because of the long right tail)
- 1 point Justifies the choice using the large-dog data: the distribution is skewed to the right with a high outlier (89 days), which pulls the mean (27.8 days) above the median (23 days). The median is resistant to outliers and is not pulled toward the long right tail, while the mean is.Earns it: "The stays are skewed right with an outlier at 89 days. That outlier pulls the mean up to 27.8, but the median of 23 is resistant to it, so the median better shows a typical stay."Does not: "The median is the middle value." (does not connect the choice to the shape or the outlier); "The mean is affected by outliers." (does not refer to the skewness or the outlier in these data)
Model response
The median is the better measure. The large-dog stays are skewed to the right and include an outlier of 89 days, and these long stays pull the mean (27.8 days) above the median (23 days). The median is resistant to the outlier and is not pulled toward the long right tail, so 23 days better describes the stay of a typical large dog than 28 days does.
D.

Biscuit, a small dog, stayed at the shelter for 19 days. Moose, a large dog, stayed for 47 days. Using the means and standard deviations in the table, determine which dog's stay was longer relative to the stays of other dogs of the same size. Justify your answer with calculations.
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Score part D
- 1 point Calculates both z-scores: $\frac{19 - 9.9}{8.0} \approx 1.14$ for Biscuit and $\frac{47 - 27.8}{20.7} \approx 0.93$ for Moose. An arithmetic error can be ignored if correct work is shown.Earns it: "Biscuit: $z = (19 - 9.9) / 8.0 = 1.14$ . Moose: $z = (47 - 27.8) / 20.7 = 0.93$ ."Does not: "Biscuit: $19 - 9.9 = 9.1$ days above the mean. Moose: $47 - 27.8 = 19.2$ days above the mean." (distances in days do not account for the different standard deviations); "Biscuit: $z = (19 - 9.9) / 20.7$ " (uses the standard deviation of the wrong group)
- 1 point Concludes that Biscuit's stay was longer relative to dogs of the same size, because Biscuit's stay is more standard deviations above the mean for small dogs than Moose's stay is above the mean for large dogs. A conclusion that correctly compares the z-scores found in the previous point also earns this point.Earns it: "Biscuit, because 1.14 > 0.93: Biscuit's stay is more standard deviations above the small-dog mean than Moose's is above the large-dog mean."Does not: "Moose, because 47 days is longer than 19 days." (compares raw stays, not relative position); "Biscuit" with no comparison of z-scores
Model response
Biscuit: $z = \frac{19 - 9.9}{8.0} \approx 1.14$ . Moose: $z = \frac{47 - 27.8}{20.7} \approx 0.93$ .

Biscuit's stay was 1.14 standard deviations above the mean stay for small dogs, while Moose's stay was only 0.93 standard deviations above the mean stay for large dogs. So Biscuit's stay was longer relative to other dogs of the same size, even though Moose stayed more days.
E.

The shelter plans to report stays in weeks instead of days. Calculate the mean and the standard deviation of the stays for large dogs in weeks.
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Score part E
- 1 point Divides both statistics by 7: mean $\frac{27.8}{7} \approx 3.97$ weeks and standard deviation $\frac{20.7}{7} \approx 2.96$ weeks. Both values must be correct.Earns it: "Mean $= 27.8 / 7 = 3.97$ weeks, SD $= 20.7 / 7 = 2.96$ weeks."Does not: "Mean 3.97 weeks, SD 20.7 weeks" (dividing every value by 7 also divides the standard deviation by 7); "SD $= 20.7 / 49 = 0.42$ weeks" (dividing by 49 would convert the variance, not the standard deviation)
Model response
Converting days to weeks divides every stay by 7, which divides both the mean and the standard deviation by 7. Mean: $\frac{27.8}{7} \approx 3.97$ weeks. Standard deviation: $\frac{20.7}{7} \approx 2.96$ weeks.

Your score: 0 of 10. Each point is earned on its own, the way the exam scores it.