AP Statistics 1.8 Graphical Representations of Summary Statistics for One Quantitative Variable- Exam Style Questions - FRQs - New Syllabus

Question

The manager of an automotive company is interested in comparing the gas mileages for cars manufactured in Country A and cars manufactured in Country B. The manager selected a random sample of 100 cars manufactured in Country A and a random sample of 100 cars manufactured in Country B. The gas mileages for each sample, in miles per gallon (mpg), are summarized in the boxplots.

A. Compare the distributions of gas mileage for the sample of cars manufactured in Country A and the sample of cars manufactured in Country B.

B. For the distribution of gas mileage for the sample of cars manufactured in Country A, would you expect the mean to be greater than 18 mpg, less than 18 mpg, or equal to 18 mpg? Justify your answer.

C. The manager will create a new boxplot with the combined data from the sample of cars manufactured in Country A and the sample of cars manufactured in Country B.

i. What is the range of the combined data set? Justify your answer.
ii. What is a possible value of the median of the combined data set? Justify your answer by referencing the boxplots shown.

Most-appropriate topic codes (AP Statistics):

• Topic $1.7$ — Summary Statistics for One Quantitative Variable (Part $ \mathrm{C} $)
• Topic $1.8$ — Graphical Representations of Summary Statistics for One Quantitative Variable (Parts $ \mathrm{B} $, $ \mathrm{C} $)
• Topic $1.9$ — Comparisons of the Distributions for One Quantitative Variable (Part $ \mathrm{A} $)

▶️ Answer/Explanation

A.
• Center: The median gas mileage for Country B ($32\,\text{mpg}$) is higher than that of Country A ($18\,\text{mpg}$).
• Spread: The overall range for Country A ($38 – 14 = 24\,\text{mpg}$) is slightly wider than Country B ($40 – 18 = 22\,\text{mpg}$), but the Interquartile Range (IQR) for Country B ($36 – 24 = 12\,\text{mpg}$) is larger than Country A ($22 – 14 = 8\,\text{mpg}$).
• Outliers: Country A has a single extreme high value plotted as an outlier at $38\,\text{mpg}$, while Country B displays no outliers.

B.
• The mean is expected to be greater than $18\,\text{mpg}$.
• Because the distribution for Country A is skewed to the right and contains a high outlier at $38\,\text{mpg}$, the mean will be pulled upward toward the long right tail while the median remains resistant.

C. i.
• The minimum value of the combined set is $14\,\text{mpg}$ (from Country A) and the maximum value is $40\,\text{mpg}$ (from Country B).
• $\text{Combined Range} = \text{Maximum} – \text{Minimum} = 40 – 14 = 26\,\text{mpg}$.

C. ii.
• A possible value for the combined median is any value satisfying $18\,\text{mpg} \le \text{Median} \le 32\,\text{mpg}$, such as $24\,\text{mpg}$.
• Since both samples contain exactly 100 cars, combining them creates a total group of 200 cars where the new median rests between the 100th and 101st ordered values, forcing it to fall cleanly between the separate sample medians of $18\,\text{mpg}$ and $32\,\text{mpg}$.

Question

A company sells a certain type of whistle. The price of the whistle varies from store to store. Julio, a statistician at the company, wants to estimate the mean price, in dollars ($\$\delta$), of this type of whistle at all stores that sell the whistle.

(a)
i. Identify the appropriate inference procedure for Julio to use.
ii. Describe the parameter for the inference procedure you identified in part (a-i) in context.

Julio called the managers of $20$ randomly selected stores that sell the whistle and recorded the price of the whistle at each store. Following is a dotplot of Julio’s data.

The summary statistics for Julio’s data are shown in the following table.

Summary Statistics for Julio’s Data

(b) Julio wants to examine some characteristics of the distribution of the sample of whistle prices.

i. Describe the shape of the distribution of the sample of whistle prices. Justify your response using appropriate values from the summary statistics table.
ii. Using the $1.5 \times \text{IQR}$ rule, determine whether there are any outliers in the sample of whistle prices. Justify your response.

It can often be difficult to determine whether the distribution of sample data is skewed by looking at a graph of the data and the summary statistics, particularly when the sample size is small. Thus, statisticians sometimes measure how skewed a data set is. One such measure is Pearson’s coefficient of skewness, which is calculated using the following formula.

$\text{Pearson’s Coefficient of Skewness} = \dfrac{3(\bar{x}-m)}{s}$

In the formula, $\bar{x}$ is the sample mean, $m$ is the sample median, and $s$ is the sample standard deviation.

(c)
i. Calculate Pearson’s coefficient of skewness for Julio’s sample of $20$ whistle prices. Show your work.

The following graph shows conclusions that can be made about the shape of the distribution of sample data based on Pearson’s coefficient of skewness and sample size.

ii. Indicate the value of the Pearson’s coefficient of skewness you calculated in part (c-i) for the appropriate sample size by marking it with an “X” on the preceding graph.

(d) Consider your work in part (c).

i. What should you conclude about the shape of the distribution of the sample of whistle prices? Justify your response.

Julio’s inference procedure in part (a-i) needs one of the following requirements to be satisfied to verify the normality condition.
• The sample size is greater than or equal to $30$.
• If the sample size is less than $30$, the distribution of the sample data is not strongly skewed and does not have outliers.

ii. Using your response to (d-i) and the preceding requirements, is the normality condition satisfied for Julio’s data? Explain your response.

Most-appropriate topic codes (AP Statistics):

• Topic $1.7$ — Summary Statistics for One Quantitative Variable (Parts $ \mathrm{b} $, $ \mathrm{c} $)
• Topic $1.8$ — Graphical Representations of Summary Statistics for One Quantitative Variable (Parts $ \mathrm{b} $, $ \mathrm{c} $)
• Topic $4.2$ — Constructing a Confidence Interval for a Population Mean or Population Mean Difference (Parts $ \mathrm{a} $, $ \mathrm{d} $)

▶️ Answer/Explanation

(a)
i. Julio should use a one-sample $t$-interval for a population mean.
ii. The parameter of interest is $\mu$, the true mean price (in dollars) of this type of whistle at all stores that sell it.

(b)
i. The distribution of the sample of whistle prices is skewed to the right. This is because the mean ($5.12$) is greater than the median ($4.885$).
ii. $\text{IQR} = Q_3 – Q_1 = 5.475 – 4.51 = 0.965$.
Lower boundary: $Q_1 – 1.5(\text{IQR}) = 4.51 – 1.5(0.965) = 3.0625$.
Upper boundary: $Q_3 + 1.5(\text{IQR}) = 5.475 + 1.5(0.965) = 6.9225$.
Since the minimum value ($4.25$) is greater than $3.0625$ and the maximum value ($6.58$) is less than $6.9225$, there are no outliers in the sample.

(c)
i. $\text{Pearson’s Coefficient} = \dfrac{3(5.12 – 4.885)}{0.743} \approx 0.949$.
ii. On the graph, you would plot an “X” at a sample size of $y = 20$ and a skewness coefficient of $x \approx 0.949$.

(d)
i. We can conclude that the distribution of the sample of whistle prices is strongly skewed. This is justified because the calculated coefficient of $0.949$ for a sample size of $20$ falls in the “strongly skewed” region of the provided graph.
ii. No, the normality condition is not satisfied. The sample size ($n = 20$) is less than $30$, and although there are no outliers, the sample data is strongly skewed, failing the second condition.

Question

The sizes, in square feet, of the $20$ rooms in a student residence hall at a certain university are summarized in the following histogram.

(a) Based on the histogram, write a few sentences describing the distribution of room size in the residence hall.

(b) Summary statistics for the sizes are given in the following table.

Determine whether there are potential outliers in the data. Then use the following grid to sketch a boxplot of room size.

(c) What characteristic of the shape of the distribution of room size is apparent from the histogram but not from the boxplot?

Most-appropriate topic codes (AP Statistics):

• Topic $1.6$ — Describing the Distribution of One Quantitative Variable (Parts $ \mathrm{a} $, $ \mathrm{c} $)
• Topic $1.7$ — Summary Statistics for One Quantitative Variable (Part $ \mathrm{b} $)
• Topic $1.8$ — Graphical Representations of Summary Statistics for One Quantitative Variable (Part $ \mathrm{b} $)

▶️ Answer/Explanation

(a)
The distribution of the sample of room sizes is bimodal and roughly symmetric.
Most room sizes fall into two clusters: $100$ to $200$ square feet and $250$ to $350$ square feet.
The center of the distribution is between $200$ and $300$ square feet.
The range of the distribution is between $150$ and $250$ square feet.
There are no apparent outliers.

(b)
The interquartile range is:
$IQR = Q_3 – Q_1 = 292 – 174 = 118$ square feet.
To find potential outliers, we check the lower and upper fences.
Lower Fence = $Q_1 – 1.5(IQR) = 174 – 1.5(118) = -3$ square feet.
Upper Fence = $Q_3 + 1.5(IQR) = 292 + 1.5(118) = 469$ square feet.
There are no potential outliers because the minimum room size of $134$ square feet does not fall below $-3$, and the maximum room size of $315$ square feet does not exceed $469$.

(c)
The histogram clearly shows the bimodal nature (two peaks or clusters) of the distribution of room sizes.
This bimodal characteristic is not apparent from the boxplot.

Question

The chemicals in clay used to make pottery can differ depending on the geographical region where the clay originated. Sometimes, archaeologists use a chemical analysis of clay to help identify where a piece of pottery originated. Such an analysis measures the amount of a chemical in the clay as a percent of the total weight of the piece of pottery. The boxplots below summarize analyses done for three chemicals—X, Y, and Z—on pieces of pottery that originated at one of three sites: I, II, or III.

(a) For chemical Z, describe how the percents found in the pieces of pottery are similar and how they differ among the three sites.

(b) Consider a piece of pottery known to have originated at one of the three sites, but the actual site is not known.

(i) Suppose an analysis of the clay reveals that the sum of the percents of the three chemicals X, Y, and Z is $20.5\%$. Based on the boxplots, which site—I, II, or III—is the most likely site where the piece of pottery originated? Justify your choice.

(ii) Suppose only one chemical could be analyzed in the piece of pottery. Which chemical—X, Y, or Z—would be the most useful in identifying the site where the piece of pottery originated? Justify your choice.

Most-appropriate topic codes (AP Statistics):

• Topic $1.9$ — Comparing the Distributions of One Quantitative Variable (Part $ \mathrm{a} $)
• Topic $1.8$ — Graphical Representations of Summary Statistics for One Quantitative Variable (Parts $ \mathrm{a} $, $ \mathrm{b} $)

▶️ Answer/Explanation

(a)
Similarity: The median percent of chemical Z is approximately the same across all three sites, at roughly $7\%$ of total weight. This means the center of the distribution of chemical Z does not help distinguish one site from another.
Difference: The spread (range) of the percent of chemical Z differs considerably among the three sites.
Site II has the smallest range, approximately $2\%$ (from about $6\%$ to $8\%$).
Site I has a moderate range, approximately $6\%$ (from about $4\%$ to $10\%$).
Site III has the largest range, approximately $8\%$ (from about $3\%$ to $11\%$).
So while the typical chemical Z content is similar at all three sites, Site II pottery is far more consistent in its Z content, while Site III pottery shows the greatest variability.

(b)(i)
The piece of pottery most likely originated at Site III.
To decide, we estimate the minimum and maximum possible sums of the percents of X, Y, and Z at each site by reading the minimum and maximum values from the boxplots:

For Site I, the possible sum ranges from $21\%$ to $33\%$ — since $20.5\%$ falls below the minimum, a sum of $20.5\%$ is unlikely for Site I.
For Site II, the possible sum ranges from $12.9\%$ to $19\%$ — since $20.5\%$ falls above the maximum, a sum of $20.5\%$ is also unlikely for Site II.
For Site III, the possible sum ranges from $14\%$ to $26.5\%$ — since $20.5\%$ falls comfortably within this interval, Site III is the most plausible origin.
$\boxed{\text{Most likely site: Site III}}$

(b)(ii)
Chemical Y would be the most useful for identifying the site of origin.
Looking at the boxplots for chemical Y across the three sites:
Site I: chemical Y ranges from approximately $11\%$ to $15\%$
Site II: chemical Y ranges from approximately $1.9\%$ to $4\%$
Site III: chemical Y ranges from approximately $6\%$ to $8\%$
These three ranges do not overlap at all — every possible value of chemical Y belongs to exactly one site’s range. So if we measure chemical Y in an unknown piece, we can pinpoint its origin with certainty.
By contrast, the distributions of chemicals X and Z show substantial overlap across the three sites, meaning a measured value of X or Z could plausibly belong to multiple sites, making those chemicals far less useful for classification.
$\boxed{\text{Most useful chemical: Y}}$

Question

A consumer organization was concerned that an automobile manufacturer was misleading customers by overstating the average fuel efficiency (measured in miles per gallon, or mpg) of a particular car model. The model was advertised to get 27 mpg. To investigate, researchers selected a random sample of 10 cars of that model. Each car was then randomly assigned a different driver. Each car was driven for 5,000 miles, and the total fuel consumption was used to compute mpg for that car.

(a) Define the parameter of interest and state the null and alternative hypotheses the consumer organization is interested in testing.

One condition for conducting a one-sample $t$-test in this situation is that the mpg measurements for the population of cars of this model should be normally distributed. However, the boxplot and histogram shown below indicate that the distribution of the 10 sample values is skewed to the right.

(b) One possible statistic that measures skewness is the ratio $\dfrac{\text{sample mean}}{\text{sample median}}$. What values of that statistic (small, large, close to one) might indicate that the population distribution of mpg values is skewed to the right? Explain.

(c) Even though the mpg values in the sample were skewed to the right, it is still possible that the population distribution of mpg values is normally distributed and that the skewness was due to sampling variability. To investigate, 100 samples, each of size 10, were taken from a normal distribution with the same mean and standard deviation as the original sample. For each of those 100 samples, the statistic $\dfrac{\text{sample mean}}{\text{sample median}}$ was calculated. A dotplot of the 100 simulated statistics is shown below.

In the original sample, the value of the statistic $\dfrac{\text{sample mean}}{\text{sample median}}$ was 1.03. Based on the value of 1.03 and the dotplot above, is it plausible that the original sample of 10 cars came from a normal population, or do the simulated results suggest the original population is really skewed to the right? Explain.

(d) The table below shows summary statistics for mpg measurements for the original sample of 10 cars.

Choosing only from the summary statistics in the table, define a formula for a different statistic that measures skewness.

What values of that statistic might indicate that the distribution is skewed to the right? Explain.

Most-appropriate topic codes (AP Statistics):

• Topic $4.4$ — Setting Up a Test for a Population Mean or Population Mean Difference (Part $\mathrm{a}$)
• Topic $1.6$ — Descriptions for One Quantitative Variable Distributions (Part $\mathrm{b}$)
• Topic $2.3$ — Estimating Probabilities Using Simulation (Part $\mathrm{c}$)
• Topic $1.8$ — Graphical Representations of Summary Statistics for One Quantitative Variable (Part $\mathrm{d}$)

▶️ Answer/Explanation

(a)
Let $\mu$ = the true population mean fuel efficiency (in miles per gallon) for all cars of this particular model.
The consumer organization suspects the manufacturer is overstating the mpg, so the alternative hypothesis is lower-tailed:
$ H_0:\ \mu = 27\ \text{mpg} $
$ H_a:\ \mu < 27\ \text{mpg} $
The parameter must be defined as a population mean — not just a sample mean — and must be stated in context (fuel efficiency of this car model) to receive full credit.

(b)
Large values (greater than 1) of the ratio $\dfrac{\text{sample mean}}{\text{sample median}}$ would indicate that the population distribution is skewed to the right.
The reason is that in a right-skewed distribution, the few unusually large values in the upper tail pull the mean upward, but the median — being a positional measure — is resistant to those extreme values and stays lower.
Therefore, when the distribution is right-skewed, we expect:
$ \text{sample mean} > \text{sample median} \implies \dfrac{\text{sample mean}}{\text{sample median}} > 1 $
The further the ratio exceeds 1, the stronger the evidence of right-skewness.

(c)
The observed value of the statistic from the original sample is $1.03$.
Looking at the dotplot of the 100 simulated statistics (all drawn from a normal population), we count that 14 out of 100 simulated values are greater than or equal to $1.03$.
This gives a simulated $p$-value of approximately:
$ \hat{p} = \dfrac{14}{100} = 0.14 $
Since $0.14$ is larger than any commonly used significance level (such as $\alpha = 0.05$ or $\alpha = 0.10$), we do not have convincing evidence that the original population is skewed to the right.
It is therefore plausible that the original sample of 10 cars came from a normally distributed population, and the observed right-skewness in the sample was simply due to random sampling variability.

(d)
Using only the five-number summary values, one reasonable skewness statistic is:
$ S = \dfrac{Q_3 – \text{Median}}{\text{Median} – Q_1} $
Using the given data:
$ S = \dfrac{28 – 25.5}{25.5 – 24} = \dfrac{2.5}{1.5} \approx 1.67 $
Values greater than 1 indicate right-skewness.
The reasoning is: in a right-skewed distribution, the data in the upper half are more spread out than in the lower half, so the distance from the median up to $Q_3$ (the upper half of the middle 50%) will be larger than the distance from $Q_1$ down to the median (the lower half of the middle 50%). This makes the numerator larger than the denominator, giving a ratio greater than 1.
Other acceptable statistics using only the five-number summary include:
$ S = \dfrac{\text{Maximum} – \text{Median}}{\text{Median} – \text{Minimum}}, \qquad S = \dfrac{\text{Maximum} – Q_3}{Q_1 – \text{Minimum}}, \qquad S = \dfrac{\frac{Q_1 + Q_3}{2}}{\text{Median}} $
For all of these, values greater than 1 indicate right-skewness.

Question

As gasoline prices have increased in recent years, many drivers have expressed concern about the taxes they pay on gasoline for their cars. In the United States, gasoline taxes are imposed by both the federal government and by individual states. The boxplot below shows the distribution of the state gasoline taxes, in cents per gallon, for all 50 states on January 1, 2006.

(a) Based on the boxplot, what are the approximate values of the median and the interquartile range of the distribution of state gasoline taxes, in cents per gallon? Mark and label the boxplot to indicate how you found the approximated values.

(b) The federal tax imposed on gasoline was $18.4$ cents per gallon at the time the state taxes were in effect. The federal gasoline tax was added to the state gasoline tax for each state to create a new distribution of combined gasoline taxes. What are approximate values, in cents per gallon, of the median and interquartile range of the new distribution of combined gasoline taxes? Justify your answer.

Most-appropriate topic codes (AP Statistics):

• Topic $1.7$ — Summary Statistics for One Quantitative Variable (Part $\mathrm{a}$)
• Topic $1.8$ — Graphical Representations of Summary Statistics for One Quantitative Variable (Part $\mathrm{a}$)
• Topic $1.7$ — Summary Statistics for One Quantitative Variable (Part $\mathrm{b}$)

▶️ Answer/Explanation

(a)

From the boxplot, identify the median as the line inside the box — it falls at approximately $21$ cents per gallon.
The first quartile $Q_1$ is the left edge of the box, reading approximately $18$ cents per gallon, and the third quartile $Q_3$ is the right edge, reading approximately $25$ cents per gallon.
The interquartile range is calculated as:
$\text{IQR} = Q_3 – Q_1 \approx 25 – 18 = 7 \text{ cents per gallon}$
$\boxed{\text{Median} \approx 21 \text{ cents per gallon}, \quad \text{IQR} \approx 7 \text{ cents per gallon}}$

(b)

When a constant value is added to every observation in a distribution, the entire distribution shifts by that constant — so the median increases by exactly $18.4$ cents per gallon.
$\text{New Median} = 21 + 18.4 = 39.4 \text{ cents per gallon}$
However, adding the same constant to every data value shifts $Q_1$ and $Q_3$ by equal amounts, so their difference remains unchanged:
$\text{New } Q_1 = 18 + 18.4 = 36.4, \quad \text{New } Q_3 = 25 + 18.4 = 43.4$
$\text{New IQR} = 43.4 – 36.4 = 7 \text{ cents per gallon}$
$\boxed{\text{New Median} \approx 39.4 \text{ cents per gallon}, \quad \text{New IQR} \approx 7 \text{ cents per gallon}}$

Question

A large regional real estate company keeps records of home sales for each of its sales agents. Each month, the company publishes the sales volume for each agent. Monthly sales volume is defined as the total sales price of all homes sold by the agent during a month. The figure below displays the cumulative relative frequency plot of the most recent monthly sales volume (in hundreds of thousands of dollars) for these agents.

(a) In the context of this question, explain what information is conveyed by the circled point.

(b) What proportion of sales agents achieved monthly sales volumes between \$700,000 and \$800,000?

(c) For values between 10 and 11 on the horizontal axis, the cumulative relative frequency plot is flat. In the context of this question, explain what this means.

(d) A bonus is to be given to 20 percent of the sales agents. Those who achieved the highest monthly sales volume during the preceding month will receive a bonus. What is the minimum monthly sales volume an agent must have achieved to qualify for the bonus?

Most-appropriate topic codes (AP Statistics):

• Topic 1.5 — Graphical Representations for One Quantitative Variable (Parts $\mathrm{a}$, $\mathrm{b}$, $\mathrm{c}$, $\mathrm{d}$)
• Topic 1.8 — Graphical Representations of Summary Statistics for One Quantitative Variable (Parts $\mathrm{a}$, $\mathrm{d}$)

▶️ Answer/Explanation

(a)
The circled point is located at approximately $(3,\ 0.40)$ on the graph.
This tells us that 40 percent of the sales agents at this real estate company had a monthly sales volume of \$300,000 or less in the month shown.
In other words, the circled point represents the 40th percentile of the distribution of most recent monthly sales volumes for all agents at the company.
$\boxed{\text{40% of sales agents had monthly sales volume} \leq \$300{,}000}$

(b)
Reading from the cumulative relative frequency plot:
Proportion of agents with sales volume $\leq \$800{,}000$ (i.e., at $x = 8$) $= 0.80$
Proportion of agents with sales volume $\leq \$700{,}000$ (i.e., at $x = 7$) $= 0.70$
So the proportion with sales volume between \$700,000 and \$800,000 is:
$P(700{,}000 < X \leq 800{,}000) = P(X \leq 800{,}000) – P(X \leq 700{,}000) = 0.80 – 0.70 = 0.10$
$\boxed{0.10 \text{ (or 10 percent) of sales agents achieved monthly sales volumes between \$700,000 and \$800,000}}$

(c)
When the cumulative relative frequency plot is flat between 10 and 11 on the horizontal axis, it means the cumulative proportion does not change in that interval.
Since no increase in cumulative frequency occurs, there were no sales agents whose monthly sales volume fell between \$1,000,000 and \$1,100,000 during that month.
$\boxed{\text{No agents had a monthly sales volume between \$1,000,000 and \$1,100,000}}$

(d)
Since the bonus goes to the top 20 percent of sales agents, we need to find the 80th percentile of the distribution (because the top 20% corresponds to a cumulative relative frequency of $1 – 0.20 = 0.80$).
Looking at the cumulative relative frequency plot, a cumulative relative frequency of $0.80$ corresponds to a monthly sales volume of \$800,000 (i.e., $x = 8$ on the horizontal axis).
Therefore, an agent must have achieved a monthly sales volume greater than \$800,000 to be in the top 20 percent and qualify for the bonus.
$\boxed{\text{Minimum monthly sales volume to qualify for bonus} = \$800{,}000}$

Question

A consumer advocate conducted a test of two popular gasoline additives, A and B. There are claims that the use of either of these additives will increase gasoline mileage in cars. A random sample of 30 cars was selected. Each car was filled with gasoline and the cars were run under the same driving conditions until the gas tanks were empty. The distance traveled was recorded for each car.

Additive A was randomly assigned to 15 of the cars and additive B was randomly assigned to the other 15 cars. The gas tank of each car was filled with gasoline and the assigned additive. The cars were again run under the same driving conditions until the tanks were empty. The distance traveled was recorded and the difference in the distance with the additive minus the distance without the additive for each car was calculated.

The following table summarizes the calculated differences. Note that negative values indicate less distance was traveled with the additive than without the additive.

(a) On the grid below, display parallel boxplots (showing outliers, if any) of the differences of the two additives.

(b) Two ways that the effectiveness of a gasoline additive can be evaluated are by looking at either

• the proportion of cars that have increased gas mileage when the additive is used in those cars

• the mean increase in gas mileage when the additive is used in those cars.

i. Which additive, A or B, would you recommend if the goal is to increase gas mileage in the highest proportion of cars? Explain your choice.

ii. Which additive, A or B, would you recommend if the goal is to have the highest mean increase in gas mileage? Explain your choice.

Most-appropriate topic codes (AP Statistics):

▶️ Answer/Explanation

(a)
First, check for outliers using the $1.5 \times \text{IQR}$ rule.

For Additive A:
$\text{IQR}_A = Q_3 – Q_1 = 4 – 1 = 3$
$1.5 \times \text{IQR}_A = 1.5 \times 3 = 4.5$
Lower fence: $Q_1 – 4.5 = 1 – 4.5 = -3.5$
Upper fence: $Q_3 + 4.5 = 4 + 4.5 = 8.5$
Values below $-3.5$: $-10, -8$ → both are outliers
Values above $8.5$: $9$ → outlier

For Additive B:
$\text{IQR}_B = Q_3 – Q_1 = 25 – (-2) = 27$
$1.5 \times \text{IQR}_B = 1.5 \times 27 = 40.5$
Lower fence: $-2 – 40.5 = -42.5$
Upper fence: $25 + 40.5 = 65.5$
No values fall outside these fences → no outliers

The parallel boxplots should be drawn as follows:

(b)(i)
Additive A is the better recommendation when the goal is to improve mileage in the highest proportion of cars. Since $Q_1 = 1 > 0$ for Additive A, we know that at least 75% of the cars in the sample showed a positive difference (i.e., improved mileage). For Additive B, $Q_1 = -2 < 0$, so we can only be certain that at least 50% of the cars showed improvement — and it could be less than 75%. Therefore, Additive A is more reliable for maximizing the proportion of cars that benefit.
$\boxed{\text{Recommend Additive A for highest proportion of cars with improved mileage}}$

(b)(ii)
Additive B is the better recommendation when the goal is to maximize the mean increase in gas mileage. Although the median for Additive B (Median $= 1$) is lower than that of Additive A (Median $= 3$), the distribution for Additive B is strongly right-skewed, with very large values above $Q_3$ such as $35, 37,$ and $40$. This skewness pulls the mean well above the median. In contrast, Additive A has two low outliers ($-10, -8$) pulling the mean downward. So the mean for Additive B is expected to be substantially higher than the mean for Additive A.
$\boxed{\text{Recommend Additive B for highest mean increase in gas mileage}}$

Question

A researcher thinks that modern Thai dogs may be descendants of golden jackals. A random sample of 16 animals was collected from each of the two populations. The length (in millimeters) of the mandible (jawbone) was measured for each animal. The lower quartile, median, and upper quartile for each sample are shown in the table below, along with all values below the lower quartile and all values above the upper quartile.

(a) Display parallel boxplots of mandible lengths (showing outliers, if any) for the modern Thai dogs and the golden jackals on the grid below.

Based on the boxplots, write a few sentences comparing the distributions of mandible lengths for the two types of dogs.

(b) Is it reasonable to use the sample of mandible lengths of modern Thai dogs to construct an interval estimate of the mean mandible length for the population of modern Thai dogs? Justify your answer. (Note: You do not have to compute the interval.)

(c) Is it reasonable to use the sample data of mandible lengths of modern Thai dogs and the sample data of mandible lengths of golden jackals to perform a two-sample $t$-test for the difference in mean mandible lengths for the two types of dogs? Justify your answer. (Note: You do not have to conduct the test.)

Most-appropriate topic codes (AP Statistics):

• Topic 1.8 — Graphical Representations of Summary Statistics for One Quantitative Variable (Part a)
• Topic 1.9 — Comparisons of the Distributions for One Quantitative Variable (Part a)
• Topic 4.2 — Constructing a Confidence Interval for a Population Mean or Population Mean Difference (Part b)
• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part c)

▶️ Answer/Explanation

(a)

First, we check for outliers in each sample using the $1.5 \times \text{IQR}$ rule.
Modern Thai Dogs:
$\text{IQR} = Q_3 – Q_1 = 128 – 121 = 7$
Lower fence: $121 – 1.5(7) = 121 – 10.5 = 110.5$
Upper fence: $128 + 1.5(7) = 128 + 10.5 = 138.5$
All values (minimum = 114, maximum = 132) fall within these fences. No outliers.
Golden Jackals:
$\text{IQR} = Q_3 – Q_1 = 112 – 107 = 5$
Lower fence: $107 – 1.5(5) = 107 – 7.5 = 99.5$
Upper fence: $112 + 1.5(5) = 112 + 7.5 = 119.5$
Values 122, 124, and 125 exceed the upper fence of 119.5. Outliers: 122, 124, 125.
The parallel boxplots (with the scale from 100 to 140 mm) are shown below:

Comparison of distributions: The distributions of mandible lengths for modern Thai dogs and golden jackals are quite different. Modern Thai dogs have a much larger typical mandible length — a median of 125 mm — compared to golden jackals, whose median is only 108 mm. The distribution for modern Thai dogs appears approximately symmetric with no outliers, whereas the distribution for golden jackals is heavily skewed to the right, with three high outliers (122, 124, and 125 mm). The variability (spread) of the two distributions is roughly similar in terms of IQR, but the overall range for golden jackals is larger once the outliers are included.

(b)

Yes, it is reasonable to construct a $t$-confidence interval for the mean mandible length of modern Thai dogs. The boxplot for this sample is roughly symmetric with no outliers, which provides support for the assumption that the underlying population distribution is approximately normal. Since the data come from a random sample and the normality condition is reasonably satisfied even with a sample size of only 16, using a one-sample $t$-interval is appropriate here.

(c)

No, it would not be reasonable to perform a two-sample $t$-test using both groups. While the modern Thai dog sample looks approximately normal, the golden jackal sample is clearly not. The boxplot for golden jackals is strongly skewed to the right and contains three high outliers (122, 124, 125) in a sample of only 16 animals — a substantial proportion of the data. With such a small sample size, the $t$-test is not robust enough to overcome this serious departure from normality, so the normality condition required for the two-sample $t$-test is not reasonably met for the golden jackal population. Therefore, performing the two-sample $t$-test with this data would not be appropriate.

Question

Since Hill Valley High School eliminated the use of bells between classes, teachers have noticed that more students seem to be arriving to class a few minutes late. One teacher decided to collect data to determine whether the students’ and teachers’ watches are displaying the correct time. At exactly $12:00$ noon, the teacher asked $9$ randomly selected students and $9$ randomly selected teachers to record the times on their watches to the nearest half minute. The ordered data showing minutes after $12:00$ as positive values and minutes before $12:00$ as negative values are shown in the table below.

(a) Construct parallel boxplots using these data.

(b) Based on the boxplots in part (a), which of the two groups, students or teachers, tends to have watch times that are closer to the true time? Explain your choice.

(c) The teacher wants to know whether individual student’s watches tend to be set correctly. She proposes to test $H_0:\mu=0$ versus $H_a:\mu\neq0$, where $\mu$ represents the mean amount by which all student watches differ from the correct time. Is this an appropriate pair of hypotheses to test to answer the teacher’s question? Explain why or why not. Do not carry out the test.

Most-appropriate topic codes (AP Statistics):

• Topic 1.8 — Graphical Representations of Summary Statistics for One Quantitative Variable (Part a)
• Topic 1.9 — Comparisons of the Distributions for One Quantitative Variable (Part b)
• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part c)

▶️ Answer/Explanation

(a)
To build a boxplot, we need the five-number summary for each group. Since the data are already sorted, we just split each set in half around the median.
For the students ($n=9$):
$ \text{Min}=-4.5 $
$ Q_1=\dfrac{-3.0+(-0.5)}{2}=-1.75 $
$ \text{Median}=0 $
$ Q_3=\dfrac{0.5+1.5}{2}=1.0 $
$ \text{Max}=5.0 $
For the teachers ($n=9$):
$ \text{Min}=-2.0 $
$ Q_1=\dfrac{-1.5+(-1.5)}{2}=-1.5 $
$ \text{Median}=-1.0 $
$ Q_3=\dfrac{0+0}{2}=0 $
$ \text{Max}=0.5 $
Neither group has any outliers, since no value falls beyond $1.5\times\text{IQR}$ from the nearer quartile. Plotting both sets of five-number summaries on the same number line (a common scale is essential so the two groups can be compared directly) gives:

Each boxplot uses the same horizontal scale, with “T” marking the teachers’ plot and “S” marking the students’ plot, so the two distributions can be lined up and compared directly.

(b)
The teachers’ watch times tend to be closer to the true noon time. Looking at the two boxplots, the teachers’ values are all squeezed into a fairly narrow band (roughly from $-2.0$ to $0.5$), while the students’ values are spread out over a much wider range (from $-4.5$ to $5.0$). Even though the teachers’ watches tend to run a little slow on average (the box is shifted slightly below $0$), their times stay much closer together and closer to zero than the students’ times do. Since “closer to the true time” is really about how small the time errors tend to be, the group with less spread — the teachers — is the better answer here.

(c)
No, this is not an appropriate pair of hypotheses for answering the teacher’s question. The teacher wants to know whether individual students’ watches tend to be set correctly, but $H_0:\mu=0$ versus $H_a:\mu\neq0$ only tests something about the average error across all student watches. It’s entirely possible for this mean to come out close to $0$ even if no individual watch is actually correct — for instance, if some students’ watches run fast by a few minutes and others run slow by a few minutes, those errors could cancel out in the average, making $\mu$ look like $0$ overall. So a test about the population mean $\mu$ doesn’t tell us anything about how far off each individual student’s watch tends to be from the true time.

AP Statistics 1.8 Graphical Representations of Summary Statistics for One Quantitative Variable- Exam Style Questions - FRQs - New Syllabus

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