AP Statistics 4.4 Setting Up a Test for a Population Mean or Population Mean Difference- Exam Style Questions - FRQs - New Syllabus

Question

According to a 2017 national survey in Country B, the mean number of bedrooms in newly built houses was 2.9. Rodney, a researcher, believes the mean number of bedrooms in newly built houses in the country was different in 2024 than it was in 2017. To investigate his belief, he took a large random sample of newly built houses in Country B in 2024 and recorded the number of bedrooms in each house. The distribution of the number of bedrooms for the sampled houses is summarized in the table.

Distribution of the Number of Bedrooms for the Houses Sampled in 2024

i. A house from the sample will be selected at random. What is the probability that the house had fewer than 3 bedrooms? Show your work.
ii. What is the mean number of bedrooms for the sample of newly built houses in 2024? Show your work.

B. Rodney will use a one-sample t-test for a population mean to test his belief.

i. In the context of Rodney’s investigation, state the hypotheses for the test.
ii. Explain, in context, what a Type I error would be for Rodney’s hypothesis test.

C. A different researcher, Keisha, suggests using a confidence interval to investigate whether the mean number of bedrooms in newly built houses in 2024 in Country B was different from 2.9. Assume the conditions for inference have been met. Using Rodney’s data, Keisha calculated a one-sample 97 percent confidence interval to estimate the population mean as $(3.01, 3.19)$. Based on the confidence interval, what conclusion can be made for Rodney’s hypothesis test in part B at $\alpha = 0.03$? Justify your answer.

Most-appropriate topic codes (AP Statistics):

• Topic $1.7$ — Summary Statistics for One Quantitative Variable (Part $ \mathrm{A} $)
• Topic $2.9$ — Parameters of Random Variables (Part $ \mathrm{A} $)
• Topic $4.3$ — Justifying a Claim Based on a Confidence Interval for a Population Mean or Population Mean Difference (Parts $ \mathrm{B} $, $ \mathrm{C} $)
• Topic $4.4$ — Setting Up a Test for a Population Mean or Population Mean Difference (Part $ \mathrm{B} $)

▶️ Answer/Explanation

A. i.
Fewer than 3 bedrooms means a house has either 1 or 2 bedrooms.
$P(\text{Bedrooms} < 3) = P(1) + P(2) = 0.12 + 0.22$
$\boxed{P(\text{Bedrooms} < 3) = 0.34}$

A. ii.
The sample mean is calculated by summing the products of the values and their corresponding proportions.
$\bar{x} = \sum x_i \cdot p_i = 1(0.12) + 2(0.22) + 3(0.28) + 4(0.22) + 5(0.14) + 6(0.02)$
$\bar{x} = 0.12 + 0.44 + 0.84 + 0.88 + 0.70 + 0.12$
$\boxed{\bar{x} = 3.10\,\text{bedrooms}}$

B. i.
Let $\mu$ represent the true mean number of bedrooms in all newly built houses in Country B in 2024.
$H_0: \mu = 2.9$
$H_a: \mu \neq 2.9$

B. ii.
• A Type I error happens if Rodney concludes that the true mean number of bedrooms in 2024 is different from 2.9 when, in reality, it is still exactly 2.9.
• In practice, this means the researcher would mistakenly declare a shift in housing layout profiles where no genuine structural trend modification occurred.

C.
• Since the significance level $\alpha = 0.03$ matches the two-sided boundary of a 97% confidence interval $(1 – 0.97 = 0.03)$, we can judge the test based on whether the null value falls inside the interval boundaries.
• The hypothesized baseline mean value $\mu_0 = 2.9$ lies completely outside Keisha’s 97% confidence interval of $(3.01, 3.19)$.
• Therefore, Rodney would reject the null hypothesis $H_0$ and conclude that there is convincing statistical evidence that the true mean number of bedrooms in newly built houses in Country B in 2024 is different from 2.9.

Question

A medical researcher completed a study comparing an omega-3 fatty acids supplement to a placebo in the treatment of irritability in patients with a certain medical condition. Nineteen patients with the medical condition volunteered to participate in the study. The study was conducted using the following weekly schedule.

• Week 1: Each patient took a randomly assigned treatment, omega-3 supplement or placebo.
• Week 2: The patients did not take either the omega-3 supplement or the placebo. This was necessary to reduce the possibility of any carryover effect from the assigned treatment taken during week 1.
• Week 3: Each patient took the treatment, omega-3 supplement or placebo, that they did not take during week 1.

At the end of week 1 and week 3, each patient’s irritability was given a score on a scale of 0 to 10, with 0 representing no irritability and 10 representing the highest level of irritability.

For each patient, the two irritability scores and the difference in their scores (placebo minus omega-3) were recorded. The results are summarized in the table and boxplots.

The researcher claims the omega-3 supplement will decrease the mean irritability score of all patients with the medical condition similar to the volunteers who participated in the study. Is there convincing statistical evidence to support the researcher’s claim at a significance level of $\alpha=0.05$? Complete the appropriate inference procedure to support your answer.

Most-appropriate topic codes (AP Statistics):

• Topic $4.4$ — Setting Up a Test for a Population Mean or Population Mean Difference (Entire Question)
• Topic $4.5$ — Carrying Out a Test for a Population Mean or Population Mean Difference (Entire Question)

▶️ Answer/Explanation

We need to perform a matched-pairs $t$-test for a population mean difference ($\mu_d = \mu_{\text{placebo}} – \mu_{\text{omega-3}}$). Our hypotheses are $H_0: \mu_d = 0$ versus $H_a: \mu_d > 0$. The conditions are met: treatments were randomly assigned, and the boxplot of the differences shows no extreme outliers or severe skewness, making the $t$-procedure appropriate for $n=19$.

Calculating the test statistic, we get $t = \frac{1.789 – 0}{2.485 / \sqrt{19}} \approx 3.138$ with $df = 18$. This yields a $p$-value of approximately $0.0028$.

Because the $p$-value ($0.0028$) is less than our significance level ($\alpha = 0.05$), we reject the null hypothesis. There is convincing statistical evidence to conclude that the omega-3 supplement decreases the true mean irritability score for patients with this medical condition.

Question

A researcher conducted a study to investigate whether local car dealers tend to charge women more than men for the same car model. Using information from the county tax collector’s records, the researcher randomly selected one man and one woman from among everyone who had purchased the same model of an identically equipped car from the same dealer. The process was repeated for a total of 8 randomly selected car models.

The purchase prices and the differences (woman − man) are shown in the table below. Summary statistics are also shown.

Dotplots of the data and the differences are shown below.

Do the data provide convincing evidence that, on average, women pay more than men in the county for the same car model?

Most-appropriate topic codes (AP Statistics):

• Topic $4.4$ — Setting Up a Test for a Population Mean Difference (full question)
• Topic $4.5$ — Carrying Out a Test for a Population Mean Difference (full question)

▶️ Answer/Explanation

Step 1: State the hypotheses.
Let $\mu_{\text{diff}}$ be the population mean difference in purchase price (woman − man) for identically equipped cars of the same model sold by the same dealer in the county.
$H_0 : \mu_{\text{diff}} = 0$ (on average, women and men pay the same)
$H_a : \mu_{\text{diff}} > 0$ (on average, women pay more than men)

Step 2: Identify the procedure and check conditions.
Since each pair consists of one man and one woman buying the same car model from the same dealer, the data are paired. The appropriate procedure is a paired $t$-test.
Random: The 8 car models were randomly selected and within each model, one man and one woman were randomly selected. The random condition is met.
Normal/Large Sample: The sample size $n = 8$ is small, so we cannot rely on the Central Limit Theorem. The dotplot of the differences shows a roughly symmetric distribution with no strong skewness or outliers, so it is reasonable to assume the population of differences is approximately normally distributed. The normal condition is considered met.

Step 3: Calculate the test statistic and $p$-value.
From the summary statistics for the differences: $\bar{x}_d = 585$, $s_d = 530.71$, $n = 8$.
The test statistic is:
$t = \dfrac{\bar{x}_d – 0}{\dfrac{s_d}{\sqrt{n}}} = \dfrac{585 – 0}{\dfrac{530.71}{\sqrt{8}}} = \dfrac{585}{187.63} \approx 3.12$
Degrees of freedom: $df = n – 1 = 8 – 1 = 7$
The $p$-value for a one-sided $t$-test with $t \approx 3.12$ and $df = 7$ is:
$p\text{-value} \approx 0.008$

Step 4: State the conclusion in context.
Since the $p$-value of $0.008$ is less than $\alpha = 0.05$, we reject $H_0$. The data provide convincing evidence that, on average, women pay more than men in the county for the same car model.

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

A bottle-filling machine is set to dispense 12.1 fluid ounces into juice bottles. To ensure that the machine is filling accurately, every hour a worker randomly selects four bottles filled by the machine during the past hour and measures the contents. If there is convincing evidence that the mean amount of juice dispensed is different from 12.1 ounces or if there is convincing evidence that the standard deviation is greater than 0.05 ounce, the machine is shut down for recalibration. It can be assumed that the amount of juice that is dispensed into bottles is normally distributed.

During one hour, the mean number of fluid ounces of four randomly selected bottles was 12.05 and the standard deviation was 0.085 ounce.

(a) Perform a test of significance to determine whether the mean amount of juice dispensed is different from 12.1 fluid ounces. Assume the conditions for inference are met.

(b) To determine whether this sample of four bottles provides convincing evidence that the standard deviation of the amount of juice dispensed is greater than 0.05 ounce, a simulation study was performed. In the simulation study, 300 samples, each of size 4, were randomly generated from a normal population with a mean of 12.1 and a standard deviation of 0.05. The sample standard deviation was computed for each of the 300 samples. The dotplot below displays the values of the sample standard deviations.

Use the results of this simulation study to explain why you think the sample provides or does not provide evidence that the standard deviation of the juice dispensed exceeds 0.05 fluid ounce.

Most-appropriate topic codes (AP Statistics):

• Topic $4.4$ — Setting Up a Test for a Population Mean or Population Mean Difference (Part $\mathrm{a}$: hypotheses and test selection)
• Topic $4.5$ — Carrying Out a Test for a Population Mean or Population Mean Difference (Part $\mathrm{a}$: test statistic, $p$-value, and conclusion)
• Topic $2.3$ — Estimating Probabilities Using Simulation (Part $\mathrm{b}$: simulation-based estimation of a $p$-value)
• Topic $3.6$ — p-Values (Part $\mathrm{b}$: interpreting simulated $p$-value evidence)

▶️ Answer/Explanation

(a)
Let $\mu$ = the true mean number of fluid ounces dispensed into all juice bottles filled by the machine in the past hour.
Step 1 — Hypotheses:
$H_0: \mu = 12.1$
$H_a: \mu \neq 12.1$

Step 2 — Test: One-sample $t$-test for a mean (conditions are given as met; population standard deviation is unknown).
$t = \frac{\bar{x} – \mu_0}{s/\sqrt{n}}$

Step 3 — Mechanics:
Given: $\bar{x} = 12.05$, $s = 0.085$, $n = 4$, $\mu_0 = 12.1$
$t = \frac{12.05 – 12.1}{0.085/\sqrt{4}} = \frac{-0.05}{0.0425} \approx -1.176$
Degrees of freedom: $df = n – 1 = 3$
Two-sided $p$-value:
$p\text{-value} = 2 \cdot P(T_3 < -1.176) \approx 0.324$

Step 4 — Conclusion:
Since the $p$-value of $0.324$ is much larger than any reasonable significance level (such as $\alpha = 0.05$), we fail to reject $H_0$.
There is not sufficient evidence to conclude that the mean amount of juice being dispensed is different from $12.1$ fluid ounces. The machine does not need to be shut down on the basis of the mean.

(b)

In the simulation, 300 samples of size 4 were drawn from a normal population with $\sigma = 0.05$. The sample standard deviation of $s = 0.085$ from our actual data falls well out in the right tail of the dotplot.
Counting the dots at or beyond $0.085$ in the dotplot, only about 12 out of 300 simulated values are as large or larger than $0.085$.
This gives an estimated (simulated) $p$-value of:
$\hat{p}\text{-value} = \frac{12}{300} = 0.04$
Since this simulated $p$-value of $0.04$ is less than $\alpha = 0.05$, the sample does provide convincing evidence that the true standard deviation of the juice dispensed exceeds $0.05$ fluid ounce. The machine should be shut down for recalibration.

Question

Administrators in a large school district wanted to determine whether students who attended a new magnet school for one year achieved greater improvement in science test performance than students who did not attend the magnet school. Knowing that more parents would want to enroll their children in the magnet school than there was space available for those children, the district administrators decided to conduct a lottery of all families who expressed interest in participating. In their data analysis, the administrators would then compare the change in test scores of those children who were selected to attend the magnet school with the change in test scores of those who applied to attend the magnet school but who were not selected.

The tables below show the scores on the same science pretest and the same science posttest for 20 students. Of the 20 students, 8 were randomly selected from the magnet school and 12 were randomly selected from those who applied to attend the magnet school but who were not selected and then attended their original school.

(a) Perform a test to determine whether students who attend the magnet school demonstrate a significantly higher mean difference in test scores $(\text{Posttest} – \text{Pretest})$ than students who applied to attend the magnet school but who were not selected and then attended their original school.

Administrators were also interested in using pretest scores on this test as a predictor of posttest scores on the test. The following computer output contains the results from separate regression analyses on the magnet school scores and on the original school scores. The accompanying graph displays the data and separate regression lines for the magnet and original schools.

(b)

(i) State the equation of the regression line for the magnet school and interpret its slope in the context of the question.

(ii) State the equation of the regression line for the original school and interpret its slope in the context of the question.

(c) To determine whether there is a significant correlation between pretest score and posttest score, a test of the following hypotheses will be performed.

$H_0$: There is no correlation between pretest score and posttest score (true slope $= 0$)

versus

$H_a$: There is a correlation between pretest score and posttest score (true slope $\neq 0$)

(i) Using the regression output, state the $p$-value and conclusion for this test at the magnet school. Assume the conditions for inference have been met.

(ii) Using the regression output, state the $p$-value and conclusion for this test at the original school. Assume the conditions for inference have been met.

(d) What additional information do the regression analyses give you about student performance on the science test at the two schools beyond the comparison of mean differences in part (a)?

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 4.5 — Carrying Out a Test for a Population Mean or Population Mean Difference (Part $\mathrm{a}$)
• Topic 5.3 — Linear Regression Models (Part $\mathrm{b}$)
• Topic 5.2 — Correlation (Part $\mathrm{c}$)
• Topic 5.5 — Least-Squares Regression (Parts $\mathrm{b}$, $\mathrm{c}$, $\mathrm{d}$)

▶️ Answer/Explanation

(a)

Step 1 — Hypotheses
Let $\mu_{\text{DiffM}}$ = the mean difference (posttest $-$ pretest) for all students at the magnet school, and $\mu_{\text{DiffO}}$ = the mean difference for all students who applied but were not selected and attended their original school.
$H_0: \mu_{\text{DiffM}} = \mu_{\text{DiffO}}$
$H_a: \mu_{\text{DiffM}} > \mu_{\text{DiffO}}$

Step 2 — Test and Conditions

We use a two-sample $t$-test for the difference of two means:
$t = \dfrac{\bar{x}_M – \bar{x}_O}{\sqrt{\dfrac{s_M^2}{n_M} + \dfrac{s_O^2}{n_O}}}$

We need to assume randomness of the sampling used. It was stated in the stem that the students from the two different schools were randomly selected.
We need to check the assumption that the distributions of differences (posttest – pretest) for each of the two schools are normally distributed. Based on histograms and boxplots of these differences, there are no outliers or extreme skewness. Because these graphs reveal no obvious departures from normality, it appears reasonable to proceed with the t-test.

Step 3 — Test Statistic and $p$-value
$t = \dfrac{11.750 – 3.000}{\sqrt{\dfrac{(9.407)^2}{8} + \dfrac{(3.977)^2}{12}}} = \dfrac{8.750}{\sqrt{11.062 + 1.318}} = \dfrac{8.750}{\sqrt{12.380}} = \dfrac{8.750}{3.518} \approx 2.487$
$df \approx 8.69$, $\quad p\text{-value} \approx 0.0177$

Step 4 — Conclusion
Since $p = 0.0177 < \alpha = 0.05$, we reject $H_0$. There is convincing evidence that students who attend the magnet school have a higher mean improvement in science test scores than students who attended their original school.

(b)(i)

The regression equation for the magnet school is:
$\hat{y} = 73.27 + 0.1811x$
where $x$ is the pretest score and $\hat{y}$ is the predicted posttest score. The slope of $0.1811$ means that for each additional point scored on the pretest by a magnet school student, the posttest score is predicted to increase by $0.1811$ points, on average. The slope is positive but very close to zero, suggesting that pretest performance has almost no predictive power for posttest performance at the magnet school.

(b)(ii)

The regression equation for the original school is:
$\hat{y} = 9.24 + 0.9204x$
where $x$ is the pretest score and $\hat{y}$ is the predicted posttest score. The slope of $0.9204$ means that for each additional point scored on the pretest by an original school student, the posttest score is predicted to increase by approximately $0.9204$ points, on average — a nearly one-for-one relationship.

(c)(i) — Magnet School
From the regression output, the test statistic is $t = 0.40$ with $p\text{-value} = 0.706$.
Since $0.706 > 0.05$, we fail to reject $H_0$. There is insufficient evidence to conclude that there is a significant correlation between pretest score and posttest score at the magnet school. Pretest score is not a useful linear predictor of posttest score for magnet school students.

(c)(ii) — Original School

From the regression output, the test statistic is $t = 6.09$ with $p\text{-value} = 0.000$.
Since $0.000 < 0.05$, we reject $H_0$. There is strong evidence of a significant correlation between pretest score and posttest score at the original school. Pretest score is a very strong linear predictor of posttest score for original school students.

(d)

The two-sample $t$-test in part (a) told us only that the magnet school group had a higher average improvement — but it didn’t explain who benefited or by how much depending on their initial ability. The regression analyses reveal something much more interesting:
• At the magnet school, the slope is nearly zero ($0.1811$), and $R^2 = 2.5\%$ — this means students score high on the posttest regardless of how they did on the pretest. A student who entered the magnet school with a low pretest score of 64 scored 89 on the posttest (an improvement of 25 points), while a student with a higher pretest score of 86 actually dropped 2 points. The magnet school appears to level the playing field and disproportionately benefits students who start with lower ability.
• At the original school, the slope is close to 1 ($0.9204$) and $R^2 = 78.8\%$ — students essentially maintained their relative ranking, with high pretest scorers also achieving high posttest scores. There is very little “boost” effect for any student regardless of their starting point.
In short, the regression analyses reveal that the magnet school benefits students with low pretest scores the most, while the original school produces predictable but modest gains proportional to where students started.

Question

A certain state’s education commissioner released a new report card for all the public schools in that state. This report card provides a new tool for comparing schools across the state. One of the key measures that can be computed from the report card is the student-to-teacher ratio, which is the number of students enrolled in a given school divided by the number of teachers at that school.

The data below give the student-to-teacher ratio at the 10 schools with the highest proportion of students meeting the state reading standards in the third grade and at the 10 schools with the lowest proportion of students meeting the state reading standards in the third grade.

(a) Display a dotplot for each group to compare the distribution of student-to-teacher ratios in the top 10 schools with the distribution in the bottom 10 schools. Comment on the similarities and differences between the two distributions.

(b) Any statistical test that is used to determine whether the mean student-to-teacher ratio is the same for the top 10 schools as it is for the bottom 10 schools would be inappropriate. Explain why in a few sentences.

Most-appropriate topic codes (AP Statistics):

• Topic 1.5 — Graphical Representations for One Quantitative Variable (Part $\mathrm{a}$)
• Topic 1.9 — Comparisons of the Distributions for One Quantitative Variable (Part $\mathrm{a}$)
• Topic 1.11 — Random Sampling (Part $\mathrm{b}$)
• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part $\mathrm{b}$)

▶️ Answer/Explanation

(a)
First, let’s organize the data for both groups:
Highest Proportion group: $7, 9, 12, 16, 16, 17, 17, 18, 21, 22$
Lowest Proportion group: $12, 12, 14, 14, 16, 16, 18, 19, 20, 20$
The dotplots, displayed on a common scale from $4$ to $24$, are shown below:

Similarities: The two distributions are centered at approximately the same place. The median for the Highest Proportion group is $\dfrac{16+17}{2} = 16.5$ and the median for the Lowest Proportion group is $\dfrac{16+16}{2} = 16$, so both centers are very close to $16$.
Differences: The distribution for the Highest Proportion group is much more spread out (variable) than the distribution for the Lowest Proportion group. The range for the Highest Proportion group is $22 – 7 = 15$, while the range for the Lowest Proportion group is only $20 – 12 = 8$. In other words, the top schools show much greater variability in their student-to-teacher ratios compared to the bottom schools.

(b)
The two groups of schools are not random samples drawn from two larger populations of interest.
The group of 10 schools with the highest proportion of students meeting the standards is itself the entire population of such schools — it is not a random sample from some larger population of high-performing schools.
Similarly, the group of 10 schools with the lowest proportion is itself the complete population of the lowest-performing schools in the state — not a random sample from a larger population.
Since statistical inference is designed to generalize conclusions from a sample to a broader population, and these two groups are not random samples but rather complete populations defined by their extreme values, applying any inferential procedure (such as a two-sample $t$-test) to these data would be inappropriate. There is no larger population to generalize to.

Question

A researcher wants to conduct a study to test whether listening to soothing music for 20 minutes helps to reduce diastolic blood pressure in patients with high blood pressure, compared to simply sitting quietly in a noise-free environment for 20 minutes. One hundred patients with high blood pressure at a large medical clinic are available to participate in this study.

(a) Propose a design for this study to compare these two treatments.

(b) The null hypothesis for this study is that there is no difference in the mean reduction of diastolic blood pressure for the two treatments and the alternative hypothesis is that the mean reduction in diastolic blood pressure is greater for the music treatment. If the null hypothesis is rejected, the clinic will offer this music therapy as a free service to their patients with high blood pressure. Describe Type I and Type II errors and the consequences of each in the context of this study, and discuss which one you think is more serious.

Most-appropriate topic codes (AP Statistics):

• Topic 1.13 — Experimental Design (Part $\mathrm{a}$)
• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part $\mathrm{b}$)
• Topic 3.8 — Potential Errors When Performing Tests (Part $\mathrm{b}$)

▶️ Answer/Explanation

(a)
Proposed Design (Completely Randomized Design):
Assign each of the 100 patients a unique number from 00 to 99. Using a random number table (or random number generator), select 50 unique numbers. The patients corresponding to the selected numbers will form Group 1 (Music Treatment); the remaining 50 patients will form Group 2 (Control — Noise-Free Environment).
Measure the diastolic blood pressure of every patient before the treatment begins. Then:
Group 1 patients sit quietly in a room where soothing music is played for 20 minutes.
Group 2 patients sit quietly in a noise-free environment for 20 minutes.
At the end of the 20-minute period, measure the diastolic blood pressure of every patient again. Compute the reduction in diastolic blood pressure (before $-$ after) for each patient. Run a two-sample $t$-test to compare the mean reduction in the two groups.

Alternatively (Paired Design):
Each patient receives both treatments on two separate occasions, with a suitable washout period in between. For each patient, flip a coin to randomly assign which treatment is administered first. Compute the difference (music reduction $-$ noise-free reduction) for each patient, and use a paired $t$-test to determine whether the mean difference is significantly greater than zero.

(b)
Let $\mu_M$ = mean reduction in diastolic blood pressure under the music treatment, and $\mu_C$ = mean reduction under the control (noise-free) treatment. The hypotheses are:
$H_0: \mu_M = \mu_C \qquad H_a: \mu_M > \mu_C$
Type I Error:
Rejecting $H_0$ when it is actually true — that is, concluding that soothing music does reduce diastolic blood pressure more than sitting quietly, when in reality it does not.
Consequence: The clinic will offer music therapy as a free service to its patients even though the therapy is ineffective. This wastes clinic resources and money, providing no real medical benefit to patients.
Type II Error:
Failing to reject $H_0$ when it is actually false — that is, concluding that soothing music does not reduce diastolic blood pressure more than sitting quietly, when in reality it does.
Consequence: The clinic will not offer music therapy, even though the therapy would genuinely help patients reduce their blood pressure. Patients are denied access to an effective, low-cost treatment.

Which error is more serious?
A reasonable case can be made for either error. One well-supported argument is that the Type II error is more serious: denying patients an effective treatment that could meaningfully improve their health — and potentially save lives — is a greater harm than wasting some clinic resources on an ineffective service. In the case of a Type I error, patients who receive music therapy are unlikely to be harmed by listening to music. But in the case of a Type II error, patients who could benefit from a real treatment are denied it altogether.

Question

The nerves that supply sensation to the front portion of a person’s foot run between the long bones of the foot. Tight-fitting shoes can squeeze these nerves between the bones, causing pain when the nerves swell. This condition is called Morton’s neuroma. Because most people have a dominant foot, muscular development is not the same in both feet. People who have Morton’s neuroma may have the condition in only one foot or they may have it in both feet.

Investigators selected a random sample of 12 adult female patients with Morton’s neuroma to study this disease further. The data below are measurements of nerve swelling as recorded by a physician. A value of 1.0 is considered “normal,” and 2.0 is considered extreme swelling. The population distribution of the swelling measurements is approximately normal for adult females who have Morton’s neuroma.

(a) A scatterplot of the ordered pairs (swelling in left foot, swelling in right foot), is shown below.

The scatterplot suggests there are two distinct groups of patients. Patients within each group share a common trait. Use the scatterplot above and the table to determine the common trait and explain how this trait differs for the two groups.

(b) A scatterplot of the ordered pairs (swelling in dominant foot, swelling in nondominant foot), is shown below.

What conclusion can be drawn from this scatterplot that is not apparent from the scatterplot in part (a)?

(c) Can you conclude that there is a difference between the mean swelling in the dominant foot and the mean swelling in the nondominant foot for adult females who have Morton’s neuroma in at least one foot? Give a statistical justification to support your answer.

(For easy reference, the table of data from above also appears at the bottom of this question.)

(d) The nerve swelling measurement is used to indicate whether a foot has Morton’s neuroma. Use the 24 measurements of nerve swelling to suggest a criterion for diagnosing Morton’s neuroma. Justify your suggestion graphically.

(For easy reference, the table of data from above also appears below.)

Most-appropriate topic codes (AP Statistics):

• Topic 5.1 — Graphical Representations Between Two Quantitative Variables (Parts $\mathrm{a}$, $\mathrm{b}$)
• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part $\mathrm{c}$)
• Topic 4.5 — Carrying Out a Test for a Population Mean or Population Mean Difference (Part $\mathrm{c}$)
• Topic 1.5 — Graphical Representations for One Quantitative Variable (Part $\mathrm{d}$)
• Topic 1.9 — Comparisons of the Distributions for One Quantitative Variable (Part $\mathrm{d}$)

▶️ Answer/Explanation

(a)

The trait that distinguishes the two groups in the scatterplot is the dominant foot (left or right). All the points in the upper-left cluster represent patients whose dominant foot is the right foot, while all the points in the lower-right cluster represent patients whose dominant foot is the left foot. The dominant foot type is the common trait, and it differs between the two groups.

(b)

Two conclusions become clear from this scatterplot that were not visible before:
First, there is a positive linear relationship between swelling in the dominant foot and swelling in the nondominant foot — as swelling in the dominant foot increases, swelling in the nondominant foot tends to increase as well.
Second, and importantly, every single point lies below the line $y = x$, which means swelling in the dominant foot is consistently greater than swelling in the nondominant foot for all patients in the sample. This pattern across both groups combined is something you simply could not see in the left-foot vs. right-foot scatterplot from part (a).

(c)

We perform a matched-pairs $t$-test on the differences $d_i = \text{(dominant swelling)} – \text{(nondominant swelling)}$.
The 12 differences are:
$0.30,\ 0.30,\ 0.45,\ 0.15,\ 0.30,\ 0.35,\ 0.25,\ 0.35,\ 0.20,\ 0.25,\ 0.40,\ 0.15$
State hypotheses (where $\mu_d$ is the mean difference, dominant minus nondominant):
$H_0: \mu_d = 0$
$H_a: \mu_d \neq 0$
Check conditions:
1. We are told a random sample was selected from the population of adult females with Morton’s neuroma.
2. A dotplot of the differences shows a roughly symmetric, unimodal distribution with no outliers — it is reasonable to treat the population of differences as approximately normal.
Compute the test statistic:
$\bar{x}_d = 0.2875, \quad s_d = 0.0932, \quad n = 12, \quad df = 11$
$t = \dfrac{\bar{x}_d – 0}{\dfrac{s_d}{\sqrt{n}}} = \dfrac{0.2875 – 0}{\dfrac{0.0932}{\sqrt{12}}} = 10.68$
$p\text{-value} \approx 0.0000004 \approx 0$
Since the $p$-value is essentially $0$, which is far less than any reasonable significance level $\alpha$, we reject $H_0$. There is very convincing statistical evidence that the mean swelling in the dominant foot is different from (and specifically greater than) the mean swelling in the nondominant foot for adult females who have Morton’s neuroma in at least one foot.

(d)

To suggest a diagnostic criterion, we separate all 24 swelling measurements into two groups: the 17 foot measurements from feet that have Morton’s neuroma and the 7 foot measurements from feet that do not have Morton’s neuroma. A stacked dotplot of the two groups is shown below:

The dotplot makes it visually clear that all 7 feet without Morton’s neuroma have swelling measurements of $1.40$ or below, while the feet with Morton’s neuroma have swelling values of $1.40$ and above (with the measurements extending up to $1.85$). Based on this graphical display, a reasonable diagnostic criterion is:
$\boxed{\text{Swelling measurement} \geq 1.4 \Rightarrow \text{diagnose Morton’s neuroma}}$
A cutoff of approximately $1.4$ or higher serves as a sensible threshold for diagnosing Morton’s neuroma, since it cleanly separates the feet with and without the condition in this dataset.

Question

Investigators at the U.S. Department of Agriculture wished to compare methods of determining the level of E. coli bacteria contamination in beef. Two different methods (A and B) of determining the level of contamination were used on each of ten randomly selected specimens of a certain type of beef. The data obtained, in millimicrobes/liter of ground beef, for each of the methods are shown in the table below.

Is there a significant difference in the mean amount of E. coli bacteria detected by the two methods for this type of beef? Provide a statistical justification to support your answer.

Most-appropriate topic codes (AP Statistics):

• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Hypotheses and Conditions)
• Topic 4.5 — Carrying Out a Test for a Population Mean or Population Mean Difference (Test Mechanics and Conclusion)
• Topic 1.5 — Graphical Representations for One Quantitative Variable (Checking the Distribution of Differences)

▶️ Answer/Explanation

We conduct a paired $t$-test for the mean difference in the level of E. coli bacteria contamination detected by the two methods.

Step 1 — Hypotheses
Let $\mu_d$ be the population mean difference in E. coli contamination levels (Method A $-$ Method B).
$H_0: \mu_d = 0$ (no difference in mean contamination detected by the two methods)
$H_a: \mu_d \neq 0$ (there is a difference in mean contamination detected by the two methods)

Step 2 — Identify Test and Check Conditions
We use a paired $t$-test with test statistic:
$t = \dfrac{\bar{x}_d – 0}{s_d / \sqrt{n_d}}$
First, compute the differences $d_i = A_i – B_i$ for each specimen:
$-0.3,\quad 0.5,\quad 0.3,\quad 0.6,\quad 0.8,\quad 0.7,\quad 1.2,\quad 0.2,\quad -0.1,\quad -1.0$

Condition 1 — Independence: The 10 specimens were randomly selected, so it is reasonable to assume the 10 pairs of measurements are independent of one another.
Condition 2 — Normality of Differences: With only $n = 10$ differences, we check a histogram or boxplot of the differences. The histogram of the differences $(A – B)$ is roughly symmetric with no apparent outliers, so it is reasonable to assume the population distribution of differences is approximately normal.

Step 3 — Mechanics
From the differences:
$\bar{x}_d = 0.29, \qquad s_d = 0.6297, \qquad n = 10$
$t = \dfrac{0.29 – 0}{0.6297/\sqrt{10}} = \dfrac{0.29}{0.1991} \approx 1.456$
$\text{degrees of freedom} = n – 1 = 9$
$p\text{-value} = 2 \times P(t_9 > 1.456) \approx 0.1793$

Step 4 — Conclusion
Since the $p$-value of $0.1793$ is greater than $\alpha = 0.05$, we fail to reject $H_0$.
We do not have statistically significant evidence to conclude that there is a difference in the mean amount of E. coli bacteria detected by the two methods for this type of beef.
$\boxed{p\text{-value} = 0.1793 > 0.05 \Rightarrow \text{Fail to reject } H_0; \text{ no significant difference between the two methods.}}$

Question

A manufacturer of thermostats is concerned that the readings of its thermostats have become less reliable (more variable). In the past, the variance has been $1.52$ degrees Fahrenheit (F) squared. A random sample of $10$ recently manufactured thermostats was selected and placed in a room that was maintained at $68^\circ\text{F}$. The readings for those 10 thermostats are given in the table below.

(a) State the null and alternative hypotheses that the manufacturer is interested in testing.

It can be shown that if the population of thermostat temperatures is normally distributed, the sampling distribution of $\dfrac{(n-1)s^2}{\sigma^2}$ follows a chi-square distribution with $n-1$ degrees of freedom.

(b) Calculate the value of $\dfrac{(n-1)s^2}{1.52}$ for these data.

(c) Assume that the population of thermostat temperatures follows a normal distribution. Use the test statistic $\dfrac{(n-1)s^2}{1.52}$ from part (b) and the chi-square distribution to test the hypotheses in part (a).

(d) For the test conducted in part (c), what is the smallest value of the test statistic that would have led to the rejection of the null hypothesis at the 5 percent significance level?

Mark this value of the test statistic on the graph of the chi-square distribution below. Indicate the region that contains all of the values that would have led to the rejection of the null hypothesis.

(e) Using simulation, 1,000 samples, each of size 10, were randomly generated from 3 populations with different variances. Each population was normally distributed with mean 68 and variance greater than 1.52. The histograms below show the simulated sampling distribution of $\dfrac{(n-1)s^2}{1.52}$ for each population.

Mark the region identified in part (d) on each of the histograms below.

(f) Based on the regions that you marked in part (e), identify the simulated sampling distribution that corresponds to the population with the largest variance. Then identify the simulated sampling distribution that corresponds to the population with the smallest variance. Justify your choices.

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 4.5 — Carrying Out a Test for a Population Mean or Population Mean Difference (Parts $\mathrm{b}$, $\mathrm{c}$, $\mathrm{d}$)
• Topic 2.12 — Sampling Distributions and the Central Limit Theorem (Parts $\mathrm{e}$, $\mathrm{f}$)

▶️ Answer/Explanation

(a)

Let $\sigma^2$ denote the true population variance of the readings of recently manufactured thermostats (in degrees Fahrenheit squared).
$H_0: \sigma^2 = 1.52 \qquad \text{(variance has not changed)}$
$H_a: \sigma^2 > 1.52 \qquad \text{(recently produced thermostats are more variable)}$

(b)

First, compute the sample standard deviation from the 10 readings:
$s^2 = 2.0383 \implies s = 1.4277$
Then compute the test statistic:
$\chi^2 = \frac{(n-1)s^2}{1.52} = \frac{9 \times 2.0383}{1.52} = \frac{18.345}{1.52}$
$\boxed{\chi^2 \approx 12.069}$

(c)

Under $H_0$, the test statistic follows a $\chi^2$ distribution with $n – 1 = 9$ degrees of freedom.
The $p$-value is the probability of obtaining a test statistic as large as or larger than the observed value:
$p\text{-value} = P\!\left(\chi^2_9 \geq 12.069\right) \approx 0.2094$
(From the table: $0.20 < p\text{-value} < 0.25$)
Since the $p$-value $\approx 0.2094 > 0.05$, we fail to reject $H_0$. There is not statistically significant evidence at the $\alpha = 0.05$ level that the recently manufactured thermostats have become more variable than in the past.
$\boxed{\text{Fail to reject } H_0;\ p\text{-value} \approx 0.209}$

(d)

The smallest value of the test statistic that leads to rejection of $H_0$ at the 5% significance level is the 95th percentile of the $\chi^2$ distribution with 9 degrees of freedom:
$\boxed{\chi^2_{0.05,\,9} = 16.92}$
The rejection region consists of all chi-square values greater than or equal to $16.92$, which corresponds to the shaded right tail of the $\chi^2_9$ curve (as marked on the graph above).

(e)

The rejection region — all simulated values to the right of $16.92$ — should be marked on each of the three histograms (Histogram I, Histogram II, and Histogram III) as indicated by the dashed vertical lines in the diagrams above.

(f)

Largest variance → Histogram III. A population with a larger variance will tend to produce larger sample variances $s^2$, and hence larger values of the test statistic $\frac{(n-1)s^2}{1.52}$. Histogram III has the greatest proportion of simulated values falling to the right of $16.92$ — meaning it has the highest probability of correctly rejecting $H_0$ — so it corresponds to the population with the largest variance.
Smallest variance → Histogram II. Histogram II has its simulated values concentrated most tightly at smaller values, with the smallest proportion of values exceeding $16.92$. This means it has the lowest probability of rejecting $H_0$, consistent with coming from the population whose variance is smallest (and closest to $1.52$ among the three).
$\boxed{\text{Largest variance: Histogram III} \qquad \text{Smallest variance: Histogram II}}$

Question

The developers of a training program designed to improve manual dexterity claim that people who complete the 6-week program will increase their manual dexterity. A random sample of 12 people enrolled in the training program was selected. A measure of each person’s dexterity on a scale from 1 (lowest) to 9 (highest) was recorded just before the start of and just after the completion of the 6-week program. The data are shown in the table below.

Can one conclude that the mean manual dexterity for people who have completed the 6-week training program has significantly increased? Support your conclusion with appropriate statistical evidence.

Most-appropriate topic codes (AP Statistics):

▶️ Answer/Explanation

Step 1: State the hypotheses.
Let $\mu_D$ denote the mean difference (after $-$ before) in dexterity scores for the population of individuals enrolled in the program.
$H_0: \mu_D = 0$
$H_a: \mu_D > 0$

Step 2: Identify the correct test and check conditions.
Since the data consist of before-and-after measurements on the same individuals, we use a one-sample paired $t$-test on the differences $d_i = \text{after}_i – \text{before}_i$.
The individual differences are: $1.1,\ 0.5,\ 0.6,\ 0,\ 0.7,\ 0.5,\ 0.5,\ -0.1,\ -0.1,\ 0.5,\ 0.3,\ 0$
Conditions:
1. Random sample: The problem states that the 12 people are a random sample from the population enrolled in the program.
2. Approximate normality: With only $n = 12$ observations, we check the distribution of differences. A dotplot or histogram of the differences shows no strong skewness or outliers, so it is reasonable to assume the differences are approximately normally distributed.

Step 3: Compute the test statistic and $p$-value.
From the differences, we compute:
$\bar{d} = \frac{85.6 – 81.1}{12} = \frac{4.5}{12} = 0.375$
$s_d = 0.367$
$\text{Degrees of freedom} = n – 1 = 12 – 1 = 11$
$t = \frac{\bar{d} – 0}{s_d / \sqrt{n}} = \frac{0.375 – 0}{0.367 / \sqrt{12}} = \frac{0.375}{0.1059} = 3.54$
$p\text{-value} = P(t_{11} > 3.54) \approx 0.002$

Step 4: State the conclusion in context.
Since the $p$-value ($\approx 0.002$) is less than any common significance level such as $\alpha = 0.05$, we reject $H_0$.
There is sufficient, statistically significant evidence to conclude that, on average, people who completed the 6-week training program have meaningfully increased their manual dexterity.
$\boxed{t = 3.54,\quad p\text{-value} \approx 0.002 \implies \text{Reject } H_0 \text{ — mean dexterity has significantly increased}}$

Question

In search of a mosquito repellent that is safer than the ones that are currently on the market, scientists have developed a new compound that is rated as less toxic than the current compound, thus making a repellent that contains this new compound safer for human use. Scientists also believe that a repellent containing the new compound will be more effective than the ones that contain the current compound. To test the effectiveness of the new compound versus that of the current compound, scientists have randomly selected 100 people from a state.

Up to 100 bins, with an equal number of mosquitoes in each bin, are available for use in the study. After a compound is applied to a participant’s forearm, the participant will insert his or her forearm into a bin for 1 minute, and the number of mosquito bites on the arm at the end of that time will be determined.

(a) Suppose this study is to be conducted using a completely randomized design. Describe a randomization process and identify an inference procedure for the study.

(b) Suppose this study is to be conducted using a matched-pairs design. Describe a randomization process and identify an inference procedure for the study.

(c) Which of the designs, the one in part (a) or the one in part (b), is better for testing the effectiveness of the new compound versus that of the current compound? Justify your answer.

Most-appropriate topic codes (AP Statistics):

• Topic 1.13 — Experimental Design (Parts a, b, c)
• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part b)
• Topic 4.9 — Setting Up a Test for the Difference Between Two Population Means (Part a)

▶️ Answer/Explanation

(a) Completely Randomized Design

Randomization process:
Assign each of the 100 participants a unique random number using a random number generator. Sort the participants from smallest to largest by their assigned random number. The first 50 people on the sorted list are assigned to the new compound group, and the remaining 50 are assigned to the current compound group. To reduce bias, the compounds should be placed in identical, unmarked tubes so that neither participants nor researchers know which compound is being applied (double-blind). Each participant in each group then applies their assigned compound to one forearm and inserts it into a randomly assigned bin for 1 minute, after which the number of mosquito bites is counted.

Inference procedure:
Use a two-sample $t$-test (or construct a two-sample confidence interval for the difference in means) to compare the mean number of mosquito bites between the new-compound group and the current-compound group: \[ H_0: \mu_{\text{new}} = \mu_{\text{current}} \quad \text{vs.} \quad H_a: \mu_{\text{new}} < \mu_{\text{current}} \] where $\mu_{\text{new}}$ and $\mu_{\text{current}}$ are the mean number of bites under each compound.

(b) Matched-Pairs Design

Randomization process:
Each of the 100 participants serves as their own pair — both compounds are applied, one to each arm. For each participant, flip a coin (or use a random number generator) to decide which arm receives the new compound; the other arm receives the current compound. Each participant then inserts both arms simultaneously into a randomly assigned bin for 1 minute, and the number of bites on each arm is recorded. The compounds should again be placed in identical, unmarked tubes to maintain blinding.

Inference procedure:
Compute the difference in bites for each participant as $d_i = (\text{bites, new compound}) – (\text{bites, current compound})$. Use a one-sample $t$-test on the differences (or a paired confidence interval for the mean difference): \[ H_0: \mu_d = 0 \quad \text{vs.} \quad H_a: \mu_d < 0 \] where $\mu_d$ is the true mean difference in bites between the new and current compounds.

(c) Which design is better?

The matched-pairs design in part (b) is the better choice.

The key reason is that people naturally vary in how attractive they are to mosquitoes — some individuals get bitten far more than others regardless of which compound is used. In a completely randomized design, this person-to-person variability shows up as noise in the data, making it harder to detect a real difference between the two compounds. The matched-pairs design controls for this source of variability by having each person test both compounds, so individual differences in susceptibility cancel out when computing the within-person difference. This leads to a more precise and more powerful comparison of the two compounds.

Question

Regulations require that product labels on containers of food that are available for sale to the public accurately state the amount of food in those containers. Specifically, if milk containers are labeled to have 128 fluid ounces and the mean number of fluid ounces of milk in the containers is at least 128, the milk processor is considered to be in compliance with the regulations. The filling machines can be set to the labeled amount. Variability in the filling process causes the actual contents of milk containers to be normally distributed. A random sample of 12 containers of milk was drawn from the milk processing line in a plant, and the amount of milk in each container was recorded.

(a) The sample mean and standard deviation of this sample of 12 containers of milk were 127.2 ounces and 2.1 ounces, respectively. Is there sufficient evidence to conclude that the packaging plant is not in compliance with the regulations? Provide statistical justification for your answer.

Inspectors decide to study a particular filling machine within this plant further. For this machine, the amount of milk in the containers has a mean of 128.0 fluid ounces and a standard deviation of 2.0 fluid ounces.

(b) What is the probability that a randomly selected container filled by this machine contains at least 125 fluid ounces?

(c) An inspector will randomly select 12 containers filled by this machine and record the amount of milk in each. What is the probability that the minimum (smallest amount of milk) recorded in the 12 containers will be at least 125 fluid ounces? (Note: In order for the minimum to be at least 125 fluid ounces, each of the 12 containers must contain at least 125 fluid ounces.)

An analyst wants to use simulation to investigate the sampling distribution of the minimum. This analyst randomly generates 150 samples, each consisting of 12 observations, from a normal distribution with mean 128 and standard deviation 2 and finds the minimum for each sample. The 150 minimums (sorted from smallest to largest) are shown in the table below.

Sample	Minimum	Sample	Minimum	Sample	Minimum
1	121.45	51	124.28	101	125.25
2	122.51	52	124.29	102	125.31
3	122.53	53	124.30	103	125.36
4	122.72	54	124.31	104	125.38
5	122.75	55	124.34	105	125.40
6	122.89	56	124.36	106	125.42
7	122.93	57	124.37	107	125.48
8	122.99	58	124.37	108	125.49
9	123.04	59	124.39	109	125.50
10	123.08	60	124.39	110	125.52
11	123.09	61	124.41	111	125.54
12	123.10	62	124.44	112	125.56
13	123.31	63	124.53	113	125.61
14	123.34	64	124.53	114	125.67
15	123.39	65	124.54	115	125.72
16	123.40	66	124.55	116	125.76
17	123.41	67	124.55	117	125.77
18	123.41	68	124.55	118	125.78
19	123.46	69	124.55	119	125.79
20	123.49	70	124.58	120	125.84
21	123.51	71	124.67	121	125.87
22	123.57	72	124.69	122	125.87
23	123.58	73	124.73	123	125.90
24	123.59	74	124.77	124	125.90
25	123.60	75	124.78	125	125.93
26	123.66	76	124.78	126	125.93
27	123.67	77	124.80	127	125.93
28	123.72	78	124.80	128	125.94
29	123.75	79	124.81	129	125.98
30	123.77	80	124.85	130	126.00
31	123.78	81	124.91	131	126.03
32	123.84	82	124.92	132	126.05
33	123.91	83	124.92	133	126.05
34	123.93	84	124.96	134	126.06
35	123.95	85	125.00	135	126.09
36	123.95	86	125.01	136	126.15
37	123.98	87	125.02	137	126.15
38	123.99	88	125.02	138	126.16
39	124.05	89	125.03	139	126.19
40	124.05	90	125.04	140	126.19
41	124.06	91	125.05	141	126.25
42	124.12	92	125.07	142	126.26
43	124.14	93	125.08	143	126.33
44	124.15	94	125.09	144	126.35
45	124.16	95	125.14	145	126.45
46	124.19	96	125.18	146	126.50
47	124.23	97	125.21	147	126.57
48	124.27	98	125.21	148	126.62
49	124.28	99	125.22	149	126.64
50	124.28	100	125.25	150	126.95

(d) Use the simulation results to estimate the probability that was requested in part (c) and compare this estimate with the theoretical value you calculated.

Most-appropriate topic codes (AP Statistics):

• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part a)
• Topic 4.5 — Carrying Out a Test for a Population Mean or Population Mean Difference (Part a)
• Topic 2.11 — The Normal Distribution (Parts b, c)
• Topic 2.3 — Estimating Probabilities Using Simulation (Part d)

▶️ Answer/Explanation

(a)

Step 1 — State the hypotheses:
$ H_0: \mu = 128 \text{ fl oz} \quad \text{vs.} \quad H_a: \mu < 128 \text{ fl oz} $
where $\mu$ is the true mean amount of milk in containers from this plant.
We test whether the mean is below 128 fl oz, since that would indicate non-compliance.
Step 2 — Identify the procedure and check conditions:
Use a one-sample $t$-test for a mean:
$ t = \frac{\bar{x} – \mu_0}{s/\sqrt{n}} $
The problem states the filling process produces a normal distribution, so the normality condition is satisfied.
The containers were randomly sampled, so independence holds.
Step 3 — Compute the test statistic and p-value:
With $\bar{x} = 127.2$, $\mu_0 = 128$, $s = 2.1$, $n = 12$:
$ t = \frac{127.2 – 128}{2.1/\sqrt{12}} = \frac{-0.8}{0.6062} \approx -1.319 $
Degrees of freedom: $df = n – 1 = 11$.
For a one-tailed test with $t = -1.319$ and $df = 11$:
$ p\text{-value} = P(T_{11} < -1.319) \approx 0.107 $
Step 4 — State the conclusion in context:
Since the p-value of $0.107$ is greater than any reasonable significance level (e.g., $\alpha = 0.05$), we fail to reject $H_0$. There is not sufficient evidence to conclude that the plant is out of compliance. The sample mean of 127.2 fl oz is below 128, but the difference is small enough that it could plausibly be due to random sampling variability alone.

(b)

Let $X$ be the amount of milk in a randomly selected container, where $X \sim N(128.0,\ 2.0)$. We want $P(X \geq 125)$.
Standardize by converting to a $z$-score: $ z = \frac{125 – 128}{2} = \frac{-3}{2} = -1.5 $
Using the standard normal table:
$ P(X \geq 125) = P(Z \geq -1.5) = 1 – P(Z < -1.5) = 1 – 0.0668 $
$ \boxed{P(X \geq 125) = 0.9332} $
So about 93.3% of containers from this machine will contain at least 125 fl oz.

(c)

Let $X_{(1)} = \min(X_1, X_2, \ldots, X_{12})$ be the smallest value among 12 randomly selected containers.
For the minimum to be at least 125 fl oz, every single one of the 12 containers must contain at least 125 fl oz.
Since the containers are independent:
$ P(X_{(1)} \geq 125) = P(X_1 \geq 125) \times P(X_2 \geq 125) \times \cdots \times P(X_{12} \geq 125) $
$ = [P(X \geq 125)]^{12} = (0.9332)^{12} $
$ \boxed{P(X_{(1)} \geq 125) \approx 0.4362} $
There is roughly a 43.6% chance that the smallest of 12 containers all meet the 125 fl oz threshold.
Even though each individual container has a 93.3% chance of passing, the probability that all 12 pass simultaneously drops considerably.

(d)

From the sorted list of 150 simulated minimums, we count how many are at least 125 fl oz. Scanning the table, the minimums first reach 125.00 at sample 85. From sample 85 through sample 150, that gives:
$ 150 – 85 + 1 = 66 \text{ minimums that are} \geq 125 \text{ fl oz} $
The simulated probability estimate is therefore:
$ \hat{p} = \frac{66}{150} \approx \boxed{0.44} $
Comparison with the theoretical value:
The theoretical probability from part (c) was $0.4362$.
The simulation estimate of $0.44$ is very close — the difference is only:
$ |0.44 – 0.4362| = 0.0038 $
This is a tiny discrepancy, which is exactly what we expect from a simulation of this size.
The simulation does an excellent job of approximating the true theoretical probability, confirming that the independence-based calculation in part (c) is correct.

Question

A pharmaceutical company has developed a new drug to reduce cholesterol. A regulatory agency will recommend the new drug for use if there is convincing evidence that the mean reduction in cholesterol level after one month of use is more than 20 milligrams/deciliter (mg/dl), because a mean reduction of this magnitude would be greater than the mean reduction for the current most widely used drug.

The pharmaceutical company collected data by giving the new drug to a random sample of 50 people from the population of people with high cholesterol. The reduction in cholesterol level after one month of use was recorded for each individual in the sample, resulting in a sample mean reduction and standard deviation of 24 mg/dl and 15 mg/dl, respectively.

(a) The regulatory agency decides to use an interval estimate for the population mean reduction in cholesterol level for the new drug. Provide this 95 percent confidence interval. Be sure to interpret this interval.

(b) Because the 95 percent confidence interval includes 20, the regulatory agency is not convinced that the new drug is better than the current best-seller. The pharmaceutical company tested the following hypotheses.

$H_0: \mu = 20$ versus $H_a: \mu > 20$,

where $\mu$ represents the population mean reduction in cholesterol level for the new drug.

The test procedure resulted in a $t$-value of 1.89 and a $p$-value of 0.033. Because the $p$-value was less than 0.05, the company believes that there is convincing evidence that the mean reduction in cholesterol level for the new drug is more than 20. Explain why the confidence interval and the hypothesis test led to different conclusions.

(c) The company would like to determine a value $L$ that would allow them to make the following statement.

We are 95 percent confident that the true mean reduction in cholesterol level is greater than $L$.

A statement of this form is called a one-sided confidence interval. The value of $L$ can be found using the following formula.

\[L = \bar{x} – t^* \dfrac{s}{\sqrt{n}}\]

This has the same form as the lower endpoint of the confidence interval in part (a), but requires a different critical value, $t^*$. What value should be used for $t^*$?

Recall that the sample mean reduction in cholesterol level and standard deviation are 24 mg/dl and 15 mg/dl, respectively. Compute the value of $L$.

(d) If the regulatory agency had used the one-sided confidence interval in part (c) rather than the interval constructed in part (a), would it have reached a different conclusion? Explain.

Most-appropriate topic codes (AP Statistics):

• Topic 4.2 — Constructing a Confidence Interval for a Population Mean or Population Mean Difference (Part a)
• Topic 4.4 — Setting Up a Test for a Population Mean or Population Mean Difference (Part b)
• Topic 4.5 — Carrying Out a Test for a Population Mean or Population Mean Difference (Part b)
• Topic 4.2 — Constructing a Confidence Interval for a Population Mean or Population Mean Difference (Part c)
• Topic 4.3 — Justifying a Claim Based on a Confidence Interval for a Population Mean or Population Mean Difference (Parts c, d)

▶️ Answer/Explanation

(a)
The appropriate procedure is a one-sample $t$-interval for the population mean $\mu$.
Conditions:
— The data come from a random sample of 50 people.
— $\sigma$ is unknown; using the sample standard deviation $s = 15$.
— $n = 50 \geq 30$, so by the Central Limit Theorem the sampling distribution of $\bar{x}$ is approximately normal.
Given: $\bar{x} = 24$, $s = 15$, $n = 50$, and $df = 49$. For a 95% confidence interval, $t^* \approx 2.009$ (using $df = 49$).
The confidence interval formula is:
$\bar{x} \pm t^* \cdot \dfrac{s}{\sqrt{n}}$
$24 \pm 2.009 \cdot \dfrac{15}{\sqrt{50}}$
$24 \pm 2.009 \times 2.121$
$24 \pm 4.262$
$\boxed{(19.738,\ 28.262) \text{ mg/dl}}$
Interpretation: We are 95% confident that the true population mean reduction in cholesterol level after one month of use of the new drug is between approximately 19.7 mg/dl and 28.3 mg/dl.

(b)
The confidence interval and the hypothesis test led to different conclusions because they are based on different types of procedures that correspond to different questions being asked.
The 95% two-sided confidence interval is equivalent to a two-sided hypothesis test at $\alpha = 0.05$. The two-sided $p$-value for testing $H_0: \mu = 20$ against $H_a: \mu \neq 20$ would be $2 \times 0.033 = 0.066$, which exceeds $\alpha = 0.05$ — hence the confidence interval (which captures values consistent with a two-sided test) includes 20 and fails to reject $H_0$ at the 0.05 level.
The hypothesis test, however, is one-sided ($H_a: \mu > 20$) with a one-sided $p$-value of $0.033 < 0.05$, which leads to rejecting $H_0$. A one-sided test is more powerful in the direction specified and uses only one tail of the distribution. The two procedures are therefore testing different things, and it is the mismatch — using a two-sided interval to evaluate a one-sided hypothesis — that creates the apparent contradiction in conclusions.
$\boxed{\text{Two-sided CI} \leftrightarrow \text{two-sided test (}p = 0.066 > 0.05\text{)}; \quad \text{one-sided test: }p = 0.033 < 0.05}$

(c)
For a one-sided 95% confidence interval, we need to find $t^*$ such that 95% of the $t$-distribution with $df = 49$ lies above $-t^*$ (i.e., only one tail of area 0.05).
This corresponds to a tail probability of $p = 0.05$ (one tail) with $df = 49$. From the $t$-table:
$\boxed{t^* = 1.676 \quad (df = 49,\ \text{one tail}, \ \alpha = 0.05)}$
Now compute $L$:
$L = \bar{x} – t^* \cdot \dfrac{s}{\sqrt{n}} = 24 – 1.676 \cdot \dfrac{15}{\sqrt{50}}$
$= 24 – 1.676 \times 2.121$
$= 24 – 3.555$
$\boxed{L \approx 20.4 \text{ mg/dl}}$
Interpretation: We are 95% confident that the true mean reduction in cholesterol level after one month of use of the new drug is greater than approximately 20.4 mg/dl.

(d)
Yes, the regulatory agency would have reached a different conclusion using the one-sided confidence interval. The one-sided interval shows that the agency can be 95% confident that the true mean reduction is greater than $L \approx 20.4$ mg/dl, which is already above the threshold of 20 mg/dl required for recommendation. Since the entire range of plausible values for $\mu$ under the one-sided interval lies above 20, the agency would have had convincing evidence that the new drug reduces cholesterol by more than 20 mg/dl on average — and would therefore have recommended the drug for use.
$\boxed{L \approx 20.4 > 20 \Rightarrow \text{Yes, different conclusion: agency would recommend the drug}}$

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 4.4 Setting Up a Test for a Population Mean or Population Mean Difference- Exam Style Questions - FRQs - New Syllabus

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