AP Statistics 4.9 Setting Up a Test for the Difference Between Two Population Mean- Exam Style Questions - FRQs - New Syllabus

Question

Stefan, a psychologist, conducted a study to investigate the effect of time of day on reading comprehension in children. One hundred children volunteered, with their parents’ consent, to participate in the study. Fifty of the children were randomly assigned to read a story at 9 a.m. and then answer 25 questions about it. The remaining 50 children were assigned to read the same story at 3 p.m. and answer the same 25 questions. The reading comprehension for each child was measured by a reading score, which was determined by the number of questions that were answered correctly about the story. Stefan is interested in comparing the mean reading scores for the two times of day. Table 1 shows the results of Stefan’s study.

Table 1: Summary Statistics of Reading Scores

Stefan found the conditions for inference were met and conducted a two-sample t-test for the difference in two population means. Let $\mu_{\text{AM}}$ represent the mean reading score for all children, similar to those in the study, who would read the story at 9 a.m. Let $\mu_{\text{PM}}$ represent the mean reading score for all children, similar to those in the study, who would read the story at 3 p.m. Stefan’s hypotheses are as shown.

$H_0: \mu_{\text{AM}} = \mu_{\text{PM}}$
$H_a: \mu_{\text{AM}} \neq \mu_{\text{PM}}$

A. The p-value for Stefan’s hypothesis test was 0.002. State an appropriate conclusion, at the 5 percent significance level, for Stefan’s test in the context of the investigation. Justify your answer.

B. Explain why it was appropriate for Stefan to conduct a two-sample t-test for the difference in two population means instead of a paired t-test for the population mean difference.

C. Researchers are usually interested in the practical importance of their results as well as the statistical significance of the hypothesis test. The practical importance of the results indicates whether the observed results are meaningful in real life. For example, in an investigation of the heights of two groups of students, a difference in the two group means of 3.8 inches is much more meaningful, or has more practical importance, than a difference in the two group means of only 0.2 inches.

One indicator of practical importance is effect size. A common method for measuring effect size for the difference in two group means is Cohen’s d coefficient. Cohen’s d coefficient compares the absolute value of the difference in the means of the two groups to the pooled variability of the observed data values from the two groups. Cohen’s d coefficient can be calculated using

$d = \dfrac{|\bar{x}_1 – \bar{x}_2|}{s_p}$

where $s_p$ represents the pooled standard deviation, $\bar{x}_1$ represents the sample mean for the first group, and $\bar{x}_2$ represents the sample mean for the second group. When the sizes of the groups are equal, $s_p$ is calculated as

$s_p = \sqrt{\dfrac{s_1^2 + s_2^2}{2}}$

where $s_1$ represents the sample standard deviation for the first group and $s_2$ represents the sample standard deviation for the second group. Consider the summary statistics from Stefan’s study in Table 1.

i. Calculate Cohen’s d coefficient for Stefan’s study. Show your work.
ii. Higher values of Cohen’s d indicate greater practical importance and lower values of Cohen’s d indicate less practical importance. Typically, we use the intervals listed in Table 2 to help interpret practical importance.

Table 2: Guidelines for Interpreting Cohen’s d Coefficient

Based on your answer to part C (i) and the information in Tables 1 and 2, describe the practical importance of Stefan’s results, in context.

D. Suppose the results of Stefan’s study, summarized in Table 1, instead had a standard deviation for the 9 a.m. reading scores, $s_1$, and a standard deviation for the 3 p.m. reading scores, $s_2$, that were both greater than 4.43. Assume the group sample sizes and the means are not changed.

i. Would the Cohen’s d coefficient in this new situation be smaller than, larger than, or the same as the Cohen’s d coefficient calculated in part C (i)? Explain your answer.
ii. Does the Cohen’s d coefficient described in part D (i) indicate that Stefan’s observed difference in the means in the new situation would have more practical importance than, less practical importance than, or the same practical importance as what was originally determined in part C (ii)? Explain your answer.

Most-appropriate topic codes (AP Statistics):

• Topic $4.9$ — Setting Up a Test for the Difference Between Two Population Means (Part $ \mathrm{B} $)
• Topic $4.10$ — Carrying Out a Test for the Difference Between Two Population Means (Parts $ \mathrm{A} $, $ \mathrm{C} $, $ \mathrm{D} $)

▶️ Answer/Explanation

A.
• Because the reported $p$-value ($0.002$) is less than the significance level $\alpha = 0.05$, we reject the null hypothesis $H_0$.
• There is convincing statistical evidence to conclude that there is a difference between the mean reading scores for all children, similar to those in the study, who read the story at 9 a.m. versus those who read at 3 p.m.

B.
• A two-sample $t$-test is appropriate here because the two groups of children are completely independent of each other rather than paired.
• The 100 individual volunteers were randomly split into separate blocks of 50, meaning there is no logical connection or unique pairing linking a specific child’s performance in the morning slot directly to another child’s performance in the afternoon slot.

C. i.
First compute the pooled standard deviation $s_p$:
$s_p = \sqrt{\dfrac{(4.12)^2 + (4.43)^2}{2}} = \sqrt{\dfrac{16.9744 + 19.6249}{2}} = \sqrt{\dfrac{36.5993}{2}} = \sqrt{18.29965} \approx 4.2778$
Now calculate the effect size $d$:
$d = \dfrac{|15.2 – 17.9|}{4.2778} = \dfrac{2.7}{4.2778} \approx 0.6311$
$\boxed{d \approx 0.63}$

C. ii.
• Based on the calculated Cohen’s $d$ value of $0.63$, which falls within the interval $(0.20, 0.80)$, the effect is classified as “somewhat meaningful in real life.”
• This indicates that the shift in mean reading performance across different times of the day is large enough to show noticeable practical consequences for children’s learning schedules, though it is not a overwhelmingly massive real-world shift.

D. i.
• The Cohen’s $d$ coefficient in this new scenario would be smaller than the value calculated originally.
• Since the separate group sample variations are now both larger than $4.43$, the pooled denominator value $s_p$ will increase, which reduces the final value of the quotient since the raw difference in means remains fixed at $2.7$.

D. ii.
• This modified index indicates that Stefan’s observed score gap would have less practical importance than what was originally determined.
• When individual scores overlap more heavily due to wider personal variation within each testing window, a constant mean gap of $2.7$ points becomes harder to detect cleanly and feels less distinct in practical environments.

Question

High cholesterol levels in people can be reduced by exercise, diet, and medication. Twenty male patients at a local hospital who had high cholesterol levels volunteered to participate in a simulated study. Ten of the twenty patients were randomly assigned to receive drug A, and the remaining ten patients received a placebo. Data on cholesterol reduction, in milligrams per deciliter (mg/dL), were recorded for the twenty patients twice: once after a 4-week period and again after an 8-week period. The results are summarized in the table below.

Do the data provide convincing statistical evidence that, at the $\alpha = 0.01$ level, the mean cholesterol reduction for all male patients who take drug A for 8 weeks is greater than the mean cholesterol reduction for all male patients who take a placebo for 8 weeks?

Most-appropriate topic codes (AP Statistics):

• Topic 4.9 — Setting Up a Test for the Difference Between Two Population Means (Entire Question)
• Topic 4.10 — Carrying Out a Test for the Difference Between Two Population Means (Entire Question)

▶️ Answer/Explanation

Step 1: State Hypotheses and Define Parameters
Let $\mu_{A}$ represent the true mean cholesterol reduction if all such male patients at this hospital receive Drug A for 8 weeks.
Let $\mu_{B}$ represent the true mean cholesterol reduction if all such male patients at this hospital receive a placebo for 8 weeks.
The hypotheses to be tested are:
$H_{0}: \mu_{A} = \mu_{B}$ (or $\mu_{A} – \mu_{B} = 0$)
$H_{a}: \mu_{A} > \mu_{B}$ (or $\mu_{A} – \mu_{B} > 0$)

Step 2: Identify the Procedure and Check Conditions

The appropriate method is a two-sample $t$-test for the difference between two population means.
• Random Assignment: The problem states that the 20 volunteers were randomly assigned to the treatment groups (10 to Group A and 10 to Group B).
• Normality: Because the sample sizes are quite small ($n_A = 10$ and $n_B = 10$), we must assume that the populations of cholesterol reductions are approximately normally distributed or that the sample data show no extreme outliers or strong skewness.

Step 3: Compute Test Statistic and $p$-value
Demonstrates correct mechanics, including the value of the test statistic and $p$-value (or the rejection region).
The test statistic is: $t = \dfrac{\bar{X}_A – \bar{X}_B}{\sqrt{\dfrac{s_A^2}{n_A} + \dfrac{s_B^2}{n_B}}} = \dfrac{10.20 – 16.40}{\sqrt{\dfrac{7.66^2}{10} + \dfrac{9.40^2}{10}}} \approx -1.62$
With $\text{df} = 17.3$, $p$-value $\approx 0.062$.

Step 4: Make a Conclusion
Since the $p$-value ($0.033$) is greater than the specified significance level $\alpha = 0.01$, we fail to reject the null hypothesis $H_{0}$.
The observed difference in mean cholesterol reduction between the two groups is not statistically large enough to rule out random chance at this strict standard. Consequently, the study does not provide convincing clinical evidence that the mean cholesterol reduction is greater for those taking Drug A for 8 weeks compared to those taking a placebo.

Question

A large pet store buys the identical species of adult tropical fish from two different suppliers—Buy-Rite Pets and Fish Friends. Several of the managers at the pet store suspect that the lengths of the fish from Fish Friends are consistently greater than the lengths of the fish from Buy-Rite Pets. Random samples of 8 adult fish of the species from Buy-Rite Pets and 10 adult fish of the same species from Fish Friends were selected and the lengths of the fish, in inches, were recorded, as shown in the table below.

Do the data provide convincing evidence that the mean length of the adult fish of the species from Fish Friends is greater than the mean length of the adult fish of the same species from Buy-Rite Pets?

Most-appropriate topic codes (AP Statistics):

• Topic 4.9 — Setting Up a Test for the Difference Between Two Population Means (Hypotheses and conditions)
• Topic 4.10 — Carrying Out a Test for the Difference Between Two Population Means (Test statistic, p-value, and conclusion)

▶️ Answer/Explanation

Solution

Step 1: States a correct pair of hypotheses

Let $\mu_{\text{B}}$ represent the population mean length of all adult fish of this species from Buy-Rite Pets, and let $\mu_{\text{F}}$ represent the population mean length of all adult fish of this species from Fish Friends.

The hypotheses to be tested are $\text{H}_0: \mu_{\text{B}} = \mu_{\text{F}}$ versus $\text{H}_{\text{a}}: \mu_{\text{B}} < \mu_{\text{F}}$.

Step 2: Identifies a correct test procedure (by name or by formula) and checks appropriate conditions

The appropriate test is a two-sample $t$-test. The first condition is that the samples are independent random samples from the two populations. This was stated in the question. The second condition is that the population distributions of fish lengths are normal. The following dotplots reveal no obvious departures from normality, so it appears reasonable to proceed with the two-sample $t$-test.

Step 3: Demonstrates correct mechanics, including the value of the test statistic, df and $p$-value (or rejection region)

The test statistic is: $t = \dfrac{\overline{x}_{\text{B}} – \overline{x}_{\text{F}}}{\sqrt{\dfrac{s_{\text{B}}^2}{n_{\text{B}}} + \dfrac{s_{\text{F}}^2}{n_{\text{F}}}}} = \dfrac{3.40 – 3.46}{\sqrt{\dfrac{0.434^2}{8} + \dfrac{0.550^2}{10}}} \approx -0.259$

With $\text{df} = 15.99999$, $p$-value $= 0.3996$.

Step 4: States a correct conclusion in the context of the problem, using the result of the statistical test

Because this $p$-value is larger than any conventional significance level (such as $\alpha = 0.10$ or $\alpha = 0.05$), we fail to reject $\text{H}_0$. The sample data do not provide convincing evidence to conclude that the mean length of the adult fish of the species from Fish Friends is greater than the mean length of the adult fish of the same species from Buy-Rite Pets.

Question

A local arcade is hosting a tournament in which contestants play an arcade game with possible scores ranging from 0 to 20. The arcade has set up multiple game tables so that all contestants can play the game at the same time; thus contestant scores are independent. Each contestant’s score will be recorded as he or she finishes, and the contestant with the highest score is the winner.

After practicing the game many times, Josephine, one of the contestants, has established the probability distribution of her scores, shown in the table below.

Crystal, another contestant, has also practiced many times. The probability distribution for her scores is shown in the table below.

(a) Calculate the expected score for each player.

(b) Suppose that Josephine scores 16 and Crystal scores 17. The difference (Josephine minus Crystal) of their scores is $-1$. List all combinations of possible scores for Josephine and Crystal that will produce a difference (Josephine minus Crystal) of $-1$, and calculate the probability for each combination.

(c) Find the probability that the difference (Josephine minus Crystal) in their scores is $-1$.

(d) The table below lists all the possible differences in the scores between Josephine and Crystal and some associated probabilities.

Complete the table and calculate the probability that Crystal’s score will be higher than Josephine’s score.

Most-appropriate topic codes (AP Statistics):

• Topic $2.8$ — Introduction to Random Variables and Probability Distributions (Parts $\mathrm{a}$, $\mathrm{b}$, $\mathrm{c}$, $\mathrm{d}$)
• Topic $2.9$ — Parameters of Random Variables (Part $\mathrm{a}$)
• Topic $2.7$ — Independent Events and Unions of Events (Parts $\mathrm{b}$, $\mathrm{c}$, $\mathrm{d}$)

▶️ Answer/Explanation

(a)

The expected value (mean) of a discrete random variable is $\mu = \sum x_i \cdot P(x_i)$. We apply this to each player’s distribution.
For Josephine:
$\mu_J = 16(0.10) + 17(0.30) + 18(0.40) + 19(0.20)$
$\mu_J = 1.6 + 5.1 + 7.2 + 3.8 = \boxed{17.7}$
For Crystal:
$\mu_C = 17(0.45) + 18(0.40) + 19(0.15)$
$\mu_C = 7.65 + 7.20 + 2.85 = \boxed{17.7}$
Both players have the same expected score of $17.7$.

(b)

We need all pairs $(J, C)$ such that $J – C = -1$, i.e., Josephine’s score is exactly 1 less than Crystal’s. Since scores are independent, the probability of each pair is the product of the individual probabilities.
$J = 16,\ C = 17$: $\quad P = (0.10)(0.45) = 0.045$
$J = 17,\ C = 18$: $\quad P = (0.30)(0.40) = 0.120$
$J = 18,\ C = 19$: $\quad P = (0.40)(0.15) = 0.060$
These are the only three combinations that produce a difference of $-1$.

(c)

Since the three combinations in part (b) are mutually exclusive, we simply add their probabilities:
$P(J – C = -1) = 0.045 + 0.120 + 0.060$
$\boxed{P(J – C = -1) = 0.225}$

(d)

First, we find the missing probability for a difference of $-2$. Since all probabilities in the distribution must sum to 1:
$P(J – C = -2) = 1 – 0.015 – 0.225 – 0.325 – 0.260 – 0.090$
$P(J – C = -2) = 1 – 0.915 = \boxed{0.085}$
The completed distribution table is:

Crystal’s score is higher than Josephine’s when the difference $J – C < 0$, i.e., when the difference is $-3$, $-2$, or $-1$.
$P(\text{Crystal} > \text{Josephine}) = P(J – C < 0) = 0.015 + 0.085 + 0.225$
$\boxed{P(\text{Crystal} > \text{Josephine}) = 0.325}$

Question

A serum cholesterol level above $250$ milligrams per deciliter (mg/dl) of blood is a risk factor for cardiovascular disease in humans. At a medical center in St. Louis, a study to test the effectiveness of a new cholesterol-lowering drug was conducted. One hundred people with cholesterol levels between $250$ mg/dl and $300$ mg/dl were available for this study. Fifty people were assigned at random to each of two treatment groups. One group received the standard cholesterol-lowering medication and the other group received the new drug. After taking the drug for three weeks, the $50$ subjects who received the standard treatment had a mean decrease in cholesterol level of $10$ mg/dl with a standard deviation of $8$ mg/dl, and the $50$ subjects who received the new drug had a mean decrease of $18$ mg/dl with a standard deviation of $12$ mg/dl.

Does the new drug appear to be more effective than the standard treatment in lowering mean cholesterol level? Give appropriate statistical evidence to support your conclusion.

Most-appropriate topic codes (AP Statistics):

• Topic 4.9 — Setting Up a Test for the Difference Between Two Population Means
• Topic 4.10 — Carrying Out a Test for the Difference Between Two Population Means

▶️ Answer/Explanation

Step 1: State the Hypotheses
Let $\mu_S$ = true mean decrease in cholesterol level for patients taking the standard drug
Let $\mu_N$ = true mean decrease in cholesterol level for patients taking the new drug
$H_0:\, \mu_S – \mu_N = 0$ (the two drugs are equally effective at lowering cholesterol
$H_a:\, \mu_S – \mu_N < 0$ (the new drug produces a greater mean decrease in cholesterol)

Step 2: Name the Test and Verify Conditions
Procedure: Two-sample $t$-test for the difference in population means.
• Random: Subjects were randomly assigned to the two treatment groups.
– Large Samples: Both sample sizes are large ($n_S = n_N = 50 \geq 30$), so by the Central Limit Theorem the sampling distribution of $\bar{x}_S – \bar{x}_N$ is approximately normal.
– Independence: The two treatment groups are independent of each other (separate random assignments, subjects cannot be in both groups).

Step 3: Calculate the Test Statistic

Given values: $\bar{x}_S = 10,\; s_S = 8,\; n_S = 50$ and $\bar{x}_N = 18,\; s_N = 12,\; n_N = 50$
$t = \dfrac{(\bar{x}_S – \bar{x}_N) – 0}{\sqrt{\dfrac{s_S^2}{n_S} + \dfrac{s_N^2}{n_N}}} = \dfrac{10 – 18}{\sqrt{\dfrac{8^2}{50} + \dfrac{12^2}{50}}}$
$= \dfrac{-8}{\sqrt{\dfrac{64}{50} + \dfrac{144}{50}}} = \dfrac{-8}{\sqrt{\dfrac{208}{50}}} = \dfrac{-8}{\sqrt{4.16}} \approx \dfrac{-8}{2.040} \approx -3.92$
Degrees of freedom (conservative): $df = \min(n_S – 1,\; n_N – 1) = \min(49,\; 49) = 49$
From the $t$-table at $df = 49$: since $|t| = 3.92$ exceeds the critical value $t^* = 3.496$ at tail probability $0.0005$,
$p\text{-value} < 0.001$

Step 4: State the Conclusion in Context
Since $p\text{-value} < 0.001 < \alpha = 0.05$, we reject $H_0$.
$\text{There is convincing statistical evidence that the new drug produces a greater mean decrease in cholesterol level than the standard treatment.}$
This question is a classic two-sample $t$-test, and the data make a strong case. The new drug group averaged a $18$ mg/dl drop in cholesterol versus only $10$ mg/dl for the standard treatment — an $8$ mg/dl difference that turns out to be far too large to chalk up to the luck of random assignment. With $t \approx -3.92$ and a $p$-value well below $0.001$, there’s less than a 1-in-1000 chance of seeing a gap this big if the drugs were truly equally effective. Crucially, because subjects were randomly assigned to treatment groups, we’re not just observing an association — we have solid grounds to conclude that the new drug is actually causing the greater cholesterol reduction.

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

There have been many studies recently concerning coffee drinking and cholesterol level. While it is known that several coffee-bean components can elevate blood cholesterol level, it is thought that a new type of paper coffee filter may reduce the presence of some of these components in coffee.

The effect of the new filter on cholesterol level will be studied over a 10-week period using 300 nonsmokers who each drink 4 cups of caffeinated coffee per day. Each of these 300 participants will be assigned to one of two groups: the experimental group, who will only drink coffee that has been made with the new filter, or the control group, who will only drink coffee that has been made with the standard filter. Each participant’s cholesterol level will be measured at the beginning and at the end of the study.

(a) Describe an appropriate method for assigning the subjects to the two groups so that each group will have an equal number of subjects.

(b) In this study, the researchers chose to include a group who only drank coffee that was made with the standard filter. Why is it important to include a control group in this study even though cholesterol levels will be measured at the beginning and at the end of the study?

(c) Which test would you conduct to determine whether the change in cholesterol level would be greater if people used the new filter rather than using the standard filter?

(d) Why would the researchers choose to use only nonsmokers in the study?

Most-appropriate topic codes (AP Statistics):

• Topic 1.13 — Experimental Design (Parts a, b, d)
• Topic 4.10 — Carrying Out a Test for the Difference Between Two Population Means (Part c)
• Topic 4.9 — Setting Up a Test for the Difference Between Two Population Means (Part c)

▶️ Answer/Explanation

(a)

Assign each of the 300 subjects a unique number from 001 to 300.
Use a random number table or a random number generator to randomly select 150 of these numbers — the subjects corresponding to those numbers will be placed in the experimental group (new filter), and the remaining 150 subjects will be assigned to the control group (standard filter).
This ensures that each subject has an equal chance of being assigned to either group, and that both groups end up with exactly 150 subjects.

(b)

Without a control group, if cholesterol levels change over the 10-week period, there is no way to know whether the change was caused by the new filter or by some other factor that changed during that time — for example, seasonal changes in diet or physical activity.
Simply measuring cholesterol at the beginning and end for one group only tells us that a change occurred, but not why it occurred.
By including a control group using the standard filter, researchers can compare the mean change in cholesterol between the two groups, allowing them to attribute any difference specifically to the new filter rather than to an outside confounding variable.
In short, the control group is what makes it possible to isolate the effect of the new filter from other changes that might naturally occur over the 10-week period.

(c)

The appropriate test is the two-sample $t$-test for means (or equivalently, the two-sample $t$-test for mean differences).
This test compares the mean change in cholesterol level for the new-filter group against the mean change in cholesterol level for the standard-filter group, using two independent samples of size 150 each.

(d)

Smoking is known to be related to cholesterol level, so if the study included both smokers and nonsmokers, smoking status could act as an extraneous source of variability — making it harder to detect the true effect of the coffee filter.
By restricting the study to nonsmokers only, the researchers create more homogeneous groups, reducing variability within each group and allowing for a more precise and sensitive comparison of the filter’s effect on cholesterol.
This is similar in spirit to blocking — controlling for a known nuisance variable (smoking) so that it does not obscure the treatment effect, though it does mean the results can only be generalized to nonsmokers.

AP Statistics 4.9 Setting Up a Test for the Difference Between Two Population Mean- Exam Style Questions - FRQs - New Syllabus

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