STAT 350 — Exam 2 — Spring 2026

Exam Information

Course: STAT 350 — Introduction to Statistics

Semester: Spring 2026

Total Points: 105

Time Allowed: 60 minutes

Problem	Total Possible	Topic
Problem 1 (True/False, 2 pts each)	12	Experimental Design, Power, Confidence Intervals, CI–HT Duality
Problem 2 (Multiple Choice, 3 pts each)	15	CLT, Confidence Bounds, Experimental Design, Sampling Distributions, Power
Problem 3	26	CLT, Sampling Distribution of \(\bar{X}\), Conditional Probability
Problem 4	26	Confidence Bound, t- vs. z-procedures, One-Sample z-test
Problem 5	26	Two-Sample Paired t-test
Total	105

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Problem 1 — True/False (12 points, 2 points each)

Question 1.1 (2 pts)

An automotive engineer is studying how tire compound (soft, medium, hard) and suspension stiffness (low, high) affect braking distance. Because vehicle weight is known to influence braking performance, the engineer groups 30 test vehicles into five blocks based on weight class (subcompact, compact, midsize, full-size, SUV). Within each weight class, the 6 vehicles are randomly assigned so that each of the six treatment combinations is applied to exactly one vehicle. The braking distance (in meters) from 100 km/h is recorded.

True or False: In this randomized block design, the blocks are defined by all combinations of tire compound, suspension stiffness, and vehicle weight class, resulting in 30 total blocks.

Solution

Answer: FALSE

The blocks are defined only by the weight class variable, giving 5 blocks (one per weight class). The tire compound and suspension stiffness are the treatment factors, not blocking variables. Within each weight class block, the six treatment combinations (3 tire compounds × 2 suspension stiffnesses) are randomly assigned to the six vehicles. The 30 “cells” are experimental units, not blocks.

Question 1.2 (2 pts)

In the context of hypothesis testing and confidence intervals, let \(C\) denote the confidence level of an interval, \(\alpha\) denote the complementary significance level (so that \(C + \alpha = 1\)), and \(\beta\) denote the probability of a Type II error.

True or False: Since \(C + \alpha = 1\) and power \(+ \beta = 1\), it follows that \(C = \text{power}\).

Solution

Answer: FALSE

Although \(C = 1 - \alpha\) and power \(= 1 - \beta\), that does not imply \(C = \text{power}\) because \(\alpha \neq \beta\) in general.

• \(\alpha\) is the probability of rejecting \(H_0\) when \(H_0\) is true (Type I error rate).

• \(\beta\) is the probability of failing to reject \(H_0\) when \(H_a\) is true (Type II error rate).

These are probabilities computed under different distributions and are generally unequal, so \(1 - \alpha \neq 1 - \beta\).

Question 1.3 (2 pts)

An aerospace engineer constructs a confidence interval for the mean drag coefficient \(\mu\) of a new wing design from a random sample of fixed size \(n\).

True or False: The confidence level \(C\) is one of the factors that determines the width of the confidence interval.

Solution

Answer: TRUE

The width of a confidence interval is \(2 \times t^*_{\alpha/2,\,\text{df}} \cdot s / \sqrt{n}\) (or with \(z^*\) for known \(\sigma\)). The confidence level \(C = 1 - \alpha\) determines the critical value \(t^*_{\alpha/2}\) (or \(z^*_{\alpha/2}\)): a higher confidence level produces a larger critical value and therefore a wider interval. Along with the sample size \(n\) and the variability \(s\) (or \(\sigma\)), the confidence level is indeed one of the factors that determines the width.

Question 1.4 (2 pts)

A quality engineer is planning a hypothesis test to detect whether the mean diameter of manufactured bolts has shifted from the specification value \(\mu_0 = 10.00\) mm. She is evaluating the statistical power of the test at a specific alternative value \(\mu_a\) that belongs to the alternative hypothesis.

True or False: The statistical power of a hypothesis test may decrease when the sample size \(n\) is increased.

Solution

Answer: FALSE

Increasing \(n\) decreases the standard error \(\sigma / \sqrt{n}\), which tightens the sampling distribution of \(\bar{X}\) under the alternative \(\mu_a\). With a tighter distribution centered at \(\mu_a\), more of the distribution falls in the rejection region, so power increases (or, at minimum, does not decrease). There is no mechanism by which a larger sample size reduces statistical power for a fixed \(\alpha\), fixed \(\sigma\), and fixed \(\mu_a\).

Question 1.5 (2 pts)

A mechanical engineer tests the fatigue life of \(n = 40\) aluminum alloy specimens and constructs a 95% confidence interval for the true mean fatigue life \(\mu\) (in cycles). The computed interval is \((22.5,\; 22.7)\).

True or False: It is incorrect to say that \(22.5 \leq \mu \leq 22.7\) with 0.95 probability, since the inequality does not involve any random variables.

Solution

Answer: TRUE

Once the interval has been computed from the observed data, the endpoints 22.5 and 22.7 are fixed numbers, and \(\mu\) is also a fixed (unknown) constant. There is no randomness left — \(\mu\) is either inside the interval or it is not. The “95% confidence” refers to the long-run procedure: if we repeated the sampling and interval construction many times, 95% of the resulting intervals would capture \(\mu\). Assigning 0.95 probability to this particular realized interval is a common misinterpretation.

Question 1.6 (2 pts)

A researcher wishes to test \(H_0\!: \mu \leq \mu_0\) versus \(H_a\!: \mu > \mu_0\) at significance level \(\alpha = 0.01\). Before conducting the one-sided test, she constructs a 99% two-sided confidence interval for \(\mu\) from the same data and observes that \(\mu_0\) falls inside the interval.

True or False: The researcher can conclude that the one-sided test at \(\alpha = 0.01\) will fail to reject \(H_0\).

Solution

Answer: FALSE

A 99% two-sided CI corresponds to a two-sided test at \(\alpha = 0.01\). If \(\mu_0\) falls inside the two-sided CI, the two-sided test fails to reject — but that does not determine the outcome of the one-sided test.

The one-sided upper test at \(\alpha = 0.01\) uses a critical value \(z_{0.01}\) (or \(t_{0.01}\)), which is less extreme than the two-sided critical value \(z_{0.005}\) (or \(t_{0.005}\)). It is therefore possible for the sample mean to exceed \(\mu_0\) by enough to reject the one-sided test while still falling within the wider two-sided interval. The correct duality is: a one-sided test at \(\alpha\) corresponds to a one-sided confidence bound at \(C = 1 - \alpha\), not to a two-sided CI.

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Problem 2 — Multiple Choice (15 points, 3 points each)

Question 2.1 (3 pts)

A random sample of size \(n\) is drawn from a population with mean \(\mu\) and finite standard deviation \(\sigma\). The population distribution is heavily right-skewed. Which of the following statements is FALSE?

(A) Regardless of sample size, \(E[\bar{X}] = \mu\).

(B) For sufficiently large \(n\), the sampling distribution of \(\bar{X}\) is approximately normal.

(C) As the sample size \(n\) becomes sufficiently large, the observations \(X_1, X_2, \ldots, X_n\) become approximately normal.

(D) The standard deviation of \(\bar{X}\) decreases as \(n\) increases, regardless of whether \(\bar{X}\) is approximately normal.

(E) A larger sample size \(n\) is needed for the CLT to provide a good approximation here than would be needed if the population were only mildly skewed.

Solution

Answer: (C)

The Central Limit Theorem applies to the sampling distribution of \(\bar{X}\), not to the individual observations. Regardless of how large the sample size is, each observation \(X_i\) is still drawn from the same heavily right-skewed population — increasing \(n\) does not change the distribution of the individual \(X_i\)’s.

The other statements are all true:

• 1. \(E[\bar{X}] = \mu\) always holds by the linearity of expectation.

• 2. The CLT guarantees approximate normality of \(\bar{X}\) for large \(n\).

• 4. \(\text{SD}(\bar{X}) = \sigma / \sqrt{n}\) decreases as \(n\) increases.

• (E) More extreme skewness requires a larger \(n\) for the CLT approximation to be adequate.

Question 2.2 (3 pts)

A Purdue scouting intern for the Indianapolis Colts is evaluating 40-yard dash times from \(n = 9\) prospective NFL running backs at the 2026 combine. Historical records indicate that 40-yard dash times for running backs are normally distributed with a known population standard deviation of \(\sigma = 0.0853\) seconds. The intern’s sample yields \(\bar{x} = 4.4639\) seconds. To establish the slowest acceptable mean sprint time for recruitment, the intern constructs a one-sided upper confidence bound at \(\alpha = 0.03\). Which R expression gives the correct critical value?

(A) qnorm(0.03, lower.tail = FALSE)

(B) qt(0.03, df = 8, lower.tail = FALSE)

(C) qnorm(0.03/2, lower.tail = FALSE)

(D) qt(0.03/2, df = 8, lower.tail = FALSE)

(E) None of the above

Solution

Answer: (A)

Two key features determine the correct R expression:

1. z or t? The population standard deviation \(\sigma = 0.0853\) is known, so we use the normal (z) distribution, not the t-distribution. This eliminates (B) and (D).

2. Full \(\alpha\) or \(\alpha/2\)? A one-sided confidence bound places the entire \(\alpha\) in one tail, so we need \(z_\alpha = z_{0.03}\), not \(z_{\alpha/2}\). This eliminates (C).

Therefore, qnorm(0.03, lower.tail = FALSE) is correct.

Question 2.3 (3 pts)

A biomedical engineer wants to compare three physical therapy protocols for post-surgical knee recovery. She recruits 60 patients and knows from prior studies that age strongly affects recovery outcomes. She divides patients into four age groups (18–30, 31–45, 46–60, 61+), and within each age group, randomly assigns an equal number of patients to each of the three protocols. Recovery is measured by range of motion (in degrees) at 8 weeks. Which of the following statements is FALSE?

(A) The experimental design is a Randomized Block Design, with age group as the blocking variable.

(B) There are three treatments and four blocks in this experiment.

(C) The purpose of blocking by age is to reduce the variability arising from age differences, making it easier to detect treatment effects.

(D) The blocking ensures that the randomization of patients to protocols is no longer necessary within each age group.

(E) If age had no effect on recovery, a Completely Randomized Design would have been equally effective.

Solution

Answer: (D)

Blocking does not replace randomization — it supplements it. Within each age block, patients are still randomly assigned to the three protocols. Randomization within blocks is essential to ensure unbiased treatment comparisons and to guard against lurking variables other than the blocking factor.

The other statements are all correct:

• 1. This is indeed an RBD with age as the blocking variable.

• 2. There are 3 treatments (the three protocols) and 4 blocks (the four age groups).

• (C) Blocking reduces the influence of the blocked variable (age) on unexplained variability, increasing the ability to detect treatment effects.

• (E) If the blocking variable has no effect on the response, blocking adds no benefit, and a CRD would perform equally well (and may even be slightly better due to the loss of degrees of freedom from blocking).

Question 2.4 (3 pts)

The table below summarizes the properties of two independent populations A and B.

	Population A	Population B
Distribution family	Normal	Normal
Mean	\(\mu_A = \mu_B = \mu\)	\(\mu_A = \mu_B = \mu\)
Standard Deviation	\(\sigma_A = 4.5\)	\(\sigma_B = 3.9\)
Sample size	\(n_A = 44\)	\(n_B = 52\)

Which of the following statements about the sampling distributions of \(\bar{X}_A\) and \(\bar{X}_B\) is FALSE?

(A) \(P(\bar{X}_A = 4.5) = P(\bar{X}_B = 3.9)\)

(B) The pdf of \(\bar{X}_A\) and \(\bar{X}_B\) have the same value when evaluated at \(\mu\). That is, \(f_{\bar{X}_A}(\mu) = f_{\bar{X}_B}(\mu)\).

(C) \(P(\bar{X}_A \leq \mu - 1) > P(\bar{X}_B \leq \mu - 1)\)

(D) \(P\!\left(\bar{X}_A > \mu + \frac{4.5}{\sqrt{44}}\right) = P\!\left(\bar{X}_B > \mu + \frac{3.9}{\sqrt{52}}\right)\)

Solution

Answer: (B)

Both \(\bar{X}_A\) and \(\bar{X}_B\) are continuous random variables (normal), so their distributions have the following properties:

• (A) TRUE. For any continuous random variable, \(P(\bar{X} = c) = 0\) for every constant \(c\). So both sides equal 0.

• (B) FALSE. The height of the normal pdf at the mean is \(f(\mu) = \dfrac{1}{\sigma_{\bar{X}}\sqrt{2\pi}}\), which depends on the standard error. We have \(\sigma_{\bar{X}_A} = 4.5/\sqrt{44} \approx 0.6784\) and \(\sigma_{\bar{X}_B} = 3.9/\sqrt{52} \approx 0.5408\). Since the standard errors are unequal, the pdf heights at \(\mu\) are unequal: \(f_{\bar{X}_A}(\mu) \approx 0.5881 \neq 0.7376 \approx f_{\bar{X}_B}(\mu)\).

• (C) TRUE. \(\bar{X}_A\) has a larger standard error, so its distribution is more spread out and places more probability in the tails. Therefore \(P(\bar{X}_A \leq \mu - 1) > P(\bar{X}_B \leq \mu - 1)\).

• (D) TRUE. Each probability standardizes to \(P(Z > 1)\) because \(\mu + \sigma_{\bar{X}} / \sigma_{\bar{X}} = \mu + 1\) standard error. Specifically, \(P\!\left(\bar{X}_A > \mu + \sigma_A/\sqrt{n_A}\right) = P(Z > 1)\) and \(P\!\left(\bar{X}_B > \mu + \sigma_B/\sqrt{n_B}\right) = P(Z > 1)\).

Question 2.5 (3 pts)

An industrial engineer tests whether a process improvement has increased the mean production rate above the current standard of \(\mu_0 = 200\) units/hour. The population standard deviation is known to be \(\sigma = 18\) units/hour. She collects a sample of \(n = 36\) observations and conducts an upper-tailed \(z\)-test at \(\alpha = 0.05\). The engineer wants to know the probability that this test will correctly detect an increase if the true mean has shifted to \(\mu_a = 206\) units/hour.

The following R outputs are provided:

> qnorm(0.05, lower.tail = FALSE)
[1] 1.644854
> qnorm(0.025, lower.tail = FALSE)
[1] 1.959964

(A) pnorm(204.9346, mean = 206, sd = 3, lower.tail = FALSE)

(B) pnorm(204.9346, mean = 206, sd = 18, lower.tail = FALSE)

(C) pnorm(205.8799, mean = 206, sd = 3, lower.tail = FALSE)

(D) pnorm(204.9346, mean = 200, sd = 3, lower.tail = FALSE)

(E) pnorm(204.9346, mean = 206, sd = 3, lower.tail = TRUE)

Solution

Answer: (A)

Step 1 — Find the rejection boundary. This is a one-sided upper-tailed test at \(\alpha = 0.05\), so the critical value is \(z_{0.05} = 1.644854\) (not \(z_{0.025}\)). The rejection boundary in the original scale is:

\[\bar{x}_\text{crit} = \mu_0 + z_\alpha \cdot \frac{\sigma}{\sqrt{n}} = 200 + 1.644854 \times \frac{18}{\sqrt{36}} = 200 + 1.644854 \times 3 = 204.9346\]

Step 2 — Compute the power. Power is the probability of rejecting \(H_0\) when \(\mu = \mu_a = 206\):

\[\text{Power} = P(\bar{X} > 204.9346 \mid \mu = 206)\]

Under \(\mu_a = 206\), \(\bar{X} \sim N(206,\; 3^2)\) where \(\sigma/\sqrt{n} = 18/\sqrt{36} = 3\). In R:

pnorm(204.9346, mean = 206, sd = 3, lower.tail = FALSE)

The other options are incorrect because:

• (B) uses \(\sigma = 18\) instead of the standard error \(\sigma/\sqrt{n} = 3\).

• 3. uses the wrong cutoff (based on \(z_{0.025}\) for a two-sided test).

• 4. centers at \(\mu_0 = 200\) instead of the alternative \(\mu_a = 206\).

• (E) computes the lower tail, which gives \(\beta\) (the probability of failing to reject), not the power.

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Problem 3 (26 points) — AirBuds Battery Life

Problem 3 Setup

PineApple is assessing the battery life consistency of their next-generation AirBuds. It is known that the battery life of a new left AirBud follows a fairly symmetric distribution with a true mean operation time of \(\mu_L = 540\) minutes and a standard deviation of \(\sigma_L = 81\) minutes.

The company will randomly select a batch of \(n_L = 53\) left AirBuds from their manufacturing lines and record their battery lifetimes while playing the same audio on repeat. Let \(\bar{X}_L\) denote the random variable representing the average battery life of such a randomly selected batch.

Question 3a (5 pts)

State the approximate distribution of \(\bar{X}_L\). Include the name of its distribution family and its parameters (mean and standard error) in both symbolic and numerical forms.

Solution

\[\bar{X}_L \;\stackrel{\text{approx}}{\sim}\; \text{Normal}\!\left(\mu_L = 540,\;\; \sigma_{\bar{X}_L} = \frac{81}{\sqrt{53}} \approx 11.1262\right)\]

• Mean: \(\mu_{\bar{X}_L} = \mu_L = 540\) minutes.

• Standard error: \(\sigma_{\bar{X}_L} = \sigma_L / \sqrt{n_L} = 81 / \sqrt{53} \approx 11.1262\) minutes.

Question 3b (3 pts)

What important theorem justifies your approximation in Part (a)?

Solution

The Central Limit Theorem (CLT). The original population is described as fairly symmetric and the sample size \(n_L = 53\) is reasonably large. By the CLT, the sampling distribution of \(\bar{X}_L\) is approximately normal regardless of the shape of the population distribution, provided \(n\) is sufficiently large and the population has finite mean and variance.

R Output for Parts (c) and (d)

The following R calculations are provided. Default behavior is lower.tail = TRUE.

> pnorm(-2.2469)
[1] 0.0123
> pnorm(-2.1424)
[1] 0.0161
> pnorm(-1.8363)
[1] 0.0332
> pnorm(-1.7976)
[1] 0.0361
> pnorm(0.6121)
[1] 0.7298
> pnorm(0.8988)
[1] 0.8157

Question 3c (6 pts)

What is the probability that the sample mean lifetime falls between 520 and 550 minutes?

Solution

Standardize both endpoints using \(\sigma_{\bar{X}_L} = 81/\sqrt{53}\):

\[P(520 < \bar{X}_L < 550) = P\!\left(\frac{520 - 540}{81/\sqrt{53}} < Z < \frac{550 - 540}{81/\sqrt{53}}\right)\]

\[= P(-1.7976 < Z < 0.8988)\]

\[= \Phi(0.8988) - \Phi(-1.7976) = 0.8157 - 0.0361 = \boxed{0.7796}\]

Question 3d (12 pts)

Suppose the testing software is programmed to automatically discard any batch of 53 AirBuds if their sample mean is lower than 515 minutes, thus a batch is successfully retained and recorded if their sample mean is at least 515 minutes. Given that a batch is successfully recorded, what is the probability that its sample mean falls between 520 and 550 minutes?

Solution

This is a conditional probability problem. We need:

\[P(520 < \bar{X}_L < 550 \mid \bar{X}_L \geq 515) = \frac{P(520 < \bar{X}_L < 550)}{P(\bar{X}_L \geq 515)}\]

The numerator was computed in part (c). Note that the event \(520 < \bar{X}_L < 550\) is a subset of \(\bar{X}_L \geq 515\), so the intersection simplifies to just the numerator event.

Numerator: \(P(520 < \bar{X}_L < 550) = 0.7796\) (from part c).

Denominator: Standardize 515:

\[P(\bar{X}_L \geq 515) = P\!\left(Z \geq \frac{515 - 540}{81/\sqrt{53}}\right) = P(Z \geq -2.2469) = 1 - \Phi(-2.2469) = 1 - 0.0123 = 0.9877\]

Conditional probability:

\[\frac{0.7796}{0.9877} = \boxed{0.7893}\]

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Problem 4 (26 points) — eBay Broken Laser Pointer

Problem 4 Setup

In an effort to consolidate the conceptual understanding of interval estimation (confidence intervals and bounds) in STAT 350, an interactive trivia-based classroom activity was implemented during the Fall semester. Students were asked to estimate the sale price (in US dollars) of the first item ever sold on eBay, which was a broken laser pointer. A random sample of \(n = 16\) student point estimates yielded a sample mean of \(\bar{x} = 13.22\) and a sample standard deviation of \(s = 4.3\). Typically, this type of guessing data tends to be mildly skewed due to psychological anchoring in human decision-making.

Question 4a (2 pts)

Does the information provided above justify the construction of standard confidence intervals or bounds? Give one statistical justification for your answer.

Solution

Multiple valid responses were accepted:

Yes, justified: The sample is described as a random sample, satisfying the independence assumption. Although the data are mildly skewed and \(n = 16\) is a small-to-moderate sample size, the \(t\)-procedure is robust to mild departures from normality. With only mild skewness, the sampling distribution of \(\bar{X}\) should be approximately normal even at this moderate sample size, so constructing a \(t\)-based confidence interval or bound is reasonable.

No (independence concern): The estimates were collected during a classroom activity, so students may have influenced each other’s guesses through verbal or nonverbal cues. Even though the problem states a random sample, the shared environment may violate the independence assumption required for \(t\)-procedures.

No (normality concern): Independence is satisfied by the random sampling, but the population is stated to be mildly skewed and \(n = 16\) may not be large enough for the CLT to guarantee an approximately normal sampling distribution.

Question 4b Setup

Suppose we assume the conditions are met to conduct statistical inference. The actual selling price of the broken laser pointer was $14.83. Suspecting that students tend to underestimate, the instructor would like to establish, with 90% confidence, a confidence upper bound for the true mean of all student guesses across STAT 350 during the fall semester.

Question 4b.i (3 pts)

Which R output below provides the correct critical value for this computation?

(A) qnorm(0.05, lower.tail = FALSE) → 1.644854

(B) qt(0.1, df = 15, lower.tail = FALSE) → 1.340606

(C) qnorm(0.1, lower.tail = FALSE) → 1.281552

(D) qt(0.05, df = 15, lower.tail = FALSE) → 1.75305

Solution

Answer: (B)

Two key features determine the correct critical value:

1. z or t? The population standard deviation \(\sigma\) is unknown (only \(s = 4.3\) is available), so we must use the \(t\)-distribution with \(\text{df} = n - 1 = 15\). This eliminates (A) and (C).

2. Full \(\alpha\) or \(\alpha/2\)? A one-sided confidence bound places the entire \(\alpha = 0.10\) in one tail, so we need \(t_{0.10,\,15}\), not \(t_{0.05,\,15}\). This eliminates (D).

Therefore, qt(0.1, df = 15, lower.tail = FALSE) = 1.340606 is correct.

Question 4b.ii (5 pts)

Compute and interpret the 90% confidence upper bound in the context of the problem.

Solution

The 90% confidence upper bound has the form:

\[\left(-\infty,\;\; \bar{x} + t_{\alpha,\,\text{df}} \cdot \frac{s}{\sqrt{n}}\right)\]

Substituting:

\[\left(-\infty,\;\; 13.22 + 1.340606 \times \frac{4.3}{\sqrt{16}}\right) = \left(-\infty,\;\; 13.22 + 1.340606 \times 1.075\right) = \left(-\infty,\;\; 14.6612\right)\]

Interpretation: We are 90% confident that the true mean \(\mu\) of all student guesses regarding the value at which the broken laser pointer was sold across STAT 350 during the fall semester is at most $14.6612.

Question 4c (2 pts)

In general, \(t\)-intervals/bounds are [______] their corresponding \(z\)-intervals/bounds (assuming the same confidence level, standard error, and sample size). Select the correct option.

(A) narrower than

(B) wider than

(C) the same width as

Solution

Answer: (B) wider than

The \(t\)-distribution has heavier tails than the standard normal, so for any given confidence level, the \(t\) critical value \(t^*_{\alpha/2,\,\text{df}}\) exceeds the corresponding \(z\) critical value \(z^*_{\alpha/2}\). This produces a wider interval or bound, reflecting the additional uncertainty from estimating \(\sigma\) with \(s\).

Question 4d Setup

Recently, a digital marketing analyst claimed that the true mean price for vintage RadioShack laser pointers is $225. The prices of these vintage pointers are known to be normally distributed with a population variance of \(\sigma^2 = 64\). A Purdue engineering historian believes the true mean price is different. He collects a random sample of \(n = 12\) prices from e-commerce platforms, yielding a sample mean of \(\bar{x} = 210\) and a sample standard deviation of \(s = 15\). Use a significance level \(\alpha = 0.02\) for all inference calculations below.

Question 4d.i (4 pts)

Provide the first two steps of the four-step hypothesis testing procedure.

Solution

Step 1 — Parameter of interest:

Let \(\mu\) represent the true mean price (in US dollars) for vintage RadioShack laser pointers.

Step 2 — Hypotheses:

\[H_0: \mu = 225\]

\[H_a: \mu \neq 225\]

In words: the null hypothesis states that the true mean price is $225 (as claimed by the analyst), and the alternative states that the true mean price differs from $225.

Question 4d.ii (3 pts)

Compute the appropriate test statistic.

Solution

Since the population variance \(\sigma^2 = 64\) is known (\(\sigma = 8\)), we use the \(z\)-test statistic — even though \(s = 15\) is provided, we use \(\sigma\) because it is given:

\[z_\text{TS} = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{210 - 225}{8 / \sqrt{12}} = \frac{-15}{2.3094} = \boxed{-6.4952}\]

Question 4d.iii (3 pts)

Which of the following R outputs is appropriate for computing the \(p\)-value for this test?

(A) 2*pt(abs(t_ts), df=11, lower.tail=FALSE) → 0.005294732

(B) 2*pnorm(abs(z_ts), lower.tail=FALSE) → 8.292839e-11

(C) pnorm(abs(z_ts), lower.tail=FALSE) → 4.14642e-11

(D) pt(abs(t_ts), df=11, lower.tail=FALSE) → 0.002647366

Solution

Answer: (B)

Since \(\sigma\) is known, the test statistic is a \(z\)-statistic, so we need pnorm, not pt. This eliminates (A) and (D). Because the test is two-sided (\(H_a: \mu \neq 225\)), we double the one-tail probability. This eliminates (C), which computes only one tail.

Therefore: 2*pnorm(abs(z_ts), lower.tail=FALSE) = \(8.2928 \times 10^{-11}\)

Question 4d.iv (4 pts)

State your decision and provide a conclusion in the context of the problem.

Solution

Decision: Since the \(p\)-value \(= 8.2928 \times 10^{-11}\) is less than \(\alpha = 0.02\), we reject \(H_0\).

Conclusion: The data give strong support (\(p\)-value \(= 8.2928 \times 10^{-11}\)) to the claim that the true mean price for vintage RadioShack laser pointers differs from the $225 claimed by the analyst.

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Problem 5 (26 points) — Soybean Yield: Soil Amendments

Problem 5 Setup

Researchers at a Purdue agricultural extension are evaluating two soil amendments, the standard treatment (S) and a new organic treatment (N), to determine whether the new treatment increases soybean yield (measured in bushels per acre). They selected 18 farm plots in Tippecanoe County. Each plot is split in half: one half receives Amendment S and the other half receives Amendment N. After the growing season, soybean yield is recorded for each half-plot.

The researchers calculate the difference for each plot: \(D = N - S\). They have verified that the distribution of these differences is approximately normal.

	Amendment N	Amendment S	\(D = N - S\)
\(n\)	18	18	18
Sample Mean	54.8	51.2	3.6
Sample Std Dev	7.3	6.9	5.4

Question 5a (2 pts)

Which testing procedure is appropriate for this experiment?

(A) Two-sample independent \(t\)-test

(B) Two-sample paired \(t\)-test

Solution

Answer: (B) Two-sample paired \(t\)-test

Question 5b (6 pts)

Explain what specific characteristic(s) in the experimental design motivated your choice of testing procedure in part (a).

Solution

Each of the 18 farm plots is split in half, with one half receiving Amendment N and the other half receiving Amendment S. The two measurements from each plot share the same underlying conditions — identical soil composition, moisture level, sunlight exposure, and other plot-specific environmental factors — creating a natural pairing.

Because the two observations within each plot are not independent of each other (they share a common plot), a paired \(t\)-test is appropriate. This design reduces the effect of plot-to-plot variability by analyzing the within-plot differences \(D = N - S\) rather than comparing the two groups as if they were independent.

Question 5c (4 pts)

Provide the first two steps of the four-step hypothesis testing procedure. Use a significance level of \(\alpha = 0.03\).

Solution

Step 1 — Parameter of interest:

Let \(\mu_D\) represent the true mean difference in soybean yield (in bushels per acre) between land treated with the new organic treatment (N) and the standard treatment (S), where \(D = N - S\).

Step 2 — Hypotheses:

Since the research question asks whether the new treatment increases yield, this is a one-sided (upper-tailed) test:

\[H_0: \mu_D \leq 0 \qquad (\text{or equivalently } H_0: \mu_D = 0)\]

\[H_a: \mu_D > 0\]

In words: the null states that the new organic treatment does not increase soybean yield compared to the standard; the alternative states that it does increase yield.

Grading Note

A two-sided formulation (\(H_0: \mu_D = 0\) vs. \(H_a: \mu_D \neq 0\)) was also accepted, with corresponding adjustments to the p-value computation in part (e).

Question 5d (6 pts)

Calculate the test statistic for this experiment. Show your work.

Solution

Using the paired differences \(D = N - S\) with \(\bar{d} = 3.6\), \(s_D = 5.4\), and \(n_D = 18\):

\[t_\text{TS} = \frac{\bar{d} - 0}{s_D / \sqrt{n_D}} = \frac{3.6}{5.4 / \sqrt{18}} = \frac{3.6}{1.2728} = \boxed{2.8284}\]

Degrees of freedom: \(\text{df} = n_D - 1 = 17\).

Question 5e (3 pts)

Select the most appropriate R command to compute the \(p\)-value for this specific test.

(A) pt(test_statistic, df = 17, lower.tail = TRUE)

(B) 2*pt(abs(test_statistic), df = 17, lower.tail = FALSE)

(C) pt(test_statistic, df = 17, lower.tail = FALSE)

(D) pt(test_statistic, df = 34, lower.tail = FALSE)

(E) pt(test_statistic, df = 30.27, lower.tail = FALSE)

(F) pnorm(test_statistic, lower.tail = FALSE)

Solution

Answer: (C) for the one-sided test, or (B) for the two-sided formulation.

The choice must be consistent with the hypotheses stated in part (c):

• If \(H_a: \mu_D > 0\) (one-sided upper): use pt(test_statistic, df = 17, lower.tail = FALSE) → (C).

• If \(H_a: \mu_D \neq 0\) (two-sided): use 2*pt(abs(test_statistic), df = 17, lower.tail = FALSE) → (B).

Key points for eliminating other options:

• 1. computes the lower tail — wrong direction for an upper-tailed test.

• (D) uses \(\text{df} = 34\), which would apply to an independent two-sample pooled test (\(n_1 + n_2 - 2\)), not a paired test.

• (E) uses \(\text{df} = 30.27\), which suggests a Welch (unpooled) test — not appropriate for paired data.

• (F) uses pnorm (normal distribution), but \(\sigma\) is unknown, so a \(t\)-distribution with \(\text{df} = 17\) is required.

Question 5f (5 pts)

The \(p\)-value for the correct test was found to be 0.0058. Using a significance level of \(\alpha = 0.03\), state your formal decision and write a conclusion in the context of the problem.

Solution

Decision: Since the \(p\)-value = 0.0058 is less than \(\alpha = 0.03\), we reject \(H_0\).

Conclusion (one-sided formulation): The data give strong support (\(p\)-value = 0.0058) to the claim that the true mean difference in soybean yield between land treated with the new organic treatment (N) and the standard treatment (S) is positive, indicating that the new organic treatment has a beneficial effect on crop yield.

Conclusion (two-sided formulation): The data give strong support (\(p\)-value = 0.0058) to the claim that the true mean difference in soybean yield between the new organic treatment and the standard treatment differs from zero.

Grading Note

The conclusion must be consistent with the hypothesis formulation chosen in part (c). If a one-sided alternative was stated, the conclusion should reference the direction of the effect; if two-sided, it should reference a difference from zero.