STAT 350 — Exam 2 — Spring 2025

Exam Information

Course: STAT 350 — Introduction to Statistics

Semester: Spring 2025

Total Points: 105

Time Allowed: 60 minutes

Problem	Total Possible	Topic
Problem 1 (True/False, 2 pts each)	12	CLT, Experimental Design, Confidence Intervals, Power
Problem 2 (Multiple Choice, 3 pts each)	15	Experimental Design, CLT, Sampling Distributions, Power
Problem 3	24	One-Sample t-test, Confidence Bound
Problem 4	23	Paired t-test
Problem 5	31	Two-Sample Independent t-test, Confidence Interval
Total	105

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

Question 1a (2 pts)

Consider a sequence \(X_1, X_2, \ldots, X_n\) of independent and identically distributed random variables drawn from a population \(f_X(x)\) with finite mean \(\mu\), and finite variance \(\sigma^2 > 0\). Define the sample sum as:

\[S_n = \sum_{i=1}^{n} X_i,\]

where the subscript \(n\) explicitly indicates that the sum is over \(n\) random variables.

True or False: According to the central limit theorem (CLT), the standardized sample sum

\[\frac{(S_n - n\mu)}{\sqrt{n} \cdot \sigma}\]

is exactly distributed as a standard Normal distribution regardless of the shape of \(f_X(x)\), provided \(n \geq 30\).

Solution

Answer: FALSE

The CLT states that the standardized sample sum converges in distribution to a standard Normal as \(n \to \infty\) — it is an approximation that improves with larger \(n\), not an exact result for any finite \(n\). For the distribution to be exactly normal for finite \(n\), the population \(f_X(x)\) would itself need to be normal. The condition \(n \geq 30\) is a common practical guideline, but it does not make the CLT exact.

Question 1.2 (2 pts)

In a randomized block design (RBD), treatments are randomly assigned to experimental units within distinct blocks.

True or False: This is done to balance rather than mitigate or remove the impact of extraneous variables on experimental results.

Solution

Answer: FALSE

The primary purpose of blocking in a randomized block design is to mitigate or remove (not merely balance) the impact of extraneous variables. By grouping experimental units that are similar on a known source of variability into the same block, blocking reduces the unexplained variation in the response — effectively removing the block effect from the error. This is a stronger action than simply balancing: it actively controls the nuisance variable.

Question 1.3 (2 pts)

Denote \(\tau_n = \dfrac{1}{\sqrt{n}} \, SD(X_1 + X_2 + \cdots + X_n)\), where the \(X_i\)’s are independent and identically distributed with finite variance \(\sigma^2\).

True or False: Then it follows that \(\tau_3 > \tau_4\).

Solution

Answer: FALSE

Since the \(X_i\)’s are i.i.d. with variance \(\sigma^2\):

\[SD(X_1 + X_2 + \cdots + X_n) = \sqrt{n}\,\sigma\]

Therefore:

\[\tau_n = \frac{1}{\sqrt{n}} \cdot \sqrt{n}\,\sigma = \sigma\]

for all \(n\). So \(\tau_3 = \tau_4 = \sigma\) — they are equal, not \(\tau_3 > \tau_4\).

Question 1.4 (2 pts)

A researcher is calculating the sample size needed for a confidence interval with a fixed margin of error. The population standard deviation is known.

True or False: The required sample size increases as the confidence coefficient decreases.

Solution

Answer: FALSE

The sample size formula for a CI with known \(\sigma\) is \(n = \left(\dfrac{z_{\alpha/2}\,\sigma}{m}\right)^2\), where \(m\) is the desired margin of error. As the confidence coefficient decreases (e.g., from 99% to 90%), the critical value \(z_{\alpha/2}\) decreases, so the required sample size decreases, not increases. A lower confidence coefficient requires a smaller sample to achieve the same margin of error.

Question 1.5 (2 pts)

The functions pnorm() and pt() are available functions in R (RStudio),

True or False: they directly provide critical values for constructing confidence intervals and bounds.

Solution

Answer: FALSE

pnorm() and pt() are cumulative distribution functions (CDFs) — they compute probabilities \(P(Z \leq z)\) and \(P(T \leq t)\), respectively. They do not directly provide critical values. To obtain critical values for constructing confidence intervals, one uses the quantile functions qnorm() and qt() instead.

Question 1.6 (2 pts)

When conducting a hypothesis test in which the alternative hypothesis is true,

True or False: the probability of rejecting the null hypothesis increases by taking a larger sample size.

Solution

Answer: TRUE

When \(H_a\) is true, the probability of rejecting \(H_0\) is the power of the test (\(1 - \beta\)). Increasing the sample size \(n\) reduces the standard error \(\sigma/\sqrt{n}\), which tightens the sampling distribution under \(H_a\) relative to the rejection region, directly increasing the power of the test.

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

Question 2.1 (3 pts)

Which of the following techniques is primarily used to control or reduce variability arising from extraneous factors in experimental design?

1. Blocking     (B) Replication     (C) Randomization     (D) Realism     (E) All of the above

Solution

Answer: (A)

Blocking is the technique specifically designed to control or reduce variability from known extraneous factors by grouping homogeneous experimental units together. Replication increases precision but does not directly control extraneous variability. Randomization distributes unknown extraneous effects evenly but does not reduce their magnitude. “Realism” is not a standard experimental design principle. Blocking is the answer that best matches “control or reduce variability from extraneous factors.”

Question 2.2 (3 pts)

Time, resources, and practicality influence statistical studies. Which of the following is essential for drawing generalizable conclusions from a statistical investigation?

1. Selecting a representative sample

2. Having a sufficiently large sample size

3. Using appropriate well-established statistical methods

4. Clearly defining the population of interest

5. All of the above are essential for drawing generalizable conclusions

Solution

Answer: (E)

All four elements are essential. Without a representative sample (A), results cannot be extended to the population. Without adequate sample size (B), estimates are too imprecise and tests too underpowered to support generalizable conclusions. Without appropriate methods (C), the analysis may be invalid. Without a clearly defined population (D), it is unclear to whom the conclusions apply. Each element is necessary; none is sufficient alone.

Question 2.3 (3 pts)

You used Glassgate, a platform where salaries are posted anonymously, to look up salary information for a small-sized consulting firm. You found salary data from 13 verified junior data scientists.

Which of the following statements is NOT always true regarding the sampling distribution of the average salary computed from these 13 junior data scientists?

(A) If the population distribution has a strong positive skew, the Central Limit Theorem cannot be applied.

(B) Without additional information about the population distribution, the exact sampling distribution cannot be determined.

(C) The mean of the sampling distribution, \(\mu_{\bar{X}}\), is equal to the population mean \(\mu_X\), regardless of sample size.

(D) The standard deviation of the sampling distribution \(\sigma_{\bar{X}}\), is always smaller than the population standard deviation \(\sigma_X\) when the sample size exceeds 2.

(E) The central limit theorem ensures that the sampling distribution of \(\bar{X}\) is approximately Normal.

Solution

Answer: (E)

With only \(n = 13\) observations from a potentially skewed salary distribution, the CLT does not guarantee approximate normality. The CLT requires \(n\) to be sufficiently large, and for heavily skewed populations, \(n = 13\) is generally not sufficient. Therefore statement (E) is not always true.

• (A) NOT always true — but in this context with \(n = 13\), CLT indeed cannot be safely applied for a strongly skewed population, making (A) true here.

• (B) TRUE — the exact sampling distribution depends on the unknown population shape, which is not specified.

• (C) TRUE — \(E[\bar{X}] = \mu_X\) always holds for any sample size and any population with finite mean.

• (D) TRUE — \(\sigma_{\bar{X}} = \sigma_X/\sqrt{n} < \sigma_X\) for \(n > 1\).

• (E) NOT always true — CLT requires sufficiently large \(n\); \(n = 13\) is not sufficient for a strongly skewed population.

Question 2.4 (3 pts)

A delivery company, CargoSwift Logistics, operates small vans that regularly transport packages between warehouses. Each trip includes a fixed set of loading equipment (metal securing racks, crates, and straps) weighing exactly 30 lbs. The remaining cargo consists of sixteen individual packages, each with weights independently and identically distributed as follows:

\[X_i \sim \text{Uniform}(a = 44,\ b = 56), \quad i = 1, 2, \ldots, 16.\]

Recall that \(E[X_i] = \dfrac{a+b}{2}\) and \(\text{Var}(X_i) = \dfrac{(b-a)^2}{12}\).

The total weight \(T\) for a typical truckload of 16 packages is given by:

\[T = 30 + \sum_{i=1}^{16} X_i.\]

CargoSwift’s delivery vans have a maximum safe weight load capacity of 850 lbs. Select the correct code to calculate the approximate probability that a randomly selected van containing 16 packages would exceed the safe weight load.

1. pt(1.4434, df = 15, lower.tail = FALSE)

2. pnorm(1.4434, lower.tail = FALSE)

3. pt(3.6084, df = 15, lower.tail = FALSE)

4. pnorm(3.6084, lower.tail = FALSE)

5. pt(5.7735, df = 15, lower.tail = FALSE)

6. pnorm(5.7735, lower.tail = FALSE)

7. punif(850, min = 704, max = 896, lower.tail = FALSE)

Solution

Answer: (B)

Step 1 — Find the distribution of \(T\):

\[E[X_i] = \frac{44 + 56}{2} = 50, \qquad \text{Var}(X_i) = \frac{(56-44)^2}{12} = \frac{144}{12} = 12\]

Since the fixed weight is a constant:

\[E[T] = 30 + 16 \times 50 = 830 \text{ lbs}\]

\[\text{Var}(T) = 16 \times 12 = 192, \qquad SD(T) = \sqrt{192} = 13.8564 \text{ lbs}\]

Step 2 — Apply the CLT:

With \(n = 16\) independent Uniform random variables, the CLT gives \(T \approx \text{Normal}(830,\ 13.8564)\). Standardize:

\[z = \frac{850 - 830}{13.8564} = \frac{20}{13.8564} = 1.4434\]

Step 3 — Select the code:

We need \(P(T > 850) \approx P(Z > 1.4434)\). Since the CLT gives a normal approximation (not a t-distribution), and \(\sigma\) is known from the Uniform distribution, the correct code is:

pnorm(1.4434, lower.tail = FALSE)

Question 2.5 (3 pts)

A nutritionist conducts a study to test whether a new dietary program significantly reduces cholesterol levels from a known baseline of 200 mg/dL. They collect data from 25 participants and observe a sample standard deviation of 18 mg/dL. They plan to perform a lower-tailed hypothesis test at significance level \(\alpha = 0.05\).

Select the correct lines of R code to calculate the approximate power of this test for detecting a reduction in the mean cholesterol to 190 mg/dL:

# (A)
cutoff <- 200 + qt(0.05, df=24, lower.tail=FALSE)*(18/sqrt(25))
pt((cutoff - 190)/(18/sqrt(25)), df=24, lower.tail=TRUE)

# (B)
cutoff <- 190 - qt(0.95, df=24, lower.tail=FALSE)*(18/sqrt(25))
pt((cutoff - 200)/(18/sqrt(25)), df=24, lower.tail=TRUE)

# (C)
cutoff <- 200 - qnorm(0.95, lower.tail=FALSE)*(18/sqrt(25))
pnorm((cutoff - 190)/(18/sqrt(25)), lower.tail=FALSE)

# (D)
cutoff <- 200 + qnorm(0.05, lower.tail=FALSE)*(18/sqrt(25))
pnorm((cutoff - 190)/(18/sqrt(25)), lower.tail=FALSE)

# (E)
cutoff <- 200 - qt(0.05, df=24, lower.tail=FALSE)*(18/sqrt(25))
pt((cutoff - 190)/(18/sqrt(25)), df=24, lower.tail=TRUE)

Solution

Answer: (E)

For a lower-tailed t-test (\(H_a: \mu < 200\)), \(\sigma\) is unknown, so we use the t-distribution with \(df = n - 1 = 24\).

The cutoff is the value \(\bar{x}_\text{cutoff}\) below which we reject \(H_0\):

\[\bar{x}_\text{cutoff} = \mu_0 - t_{0.05,\,24} \cdot \frac{s}{\sqrt{n}} = 200 - t_{0.05,\,24} \cdot \frac{18}{\sqrt{25}}\]

In R: cutoff <- 200 - qt(0.05, df=24, lower.tail=FALSE) * (18/sqrt(25))

Note: qt(0.05, df=24, lower.tail=FALSE) returns the positive value \(t_{0.05,24} = 1.7109\), so subtracting gives a cutoff below 200.

Power is \(P(\bar{X} < \bar{x}_\text{cutoff} \mid \mu = 190)\):

\[\text{Power} = P\!\left(T_{24} < \frac{\bar{x}_\text{cutoff} - 190}{18/\sqrt{25}}\right)\]

In R: pt((cutoff - 190)/(18/sqrt(25)), df=24, lower.tail=TRUE)

• (A) Wrong: adds instead of subtracts, giving a cutoff above 200.

• (B) Wrong: centers cutoff on 190 instead of 200.

• (C), (D) Wrong: use pnorm instead of pt; \(\sigma\) is unknown.

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Problem 3 (24 points) — SUV Fuel Efficiency

Problem 3 Setup

A car manufacturer advertises that its new compact SUV averages 40 miles per gallon (mpg). A consumer advocacy group wants to evaluate this claim and believes the true average may be lower. They obtain 54 SUVs from the same model year and measure their fuel efficiency under combined city/highway driving, simulating what an average driver might encounter. After recording each vehicle’s mileage under these controlled yet representative conditions, the group finds a sample mean of 37.8 mpg. The population’s standard deviation is unknown. They plan to conduct a hypothesis test at a 3% significance level to assess the manufacturer’s claim.

Question 3a (3 pts)

Identify the parameter of interest. Clearly state its symbolic notation and define it briefly in the context of this scenario.

Solution

Let \(\mu_\text{MPG}\) denote the true average miles per gallon of the new compact SUV.

Question 3b (4 pts)

Write the null and alternative hypotheses clearly in symbolic notation.

Solution

\[H_0: \mu_\text{MPG} \geq 40\]

\[H_a: \mu_\text{MPG} < 40\]

Question 3c (8 pts)

The consumer advocacy group reports the corresponding confidence bound at 38.43. Using this confidence bound and the provided critical values, deduce the value of the sample standard deviation. Provide your answer rounded to four decimal places.

qnorm(0.03, lower.tail=FALSE)      # = 1.88
qt(0.03,   df=53, lower.tail=FALSE) # = 1.92
qnorm(0.015, lower.tail=FALSE)     # = 2.17
qt(0.015,  df=53, lower.tail=FALSE) # = 2.23

Solution

Since \(\sigma\) is unknown and this is a one-sided test at \(\alpha = 0.03\) with \(df = n - 1 = 53\), the 97% upper confidence bound takes the form:

\[\text{UCB} = \bar{x} + t_{0.03,\,53} \cdot \frac{s}{\sqrt{n}}\]

The critical value is qt(0.03, df=53, lower.tail=FALSE) \(= 1.92\).

Substituting the known values:

\[38.43 = 37.8 + 1.92 \cdot \frac{s}{\sqrt{54}}\]

Solving for \(s\):

\[s = \frac{(38.43 - 37.8) \cdot \sqrt{54}}{1.92} = \frac{0.63 \times 7.3485}{1.92} = \boxed{2.4112}\]

Question 3d (2 pts)

Is the reported confidence bound of 38.43 in part (c) an upper bound or a lower bound? Please mark the correct option.

1. Upper Bound     (B) Lower Bound

Solution

Answer: (A) Upper Bound

The hypothesis test is left-tailed (\(H_a: \mu_\text{MPG} < 40\)), and duality between hypothesis tests and confidence regions requires a one-sided upper confidence bound — providing the largest plausible value for \(\mu_\text{MPG}\). The bound of 38.43 mpg lies above \(\bar{x} = 37.8\) mpg, consistent with an upper bound.

Question 3e (7 pts)

Using the results of parts (c) and (d), obtain the result of the hypothesis test and write the formal contextual conclusion.

Solution

The 97% upper confidence bound of 38.43 mpg is below the null value \(\mu_0 = 40\). By the duality between confidence bounds and hypothesis tests, this means the null value falls outside the 97% upper confidence bound, so we reject \(H_0\) at \(\alpha = 0.03\).

The data give support (\(p\text{-value} < 0.03\)) to the claim that the true average MPG of the new compact SUVs is less than 40 MPG.

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Problem 4 (23 points) — Special Agent Gibbs: Veterans Mental Stability

Problem 4 Setup

Special Agent Gibbs decided to pursue his career in academia specializing in national security, post-traumatic stress, and investigation strategies. As part of his research, Gibbs requested access to a sensitive dataset containing information on veterans. Due to privacy and security considerations, the custodians of the dataset could not release it directly to Gibbs. Instead, they provided Gibbs with detailed descriptions of the available variables and asked him to submit clearly defined research questions. Their analyst team would then run analyses on the secure data and provide Gibbs with appropriate statistical summaries, test statistics, and supporting details. Assume none of the population standard deviations are known.

Question 4a (3 pts)

Five of Gibbs’ research questions happen to be lower-tailed hypotheses, \(H_a: \mu < \mu_0\). Which one of the following test statistics would be most likely to reject \(H_0\)? Assume the same degrees of freedom for all five test statistics.

1. \(t_{ts} = -2.25\)     (B) \(t_{ts} = -1.02\)     (C) \(t_{ts} = 0.02\)     (D) \(t_{ts} = 1.56\)     (E) \(t_{ts} = 3.02\)

Solution

Answer: (A)

For a lower-tailed test, we reject \(H_0\) when the test statistic is sufficiently negative (far into the left tail). The \(p\)-value equals \(P(T \leq t_{ts})\), which is smallest when \(t_{ts}\) is most negative. Among the five options, \(t_{ts} = -2.25\) is the most negative, giving the smallest \(p\)-value and the greatest evidence to reject \(H_0\).

Question 4b (3 pts)

Gibbs believes that veterans’ financial status may vary by different factors, such as location, household composition, and health status, so he asked the analyst team if they can control these extraneous variables. Without direct access to the dataset himself, which of the following strategies is most appropriate for the analyst team to handle these extraneous variables based on Gibbs’ request?

1. Discard any cases which are deemed extreme by Gibbs to obtain a consistent dataset.

(B) Partition the data into blocks (or strata) aligned with these extraneous variables, thereby reducing their confounding influence during analysis.

(C) Randomly select a small number of veterans broadly representative of the entire population, simplifying the analysis.

(D) Ignore extraneous variables since a small bias is acceptable if it results in reduced variance.

(E) Control all extraneous variables from the beginning by requiring veterans to live according to randomly assigned conditions.

Solution

Answer: (B)

Since the analyst team is working with an existing dataset (not a new experiment), experimental interventions like random assignment (E) are not possible. The most appropriate observational approach is to partition (block/stratify) the data by the known extraneous variables — location, household composition, and health status — which controls their confounding influence during analysis. Discarding extreme cases (A) would introduce bias; a small subsample (C) would reduce precision; ignoring confounders (D) would bias results.

Question 4c (3 pts)

Gibbs identified two variables: social isolation level (SIL, categorical) and mental stability score (MSS, numerical). SIL has four categories: socially active, somewhat socially active, somewhat isolated, and completely isolated. Specifically, Gibbs wants to see if the mean difference in MSS between somewhat isolated and completely isolated groups is greater than 20 points.

The analysts, acting on Gibbs’ request, paired 50 veterans from the somewhat isolated group with 50 veterans from the completely isolated group. Pairing was done based on demographics and veteran history to control possible extraneous factors. After matching, the difference in mental stability scores was computed for each pair as \(D = \text{Somewhat Isolated} - \text{Completely Isolated}\).

Which of the following hypothesis testing procedures is appropriate to answer Gibbs’ question?

1. One-sample \(t\)-test

2. Two-sample Independent \(t\)-test

3. Two-sample Matched Pairs \(t\)-test

4. None of the above

Solution

Answer: (C) Two-sample Matched Pairs :math:`t`**-test**

The analysts explicitly paired veterans across the two groups based on demographics and veteran history, creating dependent pairs. With the difference \(D\) computed for each pair, the appropriate procedure is a matched pairs (paired) t-test. Note that while a paired t-test is mathematically equivalent to a one-sample t-test on the differences, the context here is explicitly a two-sample matched pairs design.

Question 4d (4 pts)

State clearly the null and alternative hypotheses using the appropriate mathematical notations.

Solution

With \(D = \text{Somewhat Isolated} - \text{Completely Isolated}\):

\[H_0: \mu_D \leq 20\]

\[H_a: \mu_D > 20\]

Question 4e (3 pts)

Gibbs wants to find the \(p\)-value of a test statistic, \(t_{ts} = 2.14\). Which of the following R code statements returns the correct \(p\)-value?

# (A) pt(2.14, df = 50, lower.tail = FALSE)
# (B) pt(2.14, df = 50, lower.tail = TRUE)
# (C) pt(2.14, df = 49, lower.tail = FALSE)
# (D) pt(2.14, df = 49, lower.tail = TRUE)
# (E) pt(2.14, df = 99, lower.tail = FALSE)
# (F) pt(2.14, df = 99, lower.tail = TRUE)

Solution

Answer: (C) pt(2.14, df = 49, lower.tail = FALSE)

This is a right-tailed test (\(H_a: \mu_D > 20\)), so the \(p\)-value is the upper-tail probability, requiring lower.tail = FALSE. The degrees of freedom for a paired t-test on \(n = 50\) pairs is \(df = n - 1 = 49\).

• 1. uses df = 50 — off by one.

• 4. uses lower.tail = TRUE — wrong tail direction for a right-tailed test.

• 5. uses df = 99 — that would apply to an unpaired two-sample test.

Question 4f (7 pts)

The calculated \(p\)-value is 0.01868. At a significance level of \(\alpha = 0.02\), state your formal decision and conclusion in the context of the problem.

Solution

Decision: The \(p\)-value \(= 0.01868 \leq 0.02 = \alpha\), therefore we have sufficient evidence to reject \(H_0\).

Conclusion: The data give some support (\(p\)-value \(= 0.01868\)) to the claim that the true average mental stability score (MSS) is higher for veterans classified as Somewhat Isolated versus Completely Isolated — specifically, that the mean difference exceeds 20 points.

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Problem 5 (31 points) — Delivery Courier Mileage: California vs. Non-California

Problem 5 Setup

The rapid growth of food delivery services has dramatically increased the number of two-wheeled couriers on city streets. While this surge has benefited local businesses and consumers, it has also led to increased traffic congestion and safety concerns, prompting some cities to consider stricter delivery regulations (MassLive, 2025).

California cities, in particular, are often at the forefront of adopting innovative transportation policies and stricter environmental and safety regulations. To better understand the potential impact of these regulations and inform future policy decisions, a food delivery analytics firm seeks to compare the total monthly mileage traveled by two-wheeled delivery couriers employed by a major third-party platform. Mileage is measured as the combined total distance traveled by all couriers within each city over a four-week period in late summer 2024. Using the summary statistics below, the firm aims to determine if there is a significant difference in the average total monthly mileage between California and non-California cities.

Statistic	California cities	Non-California cities
\(n\)	11	11
\(\bar{x}\)	752,962	824,387.6
\(s\)	697,033.6	918,850.2

Question 5a (4 pts)

State the assumptions necessary for the hypothesis test on the average total monthly mileage between California and non-California cities.

Solution

As this is a two-independent sample procedure, we have the following assumptions:

Independence:

• The cities in each group (California and non-California) constitute independent, random samples.

• Total monthly mileage measurements within each city are independent of one another.

Normality: The total monthly mileage in each group is approximately normally distributed. This is especially important given the small sample sizes (\(n = 11\) per group).

Question 5b (4 pts)

As part of your summer internship with the food delivery analytics firm, you have been asked to determine if there is a significant difference in average total monthly mileage between California and non-California cities. Using \(\alpha = 0.03\), perform the first two steps of the four-step hypothesis test procedure.

Solution

Step 1 — Parameters of interest:

Let \(\mu_\text{California}\) and \(\mu_\text{non-California}\) be the parameters of interest, representing the population mean total monthly mileage of two-wheeled couriers across California and non-California cities, respectively.

Step 2 — Hypotheses:

\[H_0: \mu_\text{California} - \mu_\text{non-California} = 0\]

\[H_a: \mu_\text{California} - \mu_\text{non-California} \neq 0\]

Question 5c (8 pts)

Calculate the appropriate test statistic for comparing the mean total mileage between California and non-California cities. Clearly show the steps and formula used.

Solution

Since the population standard deviations are unknown and not assumed equal, we use the Welch two-sample t-test:

\[t_\text{TS} = \frac{\bar{x}_\text{California} - \bar{x}_\text{non-California}} {\sqrt{\dfrac{s_\text{California}^2}{n_\text{California}} + \dfrac{s_\text{non-California}^2}{n_\text{non-California}}}}\]

Substituting values:

\[t_\text{TS} = \frac{752{,}962 - 824{,}387.6} {\sqrt{\dfrac{697{,}033.6^2}{11} + \dfrac{918{,}850.2^2}{11}}} = \frac{-71{,}425.6}{\sqrt{44{,}168{,}712{,}684.5 + 76{,}753{,}244{,}549.1}}\]

\[= \frac{-71{,}425.6}{\sqrt{120{,}921{,}957{,}233.6}} = \frac{-71{,}425.6}{347{,}738.3} = \boxed{-0.2054}\]

Question 5d (3 pts)

Without additional assumptions, which of the following represents the most appropriate degrees of freedom for the test statistic found in part (c)?

1. 10     (B) 20     (C) 18.646     (D) None of the above

Solution

Answer: (C) 18.646

The Welch–Satterthwaite approximation gives:

\[\nu = \frac{\left(\dfrac{s_\text{CA}^2}{n_\text{CA}} + \dfrac{s_\text{non-CA}^2}{n_\text{non-CA}}\right)^2} {\dfrac{\left(s_\text{CA}^2/n_\text{CA}\right)^2}{n_\text{CA}-1} + \dfrac{\left(s_\text{non-CA}^2/n_\text{non-CA}\right)^2}{n_\text{non-CA}-1}} = 18.646\]

• 1. 10 would be the degrees of freedom for one group alone (\(n - 1\)).

• 2. 20 would be the pooled degrees of freedom, which assumes equal variances.

• 3. 18.646 is the Welch–Satterthwaite result — correct for unequal variances.

Question 5e (3 pts)

Select the correct R code and output that provides the correct critical value for constructing a 97% confidence interval. Assume \(\nu\) represents the correct degrees of freedom if appropriate.

# (A) qnorm(p=0.03/2, lower.tail = FALSE)   -> 2.17009
# (B) qt(p=0.03/2, df=nu, lower.tail = FALSE) -> 2.349237
# (C) 2*qnorm(p=0.03, lower.tail = FALSE)   -> 3.761587
# (D) 2*qt(p=0.03, df=nu, lower.tail = FALSE) -> 4.004843

Solution

Answer: (B) qt(p=0.03/2, df=nu, lower.tail = FALSE) → 2.349237

Since \(\sigma\) is unknown we use a t-distribution, not the normal. For a 97% confidence interval, the two-sided error is \(\alpha = 0.03\), so we need the upper \(\alpha/2 = 0.015\) critical value: qt(p=0.03/2, df=nu, lower.tail=FALSE) with \(\nu = 18.646\).

• 1. uses the z-distribution — incorrect since \(\sigma\) is unknown.

• (C) and (D) double the quantile, which is not how critical values for CIs are computed.

Question 5f (6 pts)

Using the summary statistics provided and the critical value, calculate the 97% confidence interval for the difference in mean total monthly mileage (California minus non-California). Clearly show all necessary formulas and steps.

Solution

Point estimate:

\[\bar{x}_\text{CA} - \bar{x}_\text{non-CA} = 752{,}962 - 824{,}387.6 = -71{,}425.6\]

Standard error (from part c):

\[\sqrt{\frac{s_\text{CA}^2}{n_\text{CA}} + \frac{s_\text{non-CA}^2}{n_\text{non-CA}}} = 347{,}738.3\]

97% CI:

\[\left(\bar{x}_\text{CA} - \bar{x}_\text{non-CA}\right) \pm t_{0.03/2,\,\nu} \cdot \sqrt{\frac{s_\text{CA}^2}{n_\text{CA}} + \frac{s_\text{non-CA}^2}{n_\text{non-CA}}}\]

\[-71{,}425.6 \pm 2.349237 \times 347{,}738.3\]

\[\boxed{(-888{,}345.3,\; 745{,}494.1)}\]

Question 5g (3 pts)

Provide an accurate interpretation of the confidence interval in context, explaining what it indicates about the difference in average total monthly mileage between California and non-California cities.

Solution

We are 97% confident that the true difference in average total monthly mileage (California minus non-California) is captured by the interval \((-888{,}345.3,\; 745{,}494.1)\).

Since this interval includes both negative and positive values (including zero), it indicates that there is no clear evidence of a significant difference in the average monthly mileage between California and non-California cities at the \(\alpha = 0.03\) significance level.