MATH 110 Module 6 Exam Introductory Statistics Inference &
Estimation Actual Exam 2026/2027 – Complete Questions
with Detailed Rationales | 100% Verified | Pass Guaranteed –
A+ Graded
Q1: According to the Central Limit Theorem, for a sufficiently large sample size (n ≥
30), the distribution of the sample mean will be:
A. Exactly the same shape as the population, no matter what
B. Approximately normal, regardless of the shape of the population distribution
[CORRECT]
C. Uniform, since all sample means are equally likely
D. Skewed right, because means cannot be negative
Correct Answer: B
Rationale: The CLT states that sample means become approximately normally
distributed as n grows, even for skewed or non-normal populations. Options A, C, and
D contradict the theorem's guarantee of approximate normality of the sampling
distribution.
Q2: A population has a standard deviation of σ = 12. A random sample of n = 36 is
drawn. What is the standard error of the sample mean?
A. 12
B. 0.33
C. 2 [CORRECT]
D. 72
Correct Answer: C
Rationale: The standard error is σ/√n = 12/√36 = 12/6 = 2. Option A confuses
population standard deviation with standard error, option B divides by n, and option D
multiplies instead of divides.
Q3: A researcher studies household incomes in a town, a variable that is strongly
right-skewed. She collects a random sample of 50 households and computes the
sample mean. The sampling distribution of her sample mean is best described as:
A. Approximately normal by the Central Limit Theorem because n = 50 ≥ 30 [CORRECT]
B. Strongly right-skewed because the population is right-skewed
C. Exactly normal only if the population standard deviation is known
D. Impossible to describe without knowing the population mean
Correct Answer: A
, Rationale: With n = 50 exceeding the 30-observation guideline, the CLT makes the
sampling distribution of the mean approximately normal despite population skew.
Option B incorrectly transfers the population's shape to the sampling distribution, and
the population mean is irrelevant to the shape conclusion.
Q4: In a large population, 40% of voters support a ballot measure (p = 0.40). For
random samples of n = 100 voters, the sampling distribution of the sample proportion
has:
A. Mean 0.40 and standard error 0.40
B. Mean 0.04 and standard error 0.004
C. Mean 0.60 and standard error 0.24
D. Mean 0.40 and standard error √(0.40 × 0.) ≈ 0.049 [CORRECT]
Correct Answer: D
Rationale: The sampling distribution of p̂ is centered at p = 0.40 with standard error
√(p(1−p)/n) = √(0.24/100) ≈ 0.049. Option A ignores the √n reduction, option B
confuses standard error with proportion, and option C uses 1−p as the mean and
np(1−p) as the error.
Q5: A population has μ = 50 and σ = 8. A random sample of n = 16 yields a sample
mean of x̄ = 52. What is the z-score for this sample mean, and does it represent a
typical or unusual result?
A. z = 0.25; unusual because z is positive
B. z = 1.00; typical, since it lies within two standard errors of the mean [CORRECT]
C. z = 2.00; typical, because any positive z is usual
D. z = 4.00; unusual, since it exceeds one standard deviation
Correct Answer: B
Rationale: z = (52 − 50)/(8/√16) = 2/2 = 1.00, which is well within the usual range (|z| <
2) for sample means. Option A divides by n instead of √n, option C computes 2/1
ignoring √n, and option D uses the individual-score formula with n = 16 incorrectly.
Section B: Confidence Intervals for Means & Proportions (Q6–Q10)
Q6: A researcher wants a confidence interval for a population mean but does not know
the population standard deviation σ and has a small sample (n = 15). Which
distribution should be used?
A. The standard normal (z) distribution, because confidence intervals always use z
B. The chi-square distribution
C. The t distribution with 14 degrees of freedom, because σ is estimated by s
[CORRECT]
D. The binomial distribution, because the sample is small
Estimation Actual Exam 2026/2027 – Complete Questions
with Detailed Rationales | 100% Verified | Pass Guaranteed –
A+ Graded
Q1: According to the Central Limit Theorem, for a sufficiently large sample size (n ≥
30), the distribution of the sample mean will be:
A. Exactly the same shape as the population, no matter what
B. Approximately normal, regardless of the shape of the population distribution
[CORRECT]
C. Uniform, since all sample means are equally likely
D. Skewed right, because means cannot be negative
Correct Answer: B
Rationale: The CLT states that sample means become approximately normally
distributed as n grows, even for skewed or non-normal populations. Options A, C, and
D contradict the theorem's guarantee of approximate normality of the sampling
distribution.
Q2: A population has a standard deviation of σ = 12. A random sample of n = 36 is
drawn. What is the standard error of the sample mean?
A. 12
B. 0.33
C. 2 [CORRECT]
D. 72
Correct Answer: C
Rationale: The standard error is σ/√n = 12/√36 = 12/6 = 2. Option A confuses
population standard deviation with standard error, option B divides by n, and option D
multiplies instead of divides.
Q3: A researcher studies household incomes in a town, a variable that is strongly
right-skewed. She collects a random sample of 50 households and computes the
sample mean. The sampling distribution of her sample mean is best described as:
A. Approximately normal by the Central Limit Theorem because n = 50 ≥ 30 [CORRECT]
B. Strongly right-skewed because the population is right-skewed
C. Exactly normal only if the population standard deviation is known
D. Impossible to describe without knowing the population mean
Correct Answer: A
, Rationale: With n = 50 exceeding the 30-observation guideline, the CLT makes the
sampling distribution of the mean approximately normal despite population skew.
Option B incorrectly transfers the population's shape to the sampling distribution, and
the population mean is irrelevant to the shape conclusion.
Q4: In a large population, 40% of voters support a ballot measure (p = 0.40). For
random samples of n = 100 voters, the sampling distribution of the sample proportion
has:
A. Mean 0.40 and standard error 0.40
B. Mean 0.04 and standard error 0.004
C. Mean 0.60 and standard error 0.24
D. Mean 0.40 and standard error √(0.40 × 0.) ≈ 0.049 [CORRECT]
Correct Answer: D
Rationale: The sampling distribution of p̂ is centered at p = 0.40 with standard error
√(p(1−p)/n) = √(0.24/100) ≈ 0.049. Option A ignores the √n reduction, option B
confuses standard error with proportion, and option C uses 1−p as the mean and
np(1−p) as the error.
Q5: A population has μ = 50 and σ = 8. A random sample of n = 16 yields a sample
mean of x̄ = 52. What is the z-score for this sample mean, and does it represent a
typical or unusual result?
A. z = 0.25; unusual because z is positive
B. z = 1.00; typical, since it lies within two standard errors of the mean [CORRECT]
C. z = 2.00; typical, because any positive z is usual
D. z = 4.00; unusual, since it exceeds one standard deviation
Correct Answer: B
Rationale: z = (52 − 50)/(8/√16) = 2/2 = 1.00, which is well within the usual range (|z| <
2) for sample means. Option A divides by n instead of √n, option C computes 2/1
ignoring √n, and option D uses the individual-score formula with n = 16 incorrectly.
Section B: Confidence Intervals for Means & Proportions (Q6–Q10)
Q6: A researcher wants a confidence interval for a population mean but does not know
the population standard deviation σ and has a small sample (n = 15). Which
distribution should be used?
A. The standard normal (z) distribution, because confidence intervals always use z
B. The chi-square distribution
C. The t distribution with 14 degrees of freedom, because σ is estimated by s
[CORRECT]
D. The binomial distribution, because the sample is small
