MATH 225N STATISTICS EXAM with Questions
and Answers/Plus a Rationale Updated 2026
A+/Instant Download PDFEXAM COVERAGE
1. Confidence Intervals for Population Proportions
2. Confidence Intervals for Population Means (Z-intervals and T-intervals)
3. Margin of Error Calculations
4. Sample Size Determination
5. Interpretation of Confidence Levels
6. Sampling Distributions and Central Limit Theorem
7. Critical Values ($z^*$ and $t^*$)
8. Effects of Sample Size and Confidence Level on Interval Width
1. A researcher conducts a study to estimate the proportion of adults who support a new policy. A
sample of 200 adults is surveyed, and 140 support the policy. Calculate the 95% confidence
interval for the population proportion.
A. (0.62, 0.78)
B. (0.63, 0.77)
C. (0.635, 0.765)
D. (0.60, 0.80)
CORRECT ANSWER : C
, Rationale: The sample proportion $\hat{p} = 140/200 = 0.70$. The standard error is
$\sqrt{0.70(0.30)/200} \approx 0.0324$. Using $z = 1.96$ for 95% confidence, the margin of
error is $1.96 \times 0.0324 \approx 0.0635$. Adding and subtracting this from 0.70 gives
(0.6365, 0.7635), which rounds to the provided interval. Options A, B, and D are incorrect due
to miscalculation of the margin of error or incorrect Z-scores.
2. You are calculating a 90% confidence interval for a population mean with a sample size of 25
from a normally distributed population where the population standard deviation is unknown.
Which critical value should you use?
A. $z = 1.645$
B. $t = 1.708$
C. $t = 1.711$
D. $z = 1.96$
CORRECT ANSWER : C
Rationale: Because the population standard deviation is unknown and the sample size is small
($n < 30$), we must use the t-distribution with $n-1 = 24$ degrees of freedom. For a 90%
confidence interval, the critical value for 24 degrees of freedom is approximately 1.711. Option
A is incorrect because it uses the Z-score, and Option B uses the wrong degrees of freedom.
3. A study reports a 95% confidence interval for the mean height of students as (165 cm, 175 cm).
How should this be interpreted?
A. There is a 95% probability that the true population mean lies between 165 and 175 cm.
B. We are 95% confident that the true population mean falls between 165 and 175 cm.
C. 95% of all students have heights between 165 and 175 cm.
D. If we repeated this study, 95% of the sample means would fall between 165 and 175 cm.
CORRECT ANSWER : B
Rationale: Confidence intervals describe our confidence in the estimation of the population
parameter, not the probability of the parameter itself (which is fixed) nor the distribution of
individual data points. Option A is a common misconception of probability, while C and D
misinterpret the confidence level.
4. If you increase the confidence level from 95% to 99% while keeping the sample size the same,
what happens to the confidence interval?
, A. It becomes narrower.
B. It becomes wider.
C. It stays the same.
D. It becomes biased.
CORRECT ANSWER : B
Rationale: To increase the confidence level, the critical value ($z^$ or $t^*$) must increase to
capture more of the distribution. A larger critical value increases the margin of error, resulting in
a wider interval. Options A, C, and D do not accurately reflect the mathematical relationship
between confidence level and interval width.*
5. A survey of 100 students finds that 60% prefer online learning. What is the standard error of this
proportion?
A. 0.06
B. 0.049
C. 0.05
D. 0.0024
CORRECT ANSWER : B
Rationale: The standard error of a proportion is calculated as $\sqrt{\hat{p}(1-\hat{p})/n}$.
Here, $\sqrt{0.60(0.40)/100} = \sqrt{0.24/100} = \sqrt{0.0024} \approx 0.04899$, which rounds
to 0.049. Other options result from incorrect formulas or arithmetic errors.
6. To determine the required sample size to estimate a population mean with a margin of error of 5
and a standard deviation of 20 at 95% confidence, what is the correct approach?
A. Use $n = (z \cdot \sigma / E)^2$
B. Use $n = (1.96 \cdot )^2$
C. Use $n = (1.645 \cdot )^2$
D. Use $n = (z \cdot s / E)$
CORRECT ANSWER : B
, Rationale: The formula for sample size determination for a mean is $n = (z^ \cdot \sigma /
E)^2$. Plugging in 1.96 for 95% confidence, 20 for standard deviation, and 5 for the margin of
error gives the correct calculation. Option C uses the wrong Z-score, and D uses the incorrect
algebraic form.*
7. What condition must be met to use the Z-distribution for a confidence interval of a population
mean?
A. The sample size must be at least 10.
B. The population must be normally distributed or the sample size must be large ($n \ge
30$).
C. The population variance must be estimated from the sample.
D. The sample must be skewed.
CORRECT ANSWER : B
Rationale: According to the Central Limit Theorem, the sampling distribution of the mean
approaches normality if the sample is large enough ($n \ge 30$), or if the population itself is
normal. Small, non-normal samples require non-parametric methods or the t-distribution.
Options A, C, and D are not sufficient or accurate conditions.
8. When calculating the margin of error for a mean with an unknown population standard deviation,
you use the t-score. Why?
A. The t-distribution is always narrower than the Z-distribution.
B. The t-distribution accounts for the additional uncertainty of estimating the standard
deviation from the sample.
C. The t-distribution is used only for large sample sizes.
D. The Z-score cannot be calculated for small samples.
CORRECT ANSWER : B
Rationale: The t-distribution has fatter tails than the Z-distribution to account for the variability
introduced by using the sample standard deviation ($s$) instead of the true population standard
deviation ($\sigma$). This makes the intervals more conservative and accurate. Other options
mischaracterize the properties of the distributions.
9. A 95% confidence interval for a proportion is (0.10, 0.18). What is the sample proportion
$\hat{p}$?
and Answers/Plus a Rationale Updated 2026
A+/Instant Download PDFEXAM COVERAGE
1. Confidence Intervals for Population Proportions
2. Confidence Intervals for Population Means (Z-intervals and T-intervals)
3. Margin of Error Calculations
4. Sample Size Determination
5. Interpretation of Confidence Levels
6. Sampling Distributions and Central Limit Theorem
7. Critical Values ($z^*$ and $t^*$)
8. Effects of Sample Size and Confidence Level on Interval Width
1. A researcher conducts a study to estimate the proportion of adults who support a new policy. A
sample of 200 adults is surveyed, and 140 support the policy. Calculate the 95% confidence
interval for the population proportion.
A. (0.62, 0.78)
B. (0.63, 0.77)
C. (0.635, 0.765)
D. (0.60, 0.80)
CORRECT ANSWER : C
, Rationale: The sample proportion $\hat{p} = 140/200 = 0.70$. The standard error is
$\sqrt{0.70(0.30)/200} \approx 0.0324$. Using $z = 1.96$ for 95% confidence, the margin of
error is $1.96 \times 0.0324 \approx 0.0635$. Adding and subtracting this from 0.70 gives
(0.6365, 0.7635), which rounds to the provided interval. Options A, B, and D are incorrect due
to miscalculation of the margin of error or incorrect Z-scores.
2. You are calculating a 90% confidence interval for a population mean with a sample size of 25
from a normally distributed population where the population standard deviation is unknown.
Which critical value should you use?
A. $z = 1.645$
B. $t = 1.708$
C. $t = 1.711$
D. $z = 1.96$
CORRECT ANSWER : C
Rationale: Because the population standard deviation is unknown and the sample size is small
($n < 30$), we must use the t-distribution with $n-1 = 24$ degrees of freedom. For a 90%
confidence interval, the critical value for 24 degrees of freedom is approximately 1.711. Option
A is incorrect because it uses the Z-score, and Option B uses the wrong degrees of freedom.
3. A study reports a 95% confidence interval for the mean height of students as (165 cm, 175 cm).
How should this be interpreted?
A. There is a 95% probability that the true population mean lies between 165 and 175 cm.
B. We are 95% confident that the true population mean falls between 165 and 175 cm.
C. 95% of all students have heights between 165 and 175 cm.
D. If we repeated this study, 95% of the sample means would fall between 165 and 175 cm.
CORRECT ANSWER : B
Rationale: Confidence intervals describe our confidence in the estimation of the population
parameter, not the probability of the parameter itself (which is fixed) nor the distribution of
individual data points. Option A is a common misconception of probability, while C and D
misinterpret the confidence level.
4. If you increase the confidence level from 95% to 99% while keeping the sample size the same,
what happens to the confidence interval?
, A. It becomes narrower.
B. It becomes wider.
C. It stays the same.
D. It becomes biased.
CORRECT ANSWER : B
Rationale: To increase the confidence level, the critical value ($z^$ or $t^*$) must increase to
capture more of the distribution. A larger critical value increases the margin of error, resulting in
a wider interval. Options A, C, and D do not accurately reflect the mathematical relationship
between confidence level and interval width.*
5. A survey of 100 students finds that 60% prefer online learning. What is the standard error of this
proportion?
A. 0.06
B. 0.049
C. 0.05
D. 0.0024
CORRECT ANSWER : B
Rationale: The standard error of a proportion is calculated as $\sqrt{\hat{p}(1-\hat{p})/n}$.
Here, $\sqrt{0.60(0.40)/100} = \sqrt{0.24/100} = \sqrt{0.0024} \approx 0.04899$, which rounds
to 0.049. Other options result from incorrect formulas or arithmetic errors.
6. To determine the required sample size to estimate a population mean with a margin of error of 5
and a standard deviation of 20 at 95% confidence, what is the correct approach?
A. Use $n = (z \cdot \sigma / E)^2$
B. Use $n = (1.96 \cdot )^2$
C. Use $n = (1.645 \cdot )^2$
D. Use $n = (z \cdot s / E)$
CORRECT ANSWER : B
, Rationale: The formula for sample size determination for a mean is $n = (z^ \cdot \sigma /
E)^2$. Plugging in 1.96 for 95% confidence, 20 for standard deviation, and 5 for the margin of
error gives the correct calculation. Option C uses the wrong Z-score, and D uses the incorrect
algebraic form.*
7. What condition must be met to use the Z-distribution for a confidence interval of a population
mean?
A. The sample size must be at least 10.
B. The population must be normally distributed or the sample size must be large ($n \ge
30$).
C. The population variance must be estimated from the sample.
D. The sample must be skewed.
CORRECT ANSWER : B
Rationale: According to the Central Limit Theorem, the sampling distribution of the mean
approaches normality if the sample is large enough ($n \ge 30$), or if the population itself is
normal. Small, non-normal samples require non-parametric methods or the t-distribution.
Options A, C, and D are not sufficient or accurate conditions.
8. When calculating the margin of error for a mean with an unknown population standard deviation,
you use the t-score. Why?
A. The t-distribution is always narrower than the Z-distribution.
B. The t-distribution accounts for the additional uncertainty of estimating the standard
deviation from the sample.
C. The t-distribution is used only for large sample sizes.
D. The Z-score cannot be calculated for small samples.
CORRECT ANSWER : B
Rationale: The t-distribution has fatter tails than the Z-distribution to account for the variability
introduced by using the sample standard deviation ($s$) instead of the true population standard
deviation ($\sigma$). This makes the intervals more conservative and accurate. Other options
mischaracterize the properties of the distributions.
9. A 95% confidence interval for a proportion is (0.10, 0.18). What is the sample proportion
$\hat{p}$?
