Business Statistics Quiz: One And Two Sample Tests
20 questions · exam conditions
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One And Two Sample TestsQuestion 1 of 20

A quality assurance team wants to determine if a new supplier provides parts with the same defect rate as their current supplier. Current supplier: 23 defective parts out of 400 sampled. New supplier: 31 defective parts out of 450 sampled. Using α=0.05\alpha = 0.05 for a two-tailed test, which analysis approach correctly addresses the research question?

Conduct a one-proportion z-test comparing the new supplier's rate to the current supplier's rate of 0.0575 as the hypothesized population proportion
Conduct a two-proportion z-test with H0:p1=p2H_0: p_1 = p_2 vs H1:p1≠p2H_1: p_1 \neq p_2, using pooled proportion p^=54/850=0.0635\hat{p} = 54/850 = 0.0635
Conduct a chi-square goodness of fit test comparing observed frequencies (23, 31) to expected frequencies based on equal proportions
Conduct a two-proportion z-test with H0:p1−p2=0H_0: p_1 - p_2 = 0 vs H1:p1−p2≠0H_1: p_1 - p_2 \neq 0, using separate standard errors for each proportion
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Business Statistics Quiz

Business Statistics Quiz: One And Two Sample Tests

Practice One And Two Sample Tests in Business Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on One And Two Sample Tests, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A quality assurance team wants to determine if a new supplier provides parts with the same defect rate as their current supplier. Current supplier: 23 defective parts out of 400 sampled. New supplier: 31 defective parts out of 450 sampled. Using α=0.05\alpha = 0.05 for a two-tailed test, which analysis approach correctly addresses the research question?

  1. Conduct a one-proportion z-test comparing the new supplier's rate to the current supplier's rate of 0.0575 as the hypothesized population proportion
  2. Conduct a two-proportion z-test with H0:p1=p2H_0: p_1 = p_2 vs H1:p1≠p2H_1: p_1 \neq p_2, using pooled proportion p^=54/850=0.0635\hat{p} = 54/850 = 0.0635 (correct answer)
  3. Conduct a chi-square goodness of fit test comparing observed frequencies (23, 31) to expected frequencies based on equal proportions
  4. Conduct a two-proportion z-test with H0:p1−p2=0H_0: p_1 - p_2 = 0 vs H1:p1−p2≠0H_1: p_1 - p_2 \neq 0, using separate standard errors for each proportion
Explanation: This is a comparison of two independent proportions, requiring a two-proportion z-test. The null hypothesis is that the proportions are equal (H₀: p₁ = p₂), and we use the pooled proportion p^=(23+31)/(400+450)=54/850=0.0635\hat{p} = (23+31)/(400+450) = 54/850 = 0.0635 in the standard error calculation. Choice A treats this as a one-sample test. Choice C suggests chi-square, which is unnecessarily complex. Choice D describes the same test as B but mentions separate standard errors, which is incorrect for the equal-proportions test.

Question 2

A pharmaceutical company tests whether a new drug increases recovery rates compared to the standard treatment. In the control group (standard treatment), 42 out of 120 patients recovered. In the treatment group (new drug), 58 out of 110 patients recovered. If the company uses a one-tailed test at α=0.05\alpha = 0.05 to determine if the new drug has a higher recovery rate, what should they conclude?

  1. Reject H0H_0 because the test statistic z=2.18z = 2.18 exceeds the critical value of 1.645, indicating the new drug is significantly better (correct answer)
  2. Fail to reject H0H_0 because the test statistic z=1.89z = 1.89 is less than the critical value of 1.96 for a two-tailed test
  3. Reject H0H_0 because the test statistic z=2.84z = 2.84 exceeds the critical value of 1.645, indicating the new drug is significantly better
  4. Fail to reject H0H_0 because the test statistic z=1.42z = 1.42 is less than the critical value of 1.645 for a one-tailed test
Explanation: For a two-proportion z-test: p^1=42/120=0.35\hat{p}_1 = 42/120 = 0.35, p^2=58/110=0.527\hat{p}_2 = 58/110 = 0.527. The pooled proportion is p^=(42+58)/(120+110)=0.435\hat{p} = (42+58)/(120+110) = 0.435. The test statistic is z=0.527−0.350.435(1−0.435)(1120+1110)=2.18z = \frac{0.527-0.35}{\sqrt{0.435(1-0.435)(\frac{1}{120}+\frac{1}{110})}} = 2.18. Since this exceeds 1.645 (one-tailed critical value), we reject H0H_0. Choice B uses wrong critical value. Choice C has calculation error. Choice D has major calculation error.

Question 3

A financial analyst tests whether the mean daily trading volume for Stock X equals 2.5 million shares. From 25 trading days: xˉ=2.72\bar{x} = 2.72 million, s=0.45s = 0.45 million. She calculates t=2.44t = 2.44 and determines the p-value is 0.022. If she had incorrectly used α=0.01\alpha = 0.01 instead of the intended α=0.05\alpha = 0.05, how would this affect her conclusion and Type I error risk?

  1. At α=0.01\alpha = 0.01, she would fail to reject H0H_0, but this would actually increase her Type I error risk since she's being too conservative
  2. At α=0.01\alpha = 0.01, she would still reject H0H_0 since p = 0.022 > 0.01, but her Type I error risk would decrease from 5% to 1% regardless
  3. The conclusion would remain the same (reject H0H_0) since the test statistic t = 2.44 exceeds the critical values for both α=0.05\alpha = 0.05 and α=0.01\alpha = 0.01
  4. At α=0.01\alpha = 0.01, she would fail to reject H0H_0 instead of rejecting it, reducing her Type I error risk from 5% to 1% but increasing Type II error risk (correct answer)
Explanation: When you encounter hypothesis testing questions involving significance levels, focus on the relationship between p-values, alpha levels, and error types. This question tests your understanding of how changing the significance level affects both the statistical conclusion and error risks. With a p-value of 0.022, you compare this to your chosen significance level. At α=0.05\alpha = 0.05, since 0.022 < 0.05, you reject H0H_0. However, at α=0.01\alpha = 0.01, since 0.022 > 0.01, you fail to reject H0H_0. This changes the conclusion entirely. The Type I error risk always equals your chosen alpha level when the null hypothesis is true, so using α=0.01\alpha = 0.01 reduces Type I error risk from 5% to 1%. However, being more stringent (lower alpha) increases Type II error risk—the probability of failing to reject a false null hypothesis. Option A incorrectly suggests that being conservative increases Type I error risk, when it actually decreases it. Option B contains a logical contradiction, stating you'd "still reject H0H_0" while noting that p > 0.01, and misunderstands that Type I error risk depends on your chosen alpha, not the decision outcome. Option C incorrectly claims the conclusion stays the same and focuses on critical values rather than the p-value comparison method shown. Remember this key principle: lowering your significance level makes you more conservative (harder to reject H0H_0), which reduces Type I error risk but increases Type II error risk. Always compare your p-value directly to your chosen alpha level to determine your statistical conclusion.

Question 4

A marketing researcher tests whether the proportion of customers who prefer Brand X differs between two demographic groups. Group 1: 85 out of 200 prefer Brand X. Group 2: 72 out of 180 prefer Brand X. Before conducting the test, she must verify that assumptions are met. Which assumption check reveals a potential problem with this analysis?

  1. The assumption of independence is violated because the sample sizes are different (200 vs 180), requiring equal group sizes for valid inference
  2. The assumption np ≥ 5 and n(1-p) ≥ 5 is violated for Group 2 since 180(0.4) = 72 and 180(0.6) = 108, but we need np ≥ 10
  3. All assumptions appear to be satisfied: both groups have sufficient sample sizes, expected frequencies exceed 5, and independence can be reasonably assumed (correct answer)
  4. The assumption of equal population proportions is violated since p^1=0.425\hat{p}_1 = 0.425 and p^2=0.400\hat{p}_2 = 0.400 are noticeably different values
Explanation: For a two-proportion z-test, we need: (1) independent samples, (2) np ≥ 5 and n(1-p) ≥ 5 for both groups. Group 1: 85 ≥ 5 ✓, 115 ≥ 5 ✓. Group 2: 72 ≥ 5 ✓, 108 ≥ 5 ✓. Independence can be reasonably assumed. Choice A is wrong—equal sample sizes aren't required. Choice B is wrong—the threshold is 5, not 10. Choice D misunderstands—we don't assume equal population proportions; that's what we're testing.

Question 5

An economist wants to test if the average household income in City A differs from the national average of $65,000. A sample of 36 households from City A yields $xˉ=68,200\bar{x} = 68,200 andand s=12,500s = 12,500 .Afterconductingatwo−tailedtestat. After conducting a two-tailed test at α=0.05\alpha = 0.05 $, the economist calculates a p-value of 0.032. Which statement best describes the appropriate conclusion and its business interpretation?

  1. Reject H0H_0 at α=0.05\alpha = 0.05 but not at α=0.01\alpha = 0.01; there is moderate evidence that City A's income differs significantly from the national average (correct answer)
  2. Reject H0H_0 at both α=0.05\alpha = 0.05 and α=0.01\alpha = 0.01; there is strong evidence that City A's income is significantly higher than the national average
  3. Fail to reject H0H_0 because the sample mean of $68,200 is not sufficiently different from $65,000 to conclude statistical significance
  4. Reject H0H_0 at α=0.05\alpha = 0.05; however, the practical significance is questionable since the difference of $3,200 represents only a 4.9% increase
Explanation: With p-value = 0.032, we reject H0H_0 at α=0.05\alpha = 0.05 (since 0.032 < 0.05) but would fail to reject at α=0.01\alpha = 0.01 (since 0.032 > 0.01). This indicates moderate evidence against H0H_0. Choice B is wrong because 0.032 > 0.01. Choice C incorrectly fails to reject when p < α\alpha. Choice D introduces practical significance discussion not warranted by the given information and doesn't address the strength of statistical evidence.

Question 6

An environmental agency is testing whether the mean level of a certain pollutant in a lake exceeds the safety limit of 3.5 parts per million (ppm). What are the correct null (H₀) and alternative (Hₐ) hypotheses for this test?

  1. H₀: μ ≥ 3.5, Hₐ: μ < 3.5
  2. H₀: μ ≤ 3.5, Hₐ: μ > 3.5 (correct answer)
  3. H₀: μ = 3.5, Hₐ: μ ≠ 3.5
  4. H₀: x̄ = 3.5, Hₐ: x̄ > 3.5
Explanation: The agency is trying to find evidence that the pollutant level exceeds the safety limit. The claim or condition the agency wants to find evidence for should be the alternative hypothesis (Hₐ). Therefore, Hₐ: μ > 3.5. The null hypothesis (H₀) is the statement of the status quo or no violation, which includes the equality. Thus, H₀: μ ≤ 3.5 (or often simplified to H₀: μ = 3.5, as the test is conducted at the boundary value). Option D is incorrect because hypotheses are always about population parameters (μ), not sample statistics (x̄).

Question 7

A financial firm compares the mean performance of two investment portfolios. A 95% confidence interval for the difference in mean annual returns (Portfolio A - Portfolio B) is calculated to be [-0.8%, +4.2%]. Based solely on this interval, what is the outcome of a two-tailed hypothesis test of H₀: μₐ = μₑ versus Hₐ: μₐ ≠ μₑ at the α = 0.05 significance level?

  1. Reject H₀, because a large portion of the confidence interval is positive.
  2. Fail to reject H₀, because the confidence interval contains the value of zero. (correct answer)
  3. The result is inconclusive, as the p-value is needed to make a decision.
  4. Reject H₀, because the sample mean return for Portfolio A was higher than for Portfolio B.
Explanation: There is a direct correspondence between a two-sided confidence interval and a two-tailed hypothesis test. If the confidence interval for the difference between two means contains zero, it means that zero is a plausible value for the true difference. Therefore, at the corresponding significance level (α = 1 - Confidence Level = 1 - 0.95 = 0.05), we do not have sufficient evidence to conclude that the means are different. We fail to reject the null hypothesis.

Question 8

An A/B test was conducted to compare the conversion rates of two website designs, Design A (current) and Design B (new). The null hypothesis was that the conversion rates are equal (H₀: pₐ = pₑ), while the alternative was that the new design has a different conversion rate (Hₐ: pₐ ≠ pₑ). After collecting data, a two-proportion z-test yielded a p-value of 0.042.

Based on the passage, which of the following is the most appropriate interpretation of the test result at a significance level of α = 0.05?

  1. There is a 4.2% chance that the two designs have the same conversion rate in the population.
  2. The test proves that Design B is better than Design A, so the company should switch to the new design immediately.
  3. There is statistically significant evidence of a difference between the conversion rates of the two designs. (correct answer)
  4. There is not a statistically significant difference between the designs, as the p-value is greater than 1%.
Explanation: The p-value (0.042) is the probability of observing a difference in sample proportions as extreme or more extreme than the one found, assuming the null hypothesis (that the true proportions are equal) is true. Since the p-value (0.042) is less than the significance level (α = 0.05), we reject the null hypothesis. The correct interpretation is that there is statistically significant evidence to conclude that the true conversion rates of the two designs are different.

Question 9

A marketing team conducted a study with a sample of 30 customers per group to test if a new ad campaign increased average spending. The results were not statistically significant (p > 0.05). The team strongly suspects that a real, meaningful increase in spending exists, but the study failed to detect it. Which of the following actions would be most likely to increase the power of a follow-up study to detect this effect?

  1. Decreasing the significance level (α) to 0.01 to reduce the chance of a Type I error.
  2. Substantially increasing the number of customers sampled in each group. (correct answer)
  3. Ensuring the sample standard deviation in the follow-up study is larger than in the first.
  4. Switching from a one-tailed test to a two-tailed test to capture any effect.
Explanation: Statistical power is the probability of correctly rejecting a false null hypothesis. The most direct way to increase power is to increase the sample size. A larger sample size reduces the standard error, making the test more sensitive to detecting a true effect. Decreasing α (A) decreases power. A larger standard deviation (C) decreases power. Switching from a one-tailed to a two-tailed test (D) decreases power, assuming the direction of the effect is correctly hypothesized.

Question 10

A manufacturer claims its high-performance tires last for an average of 60,000 miles. A consumer advocacy group tests a random sample of 16 tires and finds a sample mean of 58,500 miles and a sample standard deviation of 2,400 miles. To test if the true mean mileage is less than claimed, what is the appropriate test statistic?

  1. t = -2.50 (correct answer)
  2. t = -0.625
  3. z = -2.50
  4. t = -10.00
Explanation: Since the population standard deviation is unknown, a one-sample t-test is appropriate. The null hypothesis is H₀: μ = 60,000 and the alternative is Hₐ: μ < 60,000. First, calculate the standard error of the mean: SE = s / √n = 2,400 / √16 = 2,400 / 4 = 600. Then, calculate the t-statistic: t = (x̄ - μ₀) / SE = (58,500 - 60,000) / 600 = -1,500 / 600 = -2.50.

Question 11

A manufacturing plant implements a new process to reduce the proportion of defective products, which is historically 4%. A random sample of 500 products made with the new process contains 12 defects. A test of H₀: p = 0.04 vs Hₐ: p < 0.04 yields a z-statistic of -2.04 and a p-value of 0.021. Using a significance level of 0.05, what is the most appropriate business conclusion?

  1. The new process caused a reduction in the defect rate from 4% to 2.4%.
  2. There is a 2.1% chance that the new process is not better than the old one.
  3. The result is not significant, so the new process is no better than the old process.
  4. There is statistically significant evidence to support the claim that the new process has a lower defect rate than 4%. (correct answer)
Explanation: Since the p-value (0.021) is less than the significance level (α = 0.05), we reject the null hypothesis. This means we have sufficient statistical evidence to conclude that the alternative hypothesis is likely true. The most appropriate conclusion is that there is significant evidence that the true proportion of defects for the new process is less than 4%. Option A makes a causal claim which may not be warranted. Option B is a misinterpretation of the p-value. Option C makes the wrong conclusion.

Question 12

Two separate one-sample t-tests are conducted to determine if a new fertilizer increases crop yield above a baseline of 50 bushels/acre. Study A uses a sample of 25 plots and finds a sample mean of 54 with a standard deviation of 10. Study B uses a sample of 25 plots and also finds a sample mean of 54, but with a standard deviation of 18. How will the p-value from Study A compare to the p-value from Study B?

  1. The p-value for Study A will be larger than for Study B.
  2. The p-values will be identical because the sample means and sizes are the same.
  3. The p-value for Study A will be smaller than for Study B. (correct answer)
  4. The comparison is impossible without knowing the chosen alpha level for the tests.
Explanation: The t-statistic is calculated as (x̄ - μ₀) / (s/√n). Since x̄, μ₀, and n are the same for both studies, the t-statistic is inversely proportional to the sample standard deviation (s). Study A has a smaller standard deviation (10 vs 18), which will result in a smaller standard error and a larger t-statistic. A larger positive t-statistic in a right-tailed test leads to a smaller p-value. Therefore, Study A provides stronger evidence against the null hypothesis.

Question 13

An analyst conducts a two-tailed hypothesis test for a single population mean, H₀: μ = 50 vs. Hₐ: μ ≠ 50, and obtains a p-value of 0.06. The sample mean was 54. If the analyst had originally planned to conduct a one-tailed test, H₀: μ = 50 vs. Hₐ: μ > 50, what would the p-value for that test have been?

  1. 0.06
  2. 0.12
  3. 0.03 (correct answer)
  4. 0.97
Explanation: A two-tailed p-value represents the probability in both tails of the distribution. Since the sample mean (54) is greater than the hypothesized mean (50), the observed result is in the direction of the one-tailed alternative hypothesis (μ > 50). Therefore, the p-value for the one-tailed test is exactly half of the two-tailed p-value. p_onetailed = p_twotailed / 2 = 0.06 / 2 = 0.03.

Question 14

A national retail chain tests a new in-store display across 1,000 of its stores. The test shows that the average daily sales for a specific product increased by $0.25 per store compared to the old display. Due to the extremely large sample size, the p-value for this increase is less than 0.001. What is the most crucial consideration for management when deciding whether to implement the new display nationwide?

  1. The difference is statistically significant, but its practical significance (a $0.25 increase) may be too small to justify the cost. (correct answer)
  2. The risk of a Type I error, even with a small p-value, is too high to justify a nationwide rollout.
  3. Since the p-value is so low, the result must be practically significant and the change should be implemented.
  4. The result is likely a statistical anomaly because very small p-values are rare in business applications.
Explanation: When you encounter questions about statistical significance versus practical significance, especially with large sample sizes, you need to distinguish between what's statistically detectable and what's actually meaningful for business decisions. With 1,000 stores, this study has enormous statistical power to detect even tiny differences. A $0.25 daily increase per store is mathematically real (hence the p-value < 0.001), but you must ask: is this difference large enough to matter practically? Consider the costs of redesigning, manufacturing, and installing new displays across thousands of stores. If implementation costs $500 per store but only generates an extra 0.25daily(0.25 daily (91.25 annually), the return on investment would be terrible. Option A correctly identifies this core issue - the result is statistically significant but may lack practical significance given the small effect size relative to likely implementation costs. Option B misunderstands Type I error risk. With p < 0.001, there's less than 0.1% chance this result occurred by random chance alone - that's actually very low risk. Option C commits the classic error of assuming statistical significance automatically means practical significance. Large samples can make trivial differences appear "significant" statistically while remaining meaningless business-wise. Option D incorrectly suggests very small p-values are anomalies. They're actually common with large sample sizes, even for small real effects. Study tip: Remember that statistical significance only tells you an effect probably exists - it doesn't tell you whether that effect is large enough to care about. Always evaluate effect size alongside p-values, especially with large samples.

Question 15

A 99% confidence interval for the difference in proportions of customers who churn between two service plans (Plan A - Plan B) is calculated as [-0.08, -0.01]. What can be concluded about the results of a two-tailed hypothesis test of H₀: pₐ = pₑ at the α = 0.01 significance level?

  1. We would reject H₀ and conclude there is evidence of a difference in churn rates. (correct answer)
  2. We would fail to reject H₀ because the interval contains negative values.
  3. We cannot draw a conclusion because the standard error calculation is different for confidence intervals and hypothesis tests.
  4. We would reject H₀, but we would need to use α = 0.005 for the test.
Explanation: When you encounter a confidence interval for a difference in proportions, you can use it to draw conclusions about hypothesis tests at corresponding significance levels. The key relationship is that a 99% confidence interval corresponds to a two-tailed test at α = 0.01. The correct reasoning starts with examining what the confidence interval tells us. The interval [-0.08, -0.01] represents plausible values for the true difference in churn rates (Plan A - Plan B). Since this entire interval lies below zero and doesn't contain zero, we have strong evidence that Plan A has a lower churn rate than Plan B. Because zero (representing no difference) is not a plausible value within our 99% confidence level, we would reject H₀: pₐ = pₑ in favor of concluding there is a significant difference. Answer B is incorrect because the presence of negative values isn't the issue—it's whether zero is included. The interval being entirely negative actually supports rejecting the null hypothesis. Answer C misunderstands the mathematical relationship between confidence intervals and hypothesis tests; when the confidence level matches the significance level (99% CI with α = 0.01), the conclusions are equivalent. Answer D incorrectly suggests changing the significance level—a 99% confidence interval directly corresponds to α = 0.01, not α = 0.005. Study tip: Remember that if a confidence interval for a difference doesn't contain zero, you can reject the null hypothesis of "no difference" at the corresponding significance level. This duality between confidence intervals and hypothesis tests is a powerful tool for statistical inference.

Question 16

An online retailer tests two website designs to see if they generate different conversion rates. Design A: 156 conversions out of 800 visitors. Design B: 198 conversions out of 900 visitors. The test yields z=1.68z = 1.68 with p-value = 0.093. The marketing director concludes: 'Since p > 0.05, the designs perform equally well.' What is the most significant flaw in this reasoning?

  1. The director failed to account for multiple comparisons; testing two designs simultaneously requires a Bonferroni correction to maintain α=0.05\alpha = 0.05
  2. The director should have used a one-tailed test since the goal was to determine if Design B (with higher sample proportion) performs better
  3. The director ignored the effect size; Design B has a conversion rate of 22% vs Design A's 19.5%, representing a meaningful business difference
  4. The director incorrectly interpreted 'fail to reject H0H_0' as proving the null hypothesis true, when it only indicates insufficient evidence of a difference (correct answer)
Explanation: This question tests your understanding of statistical hypothesis testing, specifically the critical distinction between "failing to reject the null hypothesis" and "proving the null hypothesis is true." When you encounter p-values near but above your significance threshold, this distinction becomes crucial. The marketing director's reasoning contains a fundamental logical error. The test result (p = 0.093 > 0.05) means we fail to reject the null hypothesis that the conversion rates are equal. However, failing to reject H0H_0 doesn't prove it's true—it simply means we lack sufficient evidence to conclude the rates differ significantly. The director incorrectly interpreted this as definitive proof that "the designs perform equally well." This is like a court case where "not guilty" doesn't mean innocent—it means insufficient evidence for conviction. Option A is incorrect because this isn't a multiple comparisons situation; we're making a single comparison between two groups, not testing multiple hypotheses simultaneously. Option B misses the point—whether to use a one-tailed or two-tailed test doesn't address the fundamental interpretation error, and the choice depends on whether you hypothesized a specific direction beforehand. Option C, while noting a potentially meaningful business difference, misunderstands the statistical issue. Effect size is important, but the primary flaw is the logical interpretation of the p-value, not ignoring practical significance. Study tip: Remember that hypothesis tests can only provide evidence against the null hypothesis, never proof for it. When p > α, always conclude "insufficient evidence to reject H0H_0" rather than "H0H_0 is true."

Question 17

A researcher performs a two-sample t-test on data from two independent groups (n₁=12, n₂=15). After the analysis, a histogram of each group's data reveals that both populations are strongly skewed to the right. What is the most likely consequence of this violation of assumptions?

  1. The calculated p-value from the t-test may not be accurate, compromising the validity of the conclusion. (correct answer)
  2. The Central Limit Theorem ensures the t-test results remain valid and reliable.
  3. The researcher should have used a two-sample z-test, which does not assume normality.
  4. The test will have lower statistical power, but the Type I error rate will be unaffected.
Explanation: When you encounter t-test questions involving assumption violations, focus on how departures from normality affect the reliability of your statistical inference, especially with smaller sample sizes. The two-sample t-test assumes that both populations are approximately normally distributed. When this assumption is violated—as with the strong right skew described here—the sampling distribution of the test statistic may not follow the expected t-distribution. With sample sizes of only 12 and 15, you don't have enough data points for the Central Limit Theorem to compensate for the non-normality. This means the p-value you calculate may not accurately reflect the true probability of observing your results under the null hypothesis, potentially leading to incorrect conclusions about statistical significance. Option B incorrectly assumes the Central Limit Theorem will rescue the analysis. While the CLT does help with larger samples (typically n ≥ 30), your sample sizes of 12 and 15 are too small to rely on this theorem when dealing with strongly skewed data. Option C is wrong because z-tests actually have the same normality assumption as t-tests. The choice between z and t depends on whether you know the population standard deviation, not on normality assumptions. Option D mischaracterizes the problem. The main concern isn't just reduced power—it's that the entire inferential framework becomes questionable when assumptions are severely violated. Study tip: Remember that sample size matters for assumption violations. With small samples (n < 30) and clear non-normality, consider non-parametric alternatives like the Mann-Whitney U test instead of forcing a t-test.

Question 18

Two different suppliers provide a critical component. A company tests the tensile strength of samples from each. For Supplier A, a sample of 10 components has a mean strength of 85 units with a standard deviation of 5 units. For Supplier B, a sample of 12 components has a mean strength of 81 units with a standard deviation of 4 units. Assuming equal population variances, what is the t-statistic for a test of H₀: μₐ = μₑ against Hₐ: μₐ > μₑ?

  1. t = 1.95
  2. t = 2.05
  3. t = 0.89
  4. t = 2.09 (correct answer)
Explanation: When you encounter a two-sample t-test with the assumption of equal variances, you're testing whether two population means differ significantly. This requires calculating a pooled standard error that combines information from both samples. The t-statistic formula for equal variances is: t=xˉA−xˉBsp1nA+1nBt = \frac{\bar{x}_A - \bar{x}_B}{s_p\sqrt{\frac{1}{n_A} + \frac{1}{n_B}}} First, calculate the pooled standard deviation: sp=(nA−1)sA2+(nB−1)sB2nA+nB−2s_p = \sqrt{\frac{(n_A-1)s_A^2 + (n_B-1)s_B^2}{n_A + n_B - 2}} Substituting the values: sp=(10−1)(52)+(12−1)(42)10+12−2=9(25)+11(16)20=225+17620=20.05=4.478s_p = \sqrt{\frac{(10-1)(5^2) + (12-1)(4^2)}{10 + 12 - 2}} = \sqrt{\frac{9(25) + 11(16)}{20}} = \sqrt{\frac{225 + 176}{20}} = \sqrt{20.05} = 4.478 Now calculate the t-statistic: t=85−814.478110+112=44.4780.1833=44.478(0.428)=41.916=2.09t = \frac{85 - 81}{4.478\sqrt{\frac{1}{10} + \frac{1}{12}}} = \frac{4}{4.478\sqrt{0.1833}} = \frac{4}{4.478(0.428)} = \frac{4}{1.916} = 2.09 Answer D (t = 2.09) is correct. Answer A (t = 1.95) likely used an incorrect pooled variance calculation. Answer B (t = 2.05) probably made a minor computational error in the standard error. Answer C (t = 0.89) appears to have confused the numerator and denominator or made a significant calculation mistake. Remember: equal variance t-tests always require pooling the sample variances first, then using that pooled value in your standard error calculation. Double-check your arithmetic, especially with square roots.

Question 19

A supply chain manager wants to test if the proportion of late deliveries from a new supplier is less than the historical rate of 10%. In a random sample of 80 deliveries from the new supplier, 4 were late. The manager attempts to conduct a one-sample z-test for a proportion. What is the most critical issue with the validity of this test?

  1. The sample size of 80 is not large enough for the Central Limit Theorem to apply.
  2. The population proportion is unknown, so a t-test should be used instead of a z-test.
  3. The condition for the number of expected successes under the null hypothesis is not met. (correct answer)
  4. The sample proportion (5%) is different from the hypothesized proportion (10%), which invalidates the test.
Explanation: For a one-sample z-test for a proportion to be valid, we need to check the success/failure condition using the hypothesized proportion, p₀. The conditions are np₀ ≥ 10 and n(1-p₀) ≥ 10. In this case, n = 80 and p₀ = 0.10. The number of expected 'successes' (late deliveries) is np₀ = 80 * 0.10 = 8. Since 8 is less than 10, this condition is not met, and the normal approximation to the binomial distribution, which the z-test relies on, may not be accurate.

Question 20

A digital marketer is A/B testing two email subject lines to see if there is a difference in open rates. Subject Line A was sent to 600 people and 120 opened it. Subject Line B was sent to 400 people and 92 opened it. When conducting a two-proportion z-test, what is the value of the pooled sample proportion (p-hat pooled) used to calculate the standard error?

  1. 0.215
  2. 0.200
  3. 0.230
  4. 0.212 (correct answer)
Explanation: For a two-proportion z-test, the pooled proportion is calculated by combining the successes and sample sizes from both groups. It is given by the formula p̂_pool = (x₁ + x₂) / (n₁ + n₂). Here, x₁ = 120, n₁ = 600, x₂ = 92, and n₂ = 400. So, p̂_pool = (120 + 92) / (600 + 400) = 212 / 1000 = 0.212. Distractor A (0.215) is the average of the two individual proportions ( (120/600 + 92/400)/2 = (0.20 + 0.23)/2 ), which is incorrect when sample sizes are unequal.