One & Two-Sample Tests - Business Statistics

A fitness company developed a new high-intensity training program. To test its effectiveness, they recruited 50 volunteers. They measured each volunteer's resting heart rate, had them complete the six-week program, and then measured their resting heart rate again. Which statistical test is most appropriate for determining if the program significantly lowered resting heart rates?
Answer: A paired-samples t-test, because the 'before' and 'after' measurements are from the same individuals. The data consists of two measurements taken on the same set of subjects. These are paired or dependent samples. The goal is to analyze the mean of the differences between the 'before' and 'after' measurements for each individual. Therefore, a paired-samples t-test is the most appropriate statistical method. An independent t-test would be incorrect as it assumes the two groups of measurements are from different, unrelated individuals.
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 $\alpha = 0.05$ for a two-tailed test, which analysis approach correctly addresses the research question?
Answer: Conduct a two-proportion z-test with $H_0: p_1 = p_2$ vs $H_1: p_1 \neq p_2$ , using pooled proportion $\hat{p} = 54/850 = 0.0635$ 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 $\hat{p} = (23+31)/(400+450) = 54/850 = 0.0635$ 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.
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?
Answer: 0.03 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.
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?
Answer: t = 2.09 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: $t = \frac{\bar{x}_A - \bar{x}_B}{s_p\sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}$ First, calculate the pooled standard deviation: $s_p = \sqrt{\frac{(n_A-1)s_A^2 + (n_B-1)s_B^2}{n_A + n_B - 2}}$ Substituting the values: $s_p = \sqrt{\frac{(10-1)(5^2) + (12-1)(4^2)}{10 + 12 - 2}} = \sqrt{\frac{9(25) + 11(16)}{20}} = \sqrt{\frac{225 + 176}{20}} = \sqrt{20.05} = 4.478$ Now calculate the t-statistic: $t = \frac{85 - 81}{4.478\sqrt{\frac{1}{10} + \frac{1}{12}}} = \frac{4}{4.478\sqrt{0.1833}} = \frac{4}{4.478(0.428)} = \frac{4}{1.916} = 2.09$ 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.
Two assembly lines are compared for average production time per unit. Line A (n=28): $\bar{x}_A = 4.2$ minutes, $s_A = 0.8$ minutes. Line B (n=32): $\bar{x}_B = 4.7$ minutes, $s_B = 0.6$ minutes. A consultant performs a two-sample t-test assuming unequal variances and obtains t = -2.58 with 49.3 degrees of freedom. At $\alpha = 0.05$ , what conclusion should be drawn?
Answer: Reject $H_0$ : the test statistic |t| = 2.58 exceeds t₀.₀₂₅,₄₉ ≈ 2.01, indicating a significant difference between the two production lines For a two-tailed test with α = 0.05 and df ≈ 49, the critical value is approximately 2.01. Since |t| = |-2.58| = 2.58 > 2.01, we reject H₀. Choice A assumes a one-tailed test and adds unsupported commentary about practical significance. Choice B incorrectly suggests we fail to reject. Choice D introduces confidence interval information not provided and reaches the wrong conclusion.
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?
Answer: The p-value for Study A will be smaller than for Study B. 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.
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?
Answer: We would reject H₀ and conclude there is evidence of a difference in churn rates. 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.
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?
Answer: H₀: μ ≤ 3.5, Hₐ: μ > 3.5 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).
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?
Answer: t = -2.50 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.
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 $\alpha = 0.05$ to determine if the new drug has a higher recovery rate, what should they conclude?
Answer: Reject $H_0$ because the test statistic $z = 2.18$ exceeds the critical value of 1.645, indicating the new drug is significantly better For a two-proportion z-test: $\hat{p}_1 = 42/120 = 0.35$ , $\hat{p}_2 = 58/110 = 0.527$ . The pooled proportion is $\hat{p} = (42+58)/(120+110) = 0.435$ . The test statistic is $z = \frac{0.527-0.35}{\sqrt{0.435(1-0.435)(\frac{1}{120}+\frac{1}{110})}} = 2.18$ . Since this exceeds 1.645 (one-tailed critical value), we reject $H_0$ . Choice B uses wrong critical value. Choice C has calculation error. Choice D has major calculation error.
A two-sample t-test comparing the mean sales of two store branches yields a p-value of 0.008. Which of the following statements is a common and serious misinterpretation of this result?
Answer: The probability that the null hypothesis is true, given the observed data, is 0.8%. When you encounter p-values in hypothesis testing, it's crucial to understand exactly what they represent to avoid one of statistics' most common misinterpretations. A p-value tells you the probability of observing your sample results (or more extreme results) assuming the null hypothesis is true. With a p-value of 0.008, you can conclude that if there truly were no difference between the store branches, you'd only see sample differences this large or larger about 0.8% of the time. Answer D represents the classic p-value fallacy. The p-value is not the probability that the null hypothesis is true given your data. That would be P(H₀|data), but the p-value actually gives you P(data|H₀) – the reverse conditional probability. This confusion stems from mixing up frequentist and Bayesian interpretations of probability. Answer A correctly describes what the p-value means – it's the probability of the observed difference under the null hypothesis. Answer B is a reasonable interpretation since such a small p-value suggests the null hypothesis is likely false, meaning a real difference probably exists. Answer C is also correct because 0.008 < 0.01, so the result is statistically significant at the α = 0.01 level. Study tip: Remember that p-values work "backwards" – they assume the null hypothesis is true and ask how likely your data would be. They never tell you the probability that your hypothesis is true. Watch for this reversal trap on exam questions about statistical significance.
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?
Answer: The calculated p-value from the t-test may not be accurate, compromising the validity of the conclusion. 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.
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 $z = 1.68$ 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?
Answer: The director incorrectly interpreted 'fail to reject $H_0$ ' as proving the null hypothesis true, when it only indicates insufficient evidence of a difference 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 $H_0$ 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 $H_0$ " rather than " $H_0$ is true."
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?
Answer: All assumptions appear to be satisfied: both groups have sufficient sample sizes, expected frequencies exceed 5, and independence can be reasonably assumed 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.
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?
Answer: Fail to reject H₀, because the confidence interval contains the value of zero. 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.
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 $\bar{x} = 68,200$ and $s = 12,500$ . After conducting a two-tailed test at $\alpha = 0.05$ , the economist calculates a p-value of 0.032. Which statement best describes the appropriate conclusion and its business interpretation?
Answer: Reject $H_0$ at $\alpha = 0.05$ but not at $\alpha = 0.01$ ; there is moderate evidence that City A's income differs significantly from the national average With p-value = 0.032, we reject $H_0$ at $\alpha = 0.05$ (since 0.032 < 0.05) but would fail to reject at $\alpha = 0.01$ (since 0.032 > 0.01). This indicates moderate evidence against $H_0$ . Choice B is wrong because 0.032 > 0.01. Choice C incorrectly fails to reject when p < $\alpha$ . Choice D introduces practical significance discussion not warranted by the given information and doesn't address the strength of statistical evidence.
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?
Answer: There is statistically significant evidence of a difference between the conversion rates of the two designs. 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.
A financial analyst tests whether the mean daily trading volume for Stock X equals 2.5 million shares. From 25 trading days: $\bar{x} = 2.72$ million, $s = 0.45$ million. She calculates $t = 2.44$ and determines the p-value is 0.022. If she had incorrectly used $\alpha = 0.01$ instead of the intended $\alpha = 0.05$ , how would this affect her conclusion and Type I error risk?
Answer: At $\alpha = 0.01$ , she would fail to reject $H_0$ instead of rejecting it, reducing her Type I error risk from 5% to 1% but increasing Type II error risk 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 $\alpha = 0.05$ , since 0.022 < 0.05, you reject $H_0$ . However, at $\alpha = 0.01$ , since 0.022 > 0.01, you fail to reject $H_0$ . This changes the conclusion entirely. The Type I error risk always equals your chosen alpha level when the null hypothesis is true, so using $\alpha = 0.01$ 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 $H_0$ " 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 $H_0$ ), 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.
A fast-food chain aims to keep its average customer service time under 3 minutes (180 seconds). A random sample of 49 customers reveals a sample mean service time of 172 seconds with a sample standard deviation of 28 seconds. The operations manager conducts a one-sample t-test at a 5% significance level to determine if the chain is meeting its goal. Which of the following correctly states the test statistic and conclusion?
Answer: t = -2.00; reject the null hypothesis, as there is significant evidence the average time is under 180 seconds. The hypotheses are H₀: μ ≥ 180 and Hₐ: μ < 180. The test is a one-tailed t-test. The standard error is SE = s / √n = 28 / √49 = 28 / 7 = 4. The t-statistic is t = (x̄ - μ₀) / SE = (172 - 180) / 4 = -8 / 4 = -2.00. The degrees of freedom are df = n - 1 = 48. For a one-tailed test at α = 0.05 with df=48, the critical t-value is approximately -1.677. Since the calculated t-statistic (-2.00) is less than the critical value (-1.677), we reject the null hypothesis. This indicates there is statistically significant evidence that the average service time is less than 180 seconds.
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?
Answer: The difference is statistically significant, but its practical significance (a $0.25 increase) may be too small to justify the cost. 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 $0.25 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.

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