College Statistics Quiz: Significance Level And Decision Rules
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Significance Level And Decision RulesQuestion 1 of 20

A scientific journal article states that the findings of a particular study were "statistically significant at the 1% level." Which of the following statements is a guaranteed consequence of this declaration?

If the null hypothesis were true, the probability of obtaining the observed results is exactly 0.01.
The effect size found in the study is large and of practical importance.
The null hypothesis is definitively false, and the alternative hypothesis is true.
The p-value obtained from the study's data was less than or equal to 0.01.
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College Statistics Quiz

College Statistics Quiz: Significance Level And Decision Rules

Practice Significance Level And Decision Rules in College 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 Significance Level And Decision Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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.

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Question 1

A scientific journal article states that the findings of a particular study were "statistically significant at the 1% level." Which of the following statements is a guaranteed consequence of this declaration?

  1. If the null hypothesis were true, the probability of obtaining the observed results is exactly 0.01.
  2. The effect size found in the study is large and of practical importance.
  3. The null hypothesis is definitively false, and the alternative hypothesis is true.
  4. The p-value obtained from the study's data was less than or equal to 0.01. (correct answer)
Explanation: When you encounter statements about statistical significance, you're dealing with hypothesis testing and p-values. The phrase "statistically significant at the 1% level" has a very specific meaning that directly relates to the p-value threshold used in the study. Statistical significance at the 1% level means the researchers set their significance level (α) at 0.01 and found results that met this criterion. For results to be statistically significant, the p-value must be less than or equal to the chosen significance level. Therefore, if findings are significant at the 1% level, the p-value must be ≤ 0.01, making answer D correct. Let's examine why the other options are incorrect. Option A misinterprets what a p-value represents—it's the probability of obtaining results at least as extreme as observed, assuming the null hypothesis is true, but it's not necessarily exactly 0.01. Option B confuses statistical significance with practical significance; a statistically significant result doesn't guarantee the effect size is large or practically important. Option C overstates what statistical significance means—we reject the null hypothesis, but this doesn't definitively prove it's false or that the alternative is true; there's always a chance of Type I error. Remember this key distinction: statistical significance tells you about the strength of evidence against the null hypothesis (based on your chosen α level), but it doesn't directly tell you about effect size, practical importance, or absolute truth. Always connect "significant at X% level" directly to p-value thresholds.

Question 2

A 99% confidence interval for the mean salary (in thousands of dollars) of a certain profession is calculated to be (85.2, 98.8). A professional organization wants to test the hypothesis that the mean salary is different from $100,000. That is, H0:μ=100H_0: \mu = 100 versus Ha:μ≠100H_a: \mu \neq 100. Based on the provided confidence interval, what is the appropriate significance level for an equivalent hypothesis test and what is the correct decision?

  1. The significance level is 0.05, and H0H_0 should be rejected.
  2. The significance level is 0.01, and H0H_0 should not be rejected.
  3. The significance level is 0.005, and H0H_0 should be rejected.
  4. The significance level is 0.01, and H0H_0 should be rejected. (correct answer)
Explanation: When you encounter a problem linking confidence intervals to hypothesis testing, remember that there's a direct mathematical relationship between confidence levels and significance levels: they must sum to 1. A 99% confidence interval corresponds to a significance level of α=1−0.99=0.01\alpha = 1 - 0.99 = 0.01. To make the testing decision, check whether the hypothesized value falls within the confidence interval. The given 99% confidence interval is (85.2, 98.8) thousand dollars, and we're testing whether μ=100\mu = 100 thousand dollars. Since 100 is not contained within the interval (85.2, 98.8), we reject H0H_0 at the 0.01 significance level. This happens because confidence intervals and two-tailed hypothesis tests are mathematically equivalent procedures. If the hypothesized value lies outside the confidence interval, the p-value will be less than the significance level, leading to rejection. Looking at the wrong answers: A uses the incorrect significance level of 0.05, which would correspond to a 95% confidence interval, not 99%. B correctly identifies the 0.01 significance level but incorrectly fails to reject H0H_0 - since 100 falls outside our interval, rejection is warranted. C uses 0.005, which would correspond to a 99.5% confidence interval, showing confusion about the confidence level conversion. Study tip: Always remember that confidence level + significance level = 1, and a value outside the confidence interval means you reject H0H_0 at the corresponding significance level. This dual relationship makes confidence intervals a powerful tool for quick hypothesis testing decisions.

Question 3

A geneticist is testing 20 different, independent genes for association with a certain disease. For each gene, a separate hypothesis test is conducted with a significance level of α=0.05\alpha = 0.05. Assuming that, in reality, none of the genes are associated with the disease (i.e., all 20 null hypotheses are true), what is the probability that the geneticist will find at least one statistically significant result?

  1. Exactly 0.05
  2. Exactly 1.00
  3. Approximately 0.64 (correct answer)
  4. Approximately 0.36
Explanation: This is a problem of family-wise error rate. For a single test where H0H_0 is true, the probability of a Type I error is α=0.05\alpha = 0.05. The probability of correctly not rejecting H0H_0 is 1−α=0.951 - \alpha = 0.95. Since the 20 tests are independent, the probability of correctly not rejecting H0H_0 in all 20 tests is (0.95)20≈0.358(0.95)^{20} \approx 0.358. The probability of at least one Type I error (at least one significant result) is the complement of this event: 1−(0.95)20≈1−0.358=0.6421 - (0.95)^{20} \approx 1 - 0.358 = 0.642.

Question 4

A researcher is conducting a two-tailed hypothesis test for a population proportion and calculates a standardized test statistic of z=2.10z = 2.10. The researcher wants to know at which of the standard significance levels the null hypothesis would be rejected. At which of the following significance levels is the result statistically significant?

  1. At α=0.10\alpha = 0.10 and α=0.05\alpha = 0.05, but not at α=0.01\alpha = 0.01. (correct answer)
  2. Only at α=0.10\alpha = 0.10.
  3. At α=0.10\alpha = 0.10, α=0.05\alpha = 0.05, and α=0.01\alpha = 0.01.
  4. Only at α=0.05\alpha = 0.05.
Explanation: First, calculate the p-value for the test statistic z=2.10z = 2.10. Since the test is two-tailed, the p-value is 2×P(Z>2.10)2 \times P(Z > 2.10). From a standard normal table, P(Z>2.10)≈0.0179P(Z > 2.10) \approx 0.0179. So, the p-value is 2×0.0179=0.03582 \times 0.0179 = 0.0358. The decision rule is to reject H0H_0 if the p-value is less than or equal to α\alpha. We check this for the given levels: Is 0.0358≤0.100.0358 \le 0.10? Yes. Is 0.0358≤0.050.0358 \le 0.05? Yes. Is 0.0358≤0.010.0358 \le 0.01? No. Therefore, the result is significant at the 0.10 and 0.05 levels, but not at the 0.01 level.

Question 5

An analyst conducts a hypothesis test to determine if there is a significant difference in the mean response times for two different website designs. After collecting data, she calculates a p-value of 0.028. Which of the following is the most accurate conclusion based on this p-value?

  1. The null hypothesis would be rejected at a significance level of 0.05, but not at a significance level of 0.01. (correct answer)
  2. The probability that the null hypothesis is true is 0.028, indicating that the two designs are likely different.
  3. The null hypothesis would be rejected at any significance level less than 0.028, such as 0.01.
  4. If the study were repeated many times, a Type I error would occur in approximately 2.8% of the repetitions.
Explanation: The decision rule for a p-value is to reject the null hypothesis H0H_0 if the p-value is less than or equal to the significance level α\alpha. With a p-value of 0.028, we compare it to standard significance levels. Since 0.028≤0.050.028 \le 0.05, we would reject H0H_0 at the 5% level. Since 0.028>0.010.028 > 0.01, we would fail to reject H0H_0 at the 1% level. This makes option A the most complete and accurate conclusion regarding standard practice. Option B is a common misinterpretation of the p-value. Option C has the inequality reversed. Option D misinterprets the p-value as the actual Type I error rate for future studies.

Question 6

A quality control engineer is testing whether the mean diameter of a manufactured bolt is equal to a target of 20 mm. The null hypothesis is H0:μ=20H_0: \mu = 20 and the alternative is Ha:μ≠20H_a: \mu \neq 20. From past data, the population standard deviation is known to be σ=0.4\sigma = 0.4 mm. A sample of n=64n=64 bolts will be taken. If the engineer sets the significance level to α=0.10\alpha = 0.10, what is the decision rule in terms of the sample mean diameter, xˉ\bar{x}?

  1. Reject H0H_0 if xˉ<19.918\bar{x} < 19.918 or xˉ>20.082\bar{x} > 20.082. (correct answer)
  2. Reject H0H_0 if xˉ<19.902\bar{x} < 19.902 or xˉ>20.098\bar{x} > 20.098.
  3. Reject H0H_0 if xˉ<19.936\bar{x} < 19.936 or xˉ>20.064\bar{x} > 20.064.
  4. Reject H0H_0 if xˉ<19.872\bar{x} < 19.872 or xˉ>20.128\bar{x} > 20.128.
Explanation: This is a two-tailed z-test. For α=0.10\alpha = 0.10, the significance is split into two tails, so we need the critical z-value for an area of α/2=0.05\alpha/2 = 0.05 in each tail. The critical values are zα/2=±1.645z_{\alpha/2} = \pm 1.645. The standard error of the mean is SE=σ/n=0.4/64=0.4/8=0.05SE = \sigma / \sqrt{n} = 0.4 / \sqrt{64} = 0.4 / 8 = 0.05. The critical values for the sample mean xˉ\bar{x} are calculated as μ0±zα/2×SE\mu_0 \pm z_{\alpha/2} \times SE. This gives 20±1.645×0.05=20±0.0822520 \pm 1.645 \times 0.05 = 20 \pm 0.08225. Thus, the decision rule is to reject H0H_0 if xˉ\bar{x} is less than 19.918 or greater than 20.082.

Question 7

A researcher conducts a two-sample t-test using a significance level of α=0.05\alpha=0.05, assuming the two populations have equal variances. In reality, the two populations have substantially different variances, and the sample sizes are unequal. What is the most likely consequence of this violated assumption on the actual Type I error rate of the test?

  1. The actual Type I error rate will be unaffected, but the power of the test will be substantially reduced.
  2. The actual Type I error rate may be higher or lower than 0.05, depending on the relationship between sample sizes and variances. (correct answer)
  3. The actual Type I error rate will be more conservative and thus strictly less than 0.05.
  4. The actual Type I error rate will be more liberal and thus strictly greater than 0.05.
Explanation: When you encounter questions about assumption violations in t-tests, focus on how the violation interacts with your study design—particularly the relationship between sample sizes and population variances. The two-sample t-test assumes equal population variances (homoscedasticity). When this assumption is violated with unequal sample sizes, the actual Type I error rate depends on a specific pattern: if the smaller sample comes from the population with larger variance, the test becomes liberal (Type I error rate > 0.05). Conversely, if the smaller sample comes from the population with smaller variance, the test becomes conservative (Type I error rate < 0.05). Since we don't know which scenario applies here, the actual error rate could go either direction. Choice A is incorrect because violated assumptions in t-tests typically affect both Type I error rates and power, not just power alone. The test's validity is compromised when key assumptions fail. Choice C is wrong because it claims the test will always be conservative. This only happens in one specific scenario (smaller sample from population with smaller variance), not universally. Choice D makes the opposite error, claiming the test will always be liberal. Again, this only occurs when the smaller sample comes from the population with larger variance. Study tip: Remember the key interaction for unequal variances + unequal sample sizes: smaller sample with larger variance = liberal test; smaller sample with smaller variance = conservative test. When you don't know the specifics, the direction of bias is unpredictable.

Question 8

A pharmaceutical company develops a new drug and tests the null hypothesis that the drug has no effect. To gain regulatory approval, the company must provide strong evidence that the drug is effective. The research team decides to change their significance level for the pivotal clinical trial from α=0.05\alpha = 0.05 to α=0.01\alpha = 0.01. Which of the following describes the primary trade-off of making this change?

  1. The probability of approving an ineffective drug decreases, but the probability of failing to approve an effective drug increases. (correct answer)
  2. The probability of approving an ineffective drug increases, and the probability of failing to approve an effective drug decreases.
  3. Both the probability of approving an ineffective drug and the probability of failing to approve an effective drug will decrease.
  4. The power of the test increases, making it more likely that both effective and ineffective drugs are correctly identified.
Explanation: Changing the significance level α\alpha from 0.05 to 0.01 makes the criterion for rejecting the null hypothesis stricter. This decreases the probability of a Type I error (rejecting a true null hypothesis), which in this context means approving an ineffective drug. However, there is a trade-off: decreasing the probability of a Type I error (α\alpha) generally increases the probability of a Type II error (β\beta), which is failing to reject a false null hypothesis. In this context, a Type II error means failing to approve an effective drug. The power of the test, which is 1−β1 - \beta, will therefore decrease.

Question 9

An investigator is performing a right-tailed hypothesis test for a population mean, with H0:μ=μ0H_0: \mu = \mu_0 and Ha:μ>μ0H_a: \mu > \mu_0. The decision rule is to reject H0H_0 if the sample mean xˉ\bar{x} exceeds a certain critical value, C. If the investigator keeps the significance level α\alpha constant but increases the sample size fourfold (from nn to 4n4n), how will the critical value C be affected?

  1. The value of C will move closer to the hypothesized mean μ0\mu_0. (correct answer)
  2. The value of C will move farther away from the hypothesized mean μ0\mu_0.
  3. The value of C will not change because the significance level α\alpha is held constant.
  4. The value of C will double its distance from the hypothesized mean μ0\mu_0.
Explanation: The critical value C for the sample mean xˉ\bar{x} is determined by the formula C=μ0+zασnC = \mu_0 + z_\alpha \frac{\sigma}{\sqrt{n}}. The z-score zαz_\alpha is determined by α\alpha and does not change. When the sample size nn is replaced by 4n4n, the new critical value C′C' becomes C′=μ0+zασ4n=μ0+zασ2nC' = \mu_0 + z_\alpha \frac{\sigma}{\sqrt{4n}} = \mu_0 + z_\alpha \frac{\sigma}{2\sqrt{n}}. The term added to μ0\mu_0 is now half its original size. Therefore, the new critical value C′C' is closer to the hypothesized mean μ0\mu_0.

Question 10

In many judicial systems, a defendant is presumed innocent until proven guilty. This can be framed as a hypothesis test where the null hypothesis is that the defendant is innocent. A Type I error is convicting an innocent person, while a Type II error is acquitting a guilty person. The standard of "proof beyond a reasonable doubt" is required to convict. How does this legal standard relate to the components of a hypothesis test?

  1. It corresponds to setting the probabilities of Type I and Type II errors to be approximately equal to balance societal risks.
  2. It corresponds to selecting a very large significance level α\alpha to maximize the power to convict guilty defendants.
  3. It corresponds to selecting a very small significance level α\alpha to minimize the probability of a Type I error. (correct answer)
  4. It suggests that the power of the test, 1−β1 - \beta, should be close to 1, while the significance level α\alpha is irrelevant.
Explanation: When you encounter hypothesis testing scenarios in real-world contexts, focus on how the consequences of different errors shape the test parameters. In this legal framework, the null hypothesis is innocence, making conviction equivalent to rejecting the null hypothesis. The "proof beyond a reasonable doubt" standard is specifically designed to protect innocent defendants from wrongful conviction. This directly translates to minimizing Type I error - the probability of rejecting a true null hypothesis (convicting an innocent person). To achieve this protection, you need a very small significance level α\alpha, requiring extremely strong evidence before rejecting the presumption of innocence. Answer C correctly identifies this relationship. Answer A is wrong because the legal system deliberately does not balance these error types equally. Society views convicting the innocent as far worse than acquitting the guilty, so the errors are weighted differently, not set to be approximately equal. Answer B contradicts the legal principle entirely. A large α\alpha would make it easier to convict defendants, increasing the risk of wrongful convictions - exactly what "reasonable doubt" aims to prevent. Answer D misses the point by claiming α\alpha is irrelevant. The significance level is actually the cornerstone of this legal standard. While high power (ability to convict the guilty) might be desirable, it's secondary to protecting the innocent through a small α\alpha. Remember: When analyzing hypothesis tests in applied contexts, identify which type of error has more serious consequences. The testing parameters will be designed to minimize that more critical error type.

Question 11

A marketing analyst performs a two-tailed t-test with 24 degrees of freedom to compare the mean effectiveness of two advertising campaigns. A significance level of α=0.05\alpha = 0.05 is chosen for the test. Which statement accurately describes the rejection region?

  1. The set of t-values between −t0.05,24-t_{0.05, 24} and t0.05,24t_{0.05, 24}.
  2. The set of t-values greater than t0.05,24t_{0.05, 24}.
  3. The set of t-values greater than t0.025,24t_{0.025, 24} or less than −t0.025,24-t_{0.025, 24}. (correct answer)
  4. The set of t-values for which the corresponding p-value is greater than 0.05.
Explanation: When you encounter a two-tailed t-test, you're looking for evidence that the parameter differs from the null hypothesis value in either direction. The key insight is understanding how the significance level gets distributed across both tails of the t-distribution. In a two-tailed test with α=0.05\alpha = 0.05, you split this probability equally between the two tails: 0.025 in each tail. This means you reject the null hypothesis if your test statistic falls in the most extreme 2.5% of either tail. The rejection region consists of t-values greater than t0.025,24t_{0.025, 24} (right tail) or less than −t0.025,24-t_{0.025, 24} (left tail), making option C correct. Option A incorrectly uses t0.05,24t_{0.05, 24}, which would be appropriate for a one-tailed test but puts too much probability (5%) in each tail for a two-tailed test. Option B describes only a one-tailed test's rejection region, missing the left tail entirely. Option D fundamentally misunderstands rejection regions – you reject when the p-value is less than 0.05, not greater than it. The critical values ±t0.025,24\pm t_{0.025, 24} create boundaries where exactly 2.5% of the distribution lies beyond each point, totaling the desired 5% significance level across both tails. Study tip: For two-tailed tests, always divide your alpha level by 2 when finding critical values. Remember the pattern: two-tailed test means "alpha over 2" in each tail. This applies to z-tests, t-tests, and other hypothesis tests.

Question 12

In a test of H0:μ=30H_0: \mu = 30 against Ha:μ≠30H_a: \mu \neq 30, a p-value of 0.35 is obtained. The pre-specified significance level was α=0.05\alpha = 0.05. Which of the following is the most appropriate conclusion?

  1. The null hypothesis is accepted as true because the p-value is large.
  2. The probability that the null hypothesis is true, given the data, is 0.35.
  3. The data do not provide sufficient evidence to reject the null hypothesis at the 5% significance level. (correct answer)
  4. There is a 35% probability that a Type II error has been committed.
Explanation: The decision rule is to reject H0H_0 if the p-value ≤α\le \alpha. Here, 0.35>0.050.35 > 0.05, so we fail to reject the null hypothesis. The phrase "fail to reject" is used because the absence of evidence against H0H_0 is not proof that H0H_0 is true. Therefore, the most accurate statement is that the evidence is insufficient to reject H0H_0. Option A incorrectly uses the word "accepted." Option B is a common misinterpretation of the p-value. Option D incorrectly equates the p-value with β\beta, the probability of a Type II error.

Question 13

A scientist computes a test statistic z=1.75z = 1.75 from an experiment. The original research plan called for a two-tailed test at α=0.05\alpha = 0.05. However, based on a new theoretical development, the scientist now believes a one-tailed test with Ha:μ>μ0H_a: \mu > \mu_0 is more appropriate. How does this change in the alternative hypothesis affect the conclusion of the test, assuming α\alpha remains 0.05?

  1. The conclusion does not change; H0H_0 is not rejected under either scenario.
  2. The conclusion changes from rejecting H0H_0 to failing to reject H0H_0.
  3. The conclusion does not change; H0H_0 is rejected under both scenarios.
  4. The conclusion changes from failing to reject H0H_0 to rejecting H0H_0. (correct answer)
Explanation: When you encounter questions about changing from two-tailed to one-tailed tests, focus on how the critical values shift and what this means for your test statistic's position relative to the rejection region. With a test statistic of z=1.75z = 1.75 and α=0.05\alpha = 0.05, let's examine both scenarios. In a two-tailed test, you split the alpha level between both tails, so each tail gets α/2=0.025\alpha/2 = 0.025. The critical values are ±1.96\pm 1.96. Since 1.75<1.961.75 < 1.96, you fail to reject H0H_0 in the original two-tailed test. However, when switching to a one-tailed test with Ha:μ>μ0H_a: \mu > \mu_0, the entire α=0.05\alpha = 0.05 goes into the right tail. The critical value becomes z=1.645z = 1.645. Since your test statistic z=1.75>1.645z = 1.75 > 1.645, you now reject H0H_0. The conclusion changes from failing to reject to rejecting the null hypothesis. Answer A is wrong because the conclusion does change between the two scenarios. Answer B reverses the direction of change—it suggests you initially rejected H0H_0 but then failed to reject it, which is backwards. Answer C is incorrect because H0H_0 is not rejected in the two-tailed scenario since 1.75<1.961.75 < 1.96. The correct answer is D: the conclusion changes from failing to reject H0H_0 to rejecting H0H_0. Study tip: Remember that one-tailed tests have more power to detect effects in the predicted direction because they concentrate all the alpha in one tail, making the critical value less extreme.

Question 14

A screening test is being developed for a rare but serious disease. The null hypothesis is that the person does not have the disease. A Type I error (a false positive) leads to anxiety and further expensive, invasive testing. A Type II error (a false negative) means a person with the disease is not identified, and treatment is delayed. Which statement best guides the choice of a significance level α\alpha for this test?

  1. The choice of α\alpha is less important than minimizing β\beta, so the focus should be on maximizing sample size.
  2. A small α\alpha (e.g., 0.01) should be chosen to minimize the number of healthy people who undergo unnecessary invasive testing.
  3. α\alpha should be set to 0.05 because it is the universally accepted standard for medical tests.
  4. A relatively large α\alpha (e.g., 0.10) should be chosen to minimize the chance of missing a person with the disease. (correct answer)
Explanation: When analyzing hypothesis testing in medical screening, you need to weigh the relative consequences of Type I and Type II errors. In this scenario, a Type I error means telling a healthy person they might have the disease (false positive), while a Type II error means missing someone who actually has the disease (false negative). The significance level α\alpha directly controls the probability of Type I errors, and there's an inverse relationship between α\alpha and β\beta (Type II error probability). When you decrease α\alpha, you typically increase β\beta, and vice versa. Here, the consequences differ dramatically. A false positive causes anxiety and unnecessary testing—certainly unpleasant and expensive, but not life-threatening. A false negative means someone with a serious disease goes undiagnosed and untreated, potentially leading to severe health consequences or death. The cost of missing the disease far outweighs the cost of a false alarm. Answer D is correct because using a relatively large α\alpha (like 0.10) makes the test more sensitive, reducing the chance of missing someone with the disease, even though it increases false positives. Answer A incorrectly suggests α\alpha choice is less important—both error types matter significantly in medical contexts. Answer B advocates for a small α\alpha, which would reduce false positives but dangerously increase the chance of missing actual cases. Answer C wrongly assumes 0.05 is universally appropriate regardless of context. Key strategy: In medical screening questions, always consider which error has more serious real-world consequences, then choose the significance level that minimizes the more dangerous error type.

Question 15

A researcher performs a hypothesis test with a pre-specified significance level of α=0.10\alpha = 0.10 and obtains a p-value of 0.08. Which of the following is the correct formal decision and interpretation?

  1. The null hypothesis is not rejected because the p-value is not less than the conventional 0.05 level.
  2. The null hypothesis is rejected, and the result is considered statistically significant at the 10% level. (correct answer)
  3. The result is inconclusive, and the researcher should collect more data until the p-value is below 0.05.
  4. The null hypothesis is not rejected, and there is an 8% chance that the alternative hypothesis is actually true.
Explanation: When you encounter hypothesis testing questions, the key relationship to remember is comparing your p-value to your pre-specified significance level (α). The decision rule is straightforward: reject the null hypothesis if p-value < α, and fail to reject if p-value ≥ α. In this problem, you have α = 0.10 and p-value = 0.08. Since 0.08 < 0.10, you reject the null hypothesis and conclude the result is statistically significant at the 10% level. This makes option B correct. Let's examine why the other choices are wrong: Option A makes a critical error by using 0.05 instead of the pre-specified α = 0.10. The conventional 0.05 level is irrelevant here—you must use the significance level established before conducting the test. This is a common trap that tests whether you understand the importance of pre-specification. Option C is incorrect because hypothesis tests provide definitive decisions based on the chosen α level. There's no "inconclusive" middle ground, and the suggestion to collect more data until reaching p < 0.05 violates proper statistical practice by changing criteria post-hoc. Option D contains a fundamental misinterpretation of p-values. The p-value represents the probability of observing your data (or more extreme) assuming the null hypothesis is true—not the probability that the alternative hypothesis is true. This is one of the most common p-value misconceptions. Study tip: Always use the pre-specified α level for your decision, never the conventional 0.05 unless that was your original choice. And remember: p-values are probabilities about your data given the null hypothesis, not probabilities about hypotheses themselves.

Question 16

For a left-tailed t-test with 15 degrees of freedom, the null hypothesis is rejected if the test statistic is less than the critical value tcrit=−1.753t_{crit} = -1.753. This critical value corresponds to a significance level of α=0.05\alpha = 0.05. If an experiment yields a test statistic of exactly tobs=−1.753t_{obs} = -1.753, what is the p-value of the test and the correct decision?

  1. The p-value is 0.05, and the null hypothesis is not rejected.
  2. The p-value is 0.05, and the null hypothesis is rejected. (correct answer)
  3. The p-value is 0.10, and the null hypothesis is rejected.
  4. The p-value cannot be determined, and the result is considered a borderline case with no clear decision.
Explanation: When you encounter hypothesis testing problems, understanding the relationship between test statistics, critical values, and p-values is crucial for making correct decisions. In a left-tailed t-test, you reject the null hypothesis when your test statistic falls in the rejection region (below the critical value). Here, with tobs=−1.753t_{obs} = -1.753 and tcrit=−1.753t_{crit} = -1.753, your observed test statistic equals the critical value exactly. Since the rejection rule states "reject if less than −1.753-1.753," this borderline case technically falls outside the rejection region by the strictest interpretation. However, when the test statistic equals the critical value, the p-value equals the significance level (α=0.05\alpha = 0.05), placing you exactly at the decision boundary. The correct answer is B because the p-value is indeed 0.05, and in practice, when p=αp = \alpha, we reject the null hypothesis. This follows the standard decision rule: reject when p≤αp \leq \alpha. Option A incorrectly suggests not rejecting when p=α=0.05p = \alpha = 0.05. The standard convention is to reject when the p-value equals the significance level. Option C wrongly calculates the p-value as 0.10 – this would occur if the critical value corresponded to α=0.10\alpha = 0.10, not 0.05. Option D incorrectly claims the p-value cannot be determined, when it's clearly 0.05 since the test statistic equals the critical value. Study tip: Remember that when your test statistic exactly equals the critical value, the p-value equals your chosen significance level, and the standard convention is to reject the null hypothesis in this boundary case.

Question 17

A pharmaceutical company develops a new drug and tests the null hypothesis that the drug has no effect. To gain regulatory approval, the company must provide strong evidence that the drug is effective. The research team decides to change their significance level for the pivotal clinical trial from α=0.05\alpha = 0.05 to α=0.01\alpha = 0.01. Which of the following describes the primary trade-off of making this change?

  1. The probability of approving an ineffective drug decreases, but the probability of failing to approve an effective drug increases. (correct answer)
  2. The probability of approving an ineffective drug increases, and the probability of failing to approve an effective drug decreases.
  3. Both the probability of approving an ineffective drug and the probability of failing to approve an effective drug will decrease.
  4. The power of the test increases, making it more likely that both effective and ineffective drugs are correctly identified.
Explanation: Changing the significance level α\alpha from 0.05 to 0.01 makes the criterion for rejecting the null hypothesis stricter. This decreases the probability of a Type I error (rejecting a true null hypothesis), which in this context means approving an ineffective drug. However, there is a trade-off: decreasing the probability of a Type I error (α\alpha) generally increases the probability of a Type II error (β\beta), which is failing to reject a false null hypothesis. In this context, a Type II error means failing to approve an effective drug. The power of the test, which is 1−β1 - \beta, will therefore decrease.

Question 18

A quality control engineer is testing whether the mean diameter of a manufactured bolt is equal to a target of 20 mm. The null hypothesis is H0:μ=20H_0: \mu = 20 and the alternative is Ha:μ≠20H_a: \mu \neq 20. From past data, the population standard deviation is known to be σ=0.4\sigma = 0.4 mm. A sample of n=64n=64 bolts will be taken. If the engineer sets the significance level to α=0.10\alpha = 0.10, what is the decision rule in terms of the sample mean diameter, xˉ\bar{x}?

  1. Reject H0H_0 if xˉ<19.918\bar{x} < 19.918 or xˉ>20.082\bar{x} > 20.082. (correct answer)
  2. Reject H0H_0 if xˉ<19.902\bar{x} < 19.902 or xˉ>20.098\bar{x} > 20.098.
  3. Reject H0H_0 if xˉ<19.936\bar{x} < 19.936 or xˉ>20.064\bar{x} > 20.064.
  4. Reject H0H_0 if xˉ<19.872\bar{x} < 19.872 or xˉ>20.128\bar{x} > 20.128.
Explanation: This is a two-tailed z-test. For α=0.10\alpha = 0.10, the significance is split into two tails, so we need the critical z-value for an area of α/2=0.05\alpha/2 = 0.05 in each tail. The critical values are zα/2=±1.645z_{\alpha/2} = \pm 1.645. The standard error of the mean is SE=σ/n=0.4/64=0.4/8=0.05SE = \sigma / \sqrt{n} = 0.4 / \sqrt{64} = 0.4 / 8 = 0.05. The critical values for the sample mean xˉ\bar{x} are calculated as μ0±zα/2×SE\mu_0 \pm z_{\alpha/2} \times SE. This gives 20±1.645×0.05=20±0.0822520 \pm 1.645 \times 0.05 = 20 \pm 0.08225. Thus, the decision rule is to reject H0H_0 if xˉ\bar{x} is less than 19.918 or greater than 20.082.

Question 19

An investigator is performing a right-tailed hypothesis test for a population mean, with H0:μ=μ0H_0: \mu = \mu_0 and Ha:μ>μ0H_a: \mu > \mu_0. The decision rule is to reject H0H_0 if the sample mean xˉ\bar{x} exceeds a certain critical value, C. If the investigator keeps the significance level α\alpha constant but increases the sample size fourfold (from nn to 4n4n), how will the critical value C be affected?

  1. The value of C will move closer to the hypothesized mean μ0\mu_0. (correct answer)
  2. The value of C will move farther away from the hypothesized mean μ0\mu_0.
  3. The value of C will not change because the significance level α\alpha is held constant.
  4. The value of C will double its distance from the hypothesized mean μ0\mu_0.
Explanation: The critical value C for the sample mean xˉ\bar{x} is determined by the formula C=μ0+zασnC = \mu_0 + z_\alpha \frac{\sigma}{\sqrt{n}}. The z-score zαz_\alpha is determined by α\alpha and does not change. When the sample size nn is replaced by 4n4n, the new critical value C′C' becomes C′=μ0+zασ4n=μ0+zασ2nC' = \mu_0 + z_\alpha \frac{\sigma}{\sqrt{4n}} = \mu_0 + z_\alpha \frac{\sigma}{2\sqrt{n}}. The term added to μ0\mu_0 is now half its original size. Therefore, the new critical value C′C' is closer to the hypothesized mean μ0\mu_0.

Question 20

A researcher is conducting a two-tailed hypothesis test for a population proportion and calculates a standardized test statistic of z=2.10z = 2.10. The researcher wants to know at which of the standard significance levels the null hypothesis would be rejected. At which of the following significance levels is the result statistically significant?

  1. At α=0.10\alpha = 0.10 and α=0.05\alpha = 0.05, but not at α=0.01\alpha = 0.01. (correct answer)
  2. Only at α=0.10\alpha = 0.10.
  3. At α=0.10\alpha = 0.10, α=0.05\alpha = 0.05, and α=0.01\alpha = 0.01.
  4. Only at α=0.05\alpha = 0.05.
Explanation: First, calculate the p-value for the test statistic z=2.10z = 2.10. Since the test is two-tailed, the p-value is 2×P(Z>2.10)2 \times P(Z > 2.10). From a standard normal table, P(Z>2.10)≈0.0179P(Z > 2.10) \approx 0.0179. So, the p-value is 2×0.0179=0.03582 \times 0.0179 = 0.0358. The decision rule is to reject H0H_0 if the p-value is less than or equal to α\alpha. We check this for the given levels: Is 0.0358≤0.100.0358 \le 0.10? Yes. Is 0.0358≤0.050.0358 \le 0.05? Yes. Is 0.0358≤0.010.0358 \le 0.01? No. Therefore, the result is significant at the 0.10 and 0.05 levels, but not at the 0.01 level.