Free math worksheets

- Look for a statement that begins by assuming the null claim. - The probability must describe sample results in the direction of the alternative. - Distinguish a tail probability from both a parameter estimate and a probability that a hypothesis is true.

1. Statement II is correct because it conditions on the null value and describes results at least as extreme in the direction of the left-tailed alternative. 2. Statement I assigns a probability to the truth of the null hypothesis, which a p-value does not do. 3. Statement III does not define a probability event and incorrectly treats “caused by chance” as a quantified conclusion. 4. Statement IV confuses the p-value with an estimate of the population parameter.

a) Statement II. b) I assigns probability to \(H_0\); III misstates what event has probability \(0.28\); IV confuses the p-value with the population proportion.

- Compare the event described by the student with a full tail of the null distribution. - Match the direction of “more extreme” to the right-tailed alternative. - Think about why one exact outcome would not summarize evidence in the tail.

1. The student describes the probability of exactly one sample proportion, but a p-value includes the observed result and all results more extreme in the direction of the alternative. 2. A correct interpretation is: assuming the null hypothesis is true, the probability of obtaining a sample proportion at least as large as the observed sample proportion is \(0.070\). 3. Including more extreme outcomes measures how unusual the observed result or a stronger result would be under the null model; the probability of one exact statistic alone is not the p-value.

a) The p-value is not the probability of obtaining exactly the observed sample proportion. b) Assuming \(H_0\) is true, there is a \(7.0\%\) chance of obtaining a sample proportion at least as large as the one observed. c) The p-value accumulates the observed result and all results more supportive of the alternative.

- Compare the p-value marker with the cutoff selected for the actual study. - Do not replace a preselected cutoff with a default convention. - Keep the numerical evidence fixed while changing only the decision standard.

1. The display shows that the study’s p-value is \(0.090\) and its preselected significance level is \(0.10\). Because \(0.090<0.10\), reject \(H_0\); the result is statistically significant at the preselected level. 2. The reviewer substituted a conventional \(0.05\) cutoff for the significance level actually chosen before the study. Statistical significance is determined by comparing the p-value with the study’s preselected \(\alpha\). 3. If \(\alpha=0.05\) had been preselected, then \(0.090>0.05\), so the result would not be statistically significant.

a) Reject \(H_0\); the result is significant at \(\alpha=0.10\). b) The reviewer used \(0.05\) instead of the study’s preselected significance level. c) No. At \(\alpha=0.05\), the test would fail to reject \(H_0\).

- Recall which directions are included in a two-sided alternative. - Consider symmetric test statistics with equal magnitude and opposite signs. - Separate the amount of evidence from the direction of the observed effect.

1. A two-sided p-value combines evidence from departures in both directions. 2. The same two-sided tail area can result from a positive standardized statistic or its negative mirror. 3. Therefore, the p-value gives the strength of incompatibility with the null value but not the direction of the observed difference.

No. The observed sample proportion could have been above or below the null proportion. The sign of the test statistic or the sample estimate is needed to determine direction.

- Begin the interpretation by temporarily treating the null hypothesis as true. - Match “as extreme or more extreme” to the direction of the alternative hypothesis. - Distinguish a probability about sample outcomes from a probability about a hypothesis.

1. Assuming the true on-time proportion is \(0.78\), the probability of obtaining a sample proportion at least as large as the observed one is \(0.032\). 2. Because \(0.032<0.05\), reject \(H_0\). There is convincing statistical evidence that the true on-time proportion is greater than \(0.78\). 3. A p-value is calculated under the assumption that \(H_0\) is true; it is not the probability that \(H_0\) is true.

a) If the true on-time proportion is \(0.78\), there is a \(3.2\%\) chance of obtaining a sample proportion at least as large as the one observed. b) Reject \(H_0\); there is convincing evidence that the true on-time proportion exceeds \(0.78\). c) The p-value is a probability about possible sample results under \(H_0\), not a probability assigned to \(H_0\).

- In a simulation, the relevant proportion of trials estimates the tail probability. - Use the alternative hypothesis to decide which simulated results count as at least as extreme. - Compare the estimated probability with the stated decision threshold.

1. The estimated p-value is \(\frac{17}{1000}=0.017\). 2. If the true proportion of students who regularly bike to school is \(0.30\), about \(1.7\%\) of random samples would produce a sample proportion at least as large as the observed one. 3. Because \(0.017<0.05\), reject \(H_0\). There is convincing evidence that more than \(30\%\) of students at the school regularly bike to school.

a) \(0.017\). b) Assuming \(p=0.30\), there is about a \(1.7\%\) chance of obtaining a sample proportion at least as large as the observed one. c) Reject \(H_0\); the data provide convincing evidence that the true proportion exceeds \(0.30\).

- Ask what is assumed before a p-value is calculated. - Separate “not enough evidence against” from “evidence in favor of.” - Use decision language that reflects uncertainty rather than proof.

1. The p-value is not the probability that the null hypothesis is correct. It is a probability about sample results calculated under the assumption that \(H_0\) is true. 2. A large p-value does not justify accepting \(H_0\); it leads to failing to reject \(H_0\). 3. The observed result is not unusual under the null model, so the data do not provide convincing evidence for the one-sided alternative hypothesis.

The student incorrectly treats the p-value as \(P(H_0\text{ is true})\) and incorrectly says to accept \(H_0\). The correct conclusion is to fail to reject \(H_0\); the data do not provide convincing evidence for the alternative.

- For each simulation study, compare the qualifying outcomes with the total number generated. - Small differences between independent simulation estimates are expected. - Consider how the number of repetitions affects random variability in an estimated proportion.

1. Student A's estimate is \(\frac{12}{200}=0.060\). 2. Student B's estimate is \(\frac{58}{1000}=0.058\). 3. The estimates differ by only \(0.002\), which is a plausible result of simulation variability, so they are not meaningfully inconsistent. 4. Student B's estimate should generally be more stable because it is based on more simulated samples, reducing Monte Carlo variability.

a) Student A: \(0.060\); Student B: \(0.058\). b) No. The difference of \(0.002\) is small and can result from simulation variability. c) Student B's estimate, because it uses more simulations.

- Match the tail probability to the direction stated in the alternative. - Locate the observed statistic relative to the null value. - Distinguish failure to support one alternative from support for the opposite alternative.

1. Assuming \(p=0.25\), the probability of obtaining a sample proportion at least as large as \(0.22\) is \(0.82\). 2. The observed value lies below the null value, opposite the direction specified by \(H_a:p>0.25\). Therefore, most of the null distribution lies at or above the observed value, producing a large right-tail probability. 3. The stated test was designed to gather evidence for \(p>0.25\), not for \(p<0.25\). A large p-value means the test does not provide evidence for its alternative; it does not serve as the result of a separate left-tailed test.

a) If \(p=0.25\), there is an \(82\%\) chance of obtaining a sample proportion at least as large as \(0.22\). b) The observed proportion is in the direction opposite to the right-tailed alternative. c) No. This right-tailed test does not establish evidence for \(p<0.25\); that would require a separately specified left-tailed test.

- Compare the observed marker with the marked significance cutoff. - Notice the gaps between the attainable p-values. - Keep the conditioning direction in a p-value interpretation clear.

1. The display shows that the observed p-value is \(0.041\) and \(\alpha=0.050\). Because \(0.041<0.050\), reject \(H_0\). 2. The neighboring attainable values are \(0.067\) and \(0.024\). The binomial test statistic is discrete. Increasing the observed count by one removes an entire probability mass from the right tail, so attainable p-values change in jumps rather than continuously. 3. The p-value is a tail probability calculated under the assumption that the null hypothesis is true. It does not assign a probability to the hypothesis itself.

a) Reject \(H_0\). b) Exact binomial p-values are discrete because the success count changes only by whole numbers. c) It is the probability of an observed-or-more-extreme result under \(H_0\), not \(P(H_0\text{ is true})\).

- Match the interpretation to the two-sided alternative rather than to a one-sided claim. - Separate evidence of a difference from the direction of that difference. - Locate the sample statistic relative to the null value.

1. A two-sided p-value measures evidence that the population proportion differs from \(0.55\), not specifically that it is greater. 2. Because \(0.015\) is small, the data provide evidence against \(H_0:p=0.55\) in favor of \(p\ne 0.55\). 3. The observed sample proportion, or equivalently the sign of the test statistic, identifies the direction. Since \(\hat p<0.55\), the observed departure is downward.

a) A two-sided p-value does not establish an increase, and the observed sample proportion is below \(0.55\). b) It supports the claim that the population proportion differs from \(0.55\). c) Use the sign of \(\hat p-0.55\), or the sign of the test statistic.

- Use symmetry to divide the null distribution at its center. - For the two-sided case, define extremeness by absolute distance from the center. - Determine how much of the distribution is at least zero units from its center.

1. Half of a symmetric null distribution lies at or above \(0\), so the right-tailed p-value is \(0.50\). 2. Half lies at or below \(0\), so the left-tailed p-value is \(0.50\). 3. For a two-sided test, every possible statistic is at least as far from \(0\) as the observed distance of \(0\). Therefore, the two-sided p-value is \(1.00\). 4. A two-sided p-value includes both tails based on absolute extremeness. When the observed statistic is at the center, the qualifying region is the entire null distribution.

a) \(0.50\). b) \(0.50\). c) \(1.00\). d) The observed absolute distance from the null center is zero, so all null outcomes are at least as extreme.

- Identify the exact assumption under which the reported tail probability was computed. - Match that assumption to the claim the reader wants to assess. - Consider how changing the null center changes the reference distribution.

1. The value \(0.006\) measures how extreme the sample result is under a null distribution centered at \(0.50\). It does not measure extremeness under \(p=0.45\). 2. The analyst must specify \(H_0:p=0.45\), choose the alternative that matches the scientific question, and recompute the null standard error, test statistic, and p-value. 3. Changing the null value changes both the center and, for a one-proportion test, the spread of the null distribution. Therefore, the tail probability can change substantially.

a) It was calculated under \(p=0.50\), not under the null value \(0.45\) named in the scientific question. b) Recompute the test using \(H_0:p=0.45\) and the appropriate alternative. c) The null value determines the null distribution from which extremeness is measured.

- Compute each run's qualifying proportion separately. - For a pooled estimate, combine the underlying counts rather than only the two proportions. - Consider how many simulated outcomes each run contributes.

1. Run A gives \(\frac{8}{500}=0.016\), and Run B gives \(\frac{18}{1500}=0.012\). 2. Pooling the counts gives \(\frac{8+18}{500+1500}=\frac{26}{2000}=0.013\). 3. The simple average is \(\frac{0.016+0.012}{2}=0.014\), which weights the two runs equally even though Run B has three times as many simulations. A pooled estimate must weight each simulated outcome equally.

a) Run A: \(0.016\); Run B: \(0.012\). b) \(0.013\). c) A simple average gives equal weight to unequal-sized runs; pooling must use the combined extreme count over the combined simulation count.

- Relate the renewal and nonrenewal proportions in every possible sample. - Compare which sample outcomes fall in the relevant tail for each alternative. - Decide whether relabeling outcomes changes the set of results counted as at least as extreme.

1. The proportions are complements, so \(q=1-p\) and \(\hat q=1-\hat p\). 2. A result with \(\hat p\) far above \(0.40\) corresponds exactly to a result with \(\hat q\) equally far below \(0.60\). 3. The two tests count the same sample outcomes as being at least as extreme, so they have the same p-value, \(0.032\).

The second p-value is \(0.032\). Reversing the success label changes a right-tailed test for \(p\) into the equivalent left-tailed test for the complementary proportion \(q\), without changing the evidence.

- Compare each p-value marker with the same significance-level marker. - Note whether the two comparisons lead to the same formal decision. - Consider what a decision change says about relying on the disputed record.

1. The display shows that the original p-value is \(0.041<0.050\), so the original data lead to rejection of the null hypothesis. 2. The corrected p-value is \(0.067>0.050\), so the corrected data lead to failure to reject the null hypothesis. 3. Because a plausible correction to one item reverses the decision, the statistical conclusion is not stable with respect to that classification.

The decision is not stable. The original data lead to rejection, but the plausible corrected data do not. The team should resolve the classification issue and report the sensitivity of the conclusion.

- Distinguish an inequality displayed by software from an exact numerical value. - Keep the interpretation conditional on the null hypothesis. - Separate evidence against a hypothesis from a claim that the hypothesis has zero probability.

1. The display gives an upper bound, not an exact value, so statement a) is false. 2. A p-value does not assign a probability to the null hypothesis, so statement b) is false. 3. A p-value below \(0.001\) means the null model produces results at least this extreme with probability below \(0.001\), so statement c) is correct. 4. Such a small p-value indicates strong incompatibility between the data and the null model, so statement d) is correct.

Statements c) and d) are correct. Statements a) and b) are false.

- Relate the success count and sample proportion for a fixed sample size. - Ask whether their ordering of possible sample outcomes can differ. - A p-value depends on which null outcomes are counted as at least as extreme.

1. For fixed \(n\), the count \(X\) is a one-to-one increasing transformation of \(\hat p\). 2. Because \(X=n\hat p\) with \(n>0\), any consistently defined tail or absolute-distance extremeness rule selects the same sample outcomes for \(X\) and \(\hat p\). 3. The two analyses therefore count the same null-model outcomes and produce the same p-value, \(0.027\).

The second analyst should also obtain a p-value of \(0.027\). Using the count instead of the proportion does not change which outcomes are at least as extreme when the sample size and test definition are fixed.

- Separate the size of the data set from the way participants entered it. - Ask what assumptions make a p-value meaningful for a target population. - Consider whether a small random-error measure can remove systematic selection bias.

1. A p-value describes sampling variability under the null model only when the data-collection design supports that model. 2. The respondents chose whether to participate, so they may differ systematically from readers who did not respond. 3. A large voluntary-response sample can have very small modeled random error while still having substantial selection bias. 4. Therefore, \(0.004\) does not repair the lack of representative random sampling and cannot support generalization to all readers.

The p-value addresses random variation under the assumed null model, not voluntary-response bias. Because respondents self-selected, the result cannot be generalized reliably to all readers despite the small p-value.

- Translate each test decision into a comparison between the p-value and its significance level. - Use both comparisons at the same time. - Pay attention to which endpoint can be included under the usual rejection rule.

1. Rejection at \(\alpha=0.05\) means the p-value is at most \(0.05\). 2. Failure to reject at \(\alpha=0.01\) means the p-value is greater than \(0.01\). 3. Combining the two conditions gives \(0.01<p\text{-value}\le 0.05\).

\(0.01<p\text{-value}\le 0.05\).

- Think about the possible whole-number counts of extreme simulations. - Convert one extreme result out of the total into the smallest positive step. - Check whether the reported decimal lies on that grid of possible values.

1. With \(50\) simulations, the extreme-result count must be a whole number from \(0\) through \(50\). 2. The reported proportion must therefore have the form \(\frac{k}{50}\), so possible values are spaced \(\frac{1}{50}=0.02\) apart. 3. Since \(0.013\) is not a multiple of \(0.02\), it cannot be the direct simulated proportion from exactly \(50\) runs.

No. A direct estimate from \(50\) simulations must be a multiple of \(0.02\), and \(0.013\) is not.

- Distinguish a probability for each repetition from a guaranteed count across repetitions. - Convert the probability into a long-run expected count. - Consider the natural variation in repeated independent trials.

1. Under the null hypothesis, each comparable repetition has probability \(0.030\) of producing a result at least as extreme as the observed result. 2. Across \(100\) independent repetitions, the expected number of such results is \(100(0.030)=3\). 3. The actual count is random, so it need not equal \(3\); it could be smaller or larger.

Under the null model, about \(3\) of \(100\) repetitions are expected to be at least as extreme, but the actual number is random and will not necessarily be exactly \(3\).

- Identify which side of the observed statistic is measured by a right-tailed p-value. - Use the total area of the null distribution to find the area on the other side. - Translate cumulative area below a value into a percentile.

1. A right-tailed p-value of \(0.20\) means \(20\%\) of the null distribution lies at or above the observed statistic. 2. Therefore, \(80\%\) of the null distribution lies below the observed statistic. 3. The observed statistic is at approximately the \(80\)th percentile of the null distribution.

The observed statistic is at approximately the \(80\)th percentile of the null distribution.

- Order the possible success counts by how strongly they support the right-tailed alternative. - Compare the tail set for the new count with the tail set for the previous count. - Use the known decision boundary after establishing the p-value inequality.

1. In a right-tailed test, larger success counts are more extreme in the direction of the alternative. 2. The tail event for \(16\) successes is a strict subset of the tail event for \(15\) successes. 3. Therefore, the p-value for \(16\) successes is less than \(0.050\). 4. Since it is below \(0.05\), reject the null hypothesis.

The p-value must be less than \(0.050\), so the null hypothesis is rejected at \(\alpha=0.05\).

- Convert the percentile into the area beyond the observed statistic. - Account for an equally extreme region on the opposite side of a symmetric null distribution. - Combine the two tail areas.

1. The area above the observed statistic is \(1-0.97=0.03\). 2. A symmetric two-sided test includes an equally extreme area of \(0.03\) in the opposite tail. 3. The p-value is \(2(0.03)=0.06\).

The two-sided p-value is \(0.06\).

- Identify which cumulative area a left-tailed p-value measures. - Compare that area with one-half of a symmetric distribution. - Use the location relative to the center to infer the direction of the sample result.

1. In a left-tailed test, the p-value is the null-distribution area at or below the observed statistic. 2. An area of \(0.62\) is greater than \(0.50\), so the observed statistic lies to the right of the null distribution’s center. 3. Therefore, the observed sample proportion is above the null proportion, which is opposite the direction of the left-tailed alternative.

The observed sample proportion is above the null proportion. A left-tail area greater than \(0.50\) places the observed statistic to the right of the null center.

- Identify the assumption under which a p-value is calculated. - Separate a probability about possible sample outcomes from a prediction about a future study’s decision. - Make the wording reflect both tails because the test is two-sided.

1. A p-value describes the probability of results at least as far from the null value as the observed result, assuming the null hypothesis is true. 2. It does not give the probability that a future study will reject because that probability depends on the true population proportion, the sample size, and the chosen significance level. 3. The correct interpretation is that, if \(p=0.50\), the probability of obtaining a sample result at least as far from \(0.50\) as the observed result in either direction is \(0.07\).

a) The p-value does not determine the probability that a future study will reject \(H_0\). A replication probability depends on the true proportion, sample size, and significance level. b) If the true population proportion is \(0.50\), there is a \(7\%\) chance of obtaining a sample result at least as far from \(0.50\) as the observed result, in either direction.

- Read both qualifying simulation counts from the chart. - Focus on the phrase “at least as extreme” when deciding whether to include ties. - State the interpretation under the assumption that the null hypothesis is true.

1. Reading the chart gives \(34\) simulated statistics greater than the observed statistic and \(12\) exactly equal to it. 2. Results at least as extreme as the observation include both the \(34\) larger statistics and the \(12\) ties. 3. The estimated p-value is \(\frac{34+12}{2000}=0.023\). 4. Ties are included because “at least as extreme” includes equality. 5. If the null hypothesis is true, the estimated probability of obtaining a statistic at least as far from the null value as the observed statistic is \(0.023\).

a) \(0.023\). b) A simulated statistic equal to the observed statistic is still at least as extreme as the observation. c) Assuming the null hypothesis is true, the estimated probability of obtaining a statistic at least as far from the null value as the observed statistic is \(2.3\%\).

- A p-value is built from a region of the null distribution, not from a number alone. - Match the direction of the alternative to the part of the distribution that counts as evidence. - Distinguish one-tailed and two-tailed definitions of extremeness.

1. A p-value is a tail probability, and the alternative hypothesis determines which tail or tails count as at least as extreme. 2. For \(H_a:p>0.40\), it means sample proportions at least as large as the observed sample proportion. 3. For \(H_a:p<0.40\), it means sample proportions no greater than the observed sample proportion. 4. For \(H_a:p\ne 0.40\), it means sample proportions at least as far from \(0.40\) as the observed sample proportion in either direction. 5. Without the alternative and observed direction, the numerical p-value does not identify the relevant extremeness statement.

The interpretation is incomplete because the alternative hypothesis determines the relevant tail or tails. For \(H_a:p>0.40\), “at least as extreme” means at least as large; for \(H_a:p<0.40\), it means no greater than the observed sample proportion; and for \(H_a:p\ne 0.40\), it means at least as far from \(0.40\) in either direction.

- Use the proportion of the unit interval lying below each cutoff. - Translate each long-run proportion into an expected count of studies. - Remember that unusual results are uncommon, not impossible, under a true null hypothesis.

1. For a uniform distribution, the probability of a p-value below \(0.05\) is \(0.05\), so the expected count is \(200(0.05)=10\). 2. The probability of a p-value below \(0.20\) is \(0.20\), so the expected count is \(200(0.20)=40\). 3. Even when every null hypothesis is true, random samples occasionally produce extreme results and therefore small p-values at the corresponding long-run rates.

a) About \(10\). b) About \(40\). c) Small p-values can occur by random variation even under true null hypotheses; the uniform null distribution predicts some of them.

- An expected value is a long-run average, not a guaranteed sample result. - Distinguish a point probability from a p-value. - A left-tailed p-value includes outcomes at least as unfavorable to the null hypothesis as the observed result.

1. Statement 1 is not valid. The expected value is \(np = 200(0.52) = 104\), but individual random samples commonly fall above or below the expected value. Being below \(104\) does not by itself establish a decrease. 2. Statement 2 is not valid. For a discrete distribution with many possible outcomes, a single point probability can be small even when the model is reasonable. The relevant left-tailed p-value is the probability of an outcome at least as low as the observed one, not only \(P(X = 98)\). 3. Statement 3 is valid. Under \(H_0: p = 0.52\), \(P(X \le 98) \approx 0.21808\). This p-value is not small relative to common significance levels, so the sample does not provide convincing evidence that support has decreased.

1. Not valid. A result below the expected value can occur through ordinary sampling variation. 2. Not valid. The point probability \(P(X = 98)\) is not the relevant p-value; the left-tail probability is. 3. Valid. The p-value is \(P(X \le 98) \approx 0.21808\), so the result is not statistically significant at common levels.

- A suspected increase requires a right-tailed test. - The p-value includes the observed count and all more extreme counts in the direction of the alternative. - Compare the p-value with \(\alpha\) before stating the conclusion.

1. The hypotheses are \(H_0: p \le 0.05\) and \(H_a: p > 0.05\). 2. Let \(X\) be the number of defective chips. At the null boundary \(p = 0.05\), the right-tailed p-value is \(P(X \ge 12) \approx 0.07400\). 3. Because \(0.07400 > 0.05\), fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(5\%\) significance level that the shipment’s defective proportion exceeds \(5\%\).

1. \(H_0: p \le 0.05\); \(H_a: p > 0.05\). 2. The p-value is \(P(X \ge 12) \approx 0.07400\). 3. Fail to reject \(H_0\). There is insufficient evidence that the defective proportion exceeds \(5\%\).

- Identify the number of null simulations that were at least as extreme as the observation and the total number of simulations. - Think about how the observed sample relates to the simulated samples under the null model. - Compare the two fractions before deciding which estimate is slightly larger.

1. The plus-one estimate is \(\frac{17+1}{999+1}=\frac{18}{1000}=0.018\). 2. Under the null model, the observed sample is treated as one possible result from the same process as the simulated samples. Including it with the simulations gives the plus-one estimate. 3. The naive estimate is \(\frac{17}{999}\approx0.0170\), which is slightly smaller. The plus-one method also prevents an estimated p-value of zero when no simulation is at least as extreme as the observation.

a) The estimated p-value is \(0.018\). b) The added case represents the observed sample as one result to be ranked with the null simulations. c) The naive estimate is approximately \(0.0170\), which is slightly smaller than \(0.018\).

- The alternative is two-sided, so “as extreme” includes both directions. - Smaller values correspond to stronger incompatibility with the null model. - A single p-value can lead to different decisions under different significance levels.

1. If the true preference proportion is \(0.50\), the probability of obtaining a sample proportion at least as far from \(0.50\) as the observed sample proportion, in either direction, is \(0.084\). 2. The result provides some evidence against \(H_0\), but not strong evidence at the conventional \(0.05\) level. 3. Because \(0.084<0.10\), reject \(H_0\) at \(\alpha=0.10\). Because \(0.084>0.05\), fail to reject \(H_0\) at \(\alpha=0.05\).

a) Assuming \(p=0.50\), there is an \(8.4\%\) chance of a sample proportion at least as far from \(0.50\) as the observed one, in either direction. b) The evidence against \(H_0\) is modest, not strong. c) Reject at \(\alpha=0.10\); fail to reject at \(\alpha=0.05\).

- Use the direction of the alternative to decide which tail areas count. - For a symmetric null distribution, equally extreme results can occur on both sides. - Interpret each probability under the assumption that the null model is correct.

1. For the right-tailed alternative, \(P(Z\ge 1.80)\approx 0.0359\). Assuming \(H_0\) is true, this is the probability of a test statistic at least as large as the observed one. 2. For the two-sided alternative, include equally extreme values in both tails: \(2P(Z\ge 1.80)\approx 0.0719\). 3. The two-sided test counts departures from the null value in either direction, while the right-tailed test counts only large positive departures.

a) The p-value is approximately \(0.0359\); it is the upper-tail probability under \(H_0\). b) The p-value is approximately \(0.0719\); it is the probability of a test statistic at least as far from \(0\) in either direction. c) The two-sided alternative treats both directions as evidence against \(H_0\).

- Separate evidence about whether a difference exists from the size of that difference. - A very large sample can make a small departure statistically detectable. - State the conclusion in terms of the population proportion without exaggerating the effect.

1. The very small p-value indicates that a sample proportion at least as far from \(0.500\) as \(0.504\) would be very unusual under the null model. 2. The p-value measures statistical evidence, not practical importance. The observed difference is only \(0.004\), so its real-world importance must be judged separately. 3. Reject \(H_0\). There is convincing statistical evidence that the population proportion differs from \(0.500\), but the size of the difference may be practically small.

a) The data provide very strong statistical evidence against \(H_0\). b) No. A p-value does not measure practical importance; the effect size is only \(0.004\). c) Reject \(H_0\) and conclude that the population proportion differs from \(0.500\), while noting that the difference may be practically minor.

- Compare each study marker with the marked significance threshold. - Notice the numerical distance between the two p-values before judging the evidence. - Avoid equating failure to reject with proof of no effect.

1. Study A rejects \(H_0\) because \(0.049<0.050\). Study B fails to reject \(H_0\) because \(0.051>0.050\). 2. The displayed p-values are nearly identical, so the strength of evidence is also nearly identical despite the different formal decisions. 3. The significance threshold is a decision rule, not a sharp boundary between a real effect and no effect. Failing to reject does not establish that the effect is zero.

a) Study A rejects \(H_0\); Study B fails to reject \(H_0\). b) No. The evidence is nearly the same because the p-values are almost equal. c) Statistical significance is a thresholded decision, not proof that an effect exists on one side and is absent on the other.

- Identify which statement is assumed and which event is measured. - Conditional probabilities generally cannot be reversed without more information. - Keep the p-value tied to the null model’s distribution of possible sample results.

1. The p-value is \(P(E\mid H_0)\), not \(P(H_0\mid E)\). 2. The p-value asks how likely an extreme result is assuming the null hypothesis is true. The student’s expression asks how likely the null hypothesis is after observing an extreme result. 3. Reversing a conditional probability requires additional information about prior probabilities and the behavior of alternatives, which the p-value does not provide.

a) \(P(E\mid H_0)\). b) The p-value conditions on \(H_0\); the student’s probability conditions on the observed extremeness. c) The reverse conditional probability requires information not contained in a p-value.

- Recall that a test statistic compares an observed difference with its standard error. - Larger samples reduce standard errors. - Separate standardized evidence from the real-world size of a difference.

1. Equal p-values do not imply equal raw differences from the null proportion. The larger sample can produce the same standardized evidence from a much smaller difference. 2. The p-values indicate similar strength of statistical evidence against their respective null models, assuming comparable test directions. 3. Practical importance depends on the size and context of the observed difference, not only on the p-value. A very large sample can make a tiny difference statistically significant.

a) No. b) They indicate comparable statistical evidence against the null models. c) The larger study may detect a much smaller effect, so the observed effect sizes must be examined separately.

- Separate what happened in this finite set of simulations from the underlying probability the simulation is estimating. - Ask whether failing to observe an event in finitely many trials proves that its probability is zero. - Use the simulation result to describe the strength of evidence without claiming more numerical precision than the run provides.

1. The direct proportion of simulated results at least as extreme as the observation is \(\frac{0}{5000}=0\), but this is only a finite-simulation estimate of the null-tail probability. 2. Observing no equally extreme results in \(5000\) simulations does not show that the actual p-value is exactly \(0\); a very small positive tail probability could also produce zero such results in a finite run. 3. Report that \(0\) of the \(5000\) null simulations were at least as extreme as the observed statistic and that the p-value is too small for this simulation run to estimate precisely. More simulations would be needed for a more precise simulation-based numerical estimate. 4. The observed statistic is extremely incompatible with the null model and provides very strong evidence for the alternative hypothesis.

a) No. The simulation proportion is \(0\), but the finite simulation does not establish that the actual p-value is exactly \(0\). b) Report that \(0\) of \(5000\) null simulations were at least as extreme and that the p-value is too small for this run to estimate precisely. c) The observed result is extremely unusual under \(H_0\) and provides very strong evidence against the null hypothesis.

- Base the interpretation on a world in which the null claim is true. - Compare the reported probability separately with each proposed cutoff. - Consider when a decision rule should be fixed so that its long-run behavior is meaningful.

1. Assuming the true support proportion is \(0.50\), the probability of obtaining a sample result at least as far from \(0.50\) as the observed result, in either direction, is \(0.047\). 2. Because \(0.047<0.05\), the test rejects \(H_0\) at \(\alpha=0.05\). Because \(0.047>0.01\), it fails to reject \(H_0\) at \(\alpha=0.01\). 3. Selecting \(\alpha\) after observing the p-value allows the decision rule to be tailored to the data. The significance level should be chosen in advance to preserve the intended long-run error rate and avoid selectively labeling a result significant.

a) If the true support proportion were \(0.50\), there would be a \(4.7\%\) chance of obtaining a result at least as far from \(0.50\) as the observed result, in either direction. b) Reject \(H_0\) at \(\alpha=0.05\); fail to reject \(H_0\) at \(\alpha=0.01\). c) Choosing \(\alpha\) after seeing the p-value makes the rule data-dependent and undermines its stated error control.

- Read the counts for both qualifying tails from the chart. - Combine the qualifying outcomes before dividing by the total number of simulations. - Match the tails included in the calculation to the direction of the alternative hypothesis.

1. The two-sided p-value includes both sets of outcomes at least as far from the null expectation as the observed result. 2. Reading the chart gives \(37\) lower-tail outcomes and \(45\) upper-tail outcomes, so the estimated p-value is \(\frac{37+45}{1000}=0.082\). 3. Assuming \(p=0.30\), about \(8.2\%\) of random samples of size \(50\) would produce a success count at least as far from \(15\) as \(10\), in either direction. 4. Counting only the lower tail would estimate a left-tailed p-value and would ignore extreme outcomes above the null expectation.

a) \(0.082\). b) If \(p=0.30\), there is about an \(8.2\%\) chance of obtaining a result at least as far from the null expectation as the observed count, in either direction. c) A two-sided alternative requires extreme outcomes in both tails, not only outcomes below the null expectation.

- Compare each method’s marker with the marked significance level. - Consider how a discrete distribution differs from a continuous approximation. - Follow the analysis method selected before the data were evaluated.

1. The display shows that the exact calculation gives \(0.052>0.050\), so it fails to reject \(H_0\). 2. The normal approximation gives \(0.048<0.050\), so it rejects \(H_0\). 3. The exact binomial distribution is discrete, while the normal model is continuous and approximates the null distribution. Small numerical differences are possible, especially near a decision cutoff. 4. If the exact method was designated in advance, the reported conclusion should follow it: fail to reject \(H_0\).

a) Exact method: fail to reject \(H_0\). Normal approximation: reject \(H_0\). b) One method uses the discrete null distribution exactly, while the other uses a continuous approximation. c) Report the exact-method conclusion: fail to reject \(H_0\).

- Compare the p-value with the significance level paired with the stated confidence level. - An interval obtained by inverting tests collects the null values that those tests do not reject. - Consider how changing from \(95\%\) to \(99\%\) confidence changes the paired significance level.

1. Because \(0.032<0.05\), the two-sided test rejects \(H_0:p=0.45\) at significance level \(0.05\). 2. A confidence interval formed by inverting matching two-sided tests contains exactly the null values that are not rejected. Therefore, \(0.45\) is outside the corresponding \(95\%\) confidence interval. 3. A \(99\%\) confidence interval corresponds to two-sided tests with significance level \(0.01\). Since \(0.032>0.01\), \(0.45\) would not be rejected at that level and therefore would be contained in the corresponding \(99\%\) interval.

a) The value \(0.45\) is outside the corresponding \(95\%\) confidence interval. b) Inverting matching two-sided tests produces an interval containing exactly the null values not rejected at the paired significance level. c) No. Because \(0.032>0.01\), \(0.45\) is contained in the corresponding \(99\%\) confidence interval.

- Separate the observed effect size from the uncertainty around it. - Consider how sample size changes the spread of a null sampling distribution. - Compare what the two sample proportions estimate before interpreting their p-values.

1. Study B provides stronger evidence against \(H_0\) because its p-value is much smaller. 2. Both studies estimate the same departure, \(0.44-0.40=0.04\), but the standard error decreases as sample size increases. The much larger sample makes the same departure many more standard errors from the null value. 3. Study B does not show a larger estimated effect. Both studies estimate the same \(0.04\) difference; Study B estimates it with greater precision.

a) Study B. b) Its much larger sample size produces a smaller standard error, making the same observed difference more unusual under \(H_0\). c) No. Both studies estimate a departure of \(0.04\); Study B only has greater precision.

- Follow the analysis plan that was fixed before the result was known. - Consider what advantage is created by selecting a tail after seeing the sign of the statistic. - Identify when a directional research question legitimately supports a one-sided test.

1. For the preregistered two-sided test, \(0.080>0.05\), so fail to reject \(H_0\). 2. Choosing the direction after observing the data gives the researcher an opportunity to select the more favorable tail and changes the intended long-run error rate. 3. A one-sided p-value would be appropriate if a scientifically justified directional alternative had been specified before examining the sample result.

a) Fail to reject \(H_0\). b) The direction was selected after observing the data, making the analysis data-dependent and invalidating the planned error control. c) A one-sided test is appropriate when its directional alternative is justified and chosen in advance.

- Determine how much probability belongs in one tail for each test. - Match each tail area to a standard normal cutoff. - Account for the number of tails before comparing the extremeness of the statistics.

1. For Study A, a one-sided upper-tail area of \(0.030\) corresponds to \(|z|\approx 1.881\). 2. For Study B, a two-sided p-value of \(0.030\) places \(0.015\) in each tail, corresponding to \(|z|\approx 2.170\). 3. Study B's statistic is farther into a tail because its p-value includes two tails, so each tail must contain only half of the total p-value.

a) Study A: \(|z|\approx 1.881\); Study B: \(|z|\approx 2.170\). b) Study B. c) A two-sided p-value divides its probability across both tails, while a one-sided p-value uses one tail.

- Treat the extreme-simulation indicator as a success or failure in repeated trials. - Quantify how much a simulated proportion can vary from run to run. - Compare the distance from the decision cutoff with the simulation uncertainty.

1. The simulated p-value estimate is \(\hat p_{\text{sim}}=\frac{43}{1000}=0.043\). 2. Its simulation standard error is \(\sqrt{\frac{0.043(0.957)}{1000}}\approx 0.00641\). 3. The estimate lies only \(0.007\) below \(0.05\), about one simulation standard error. Additional simulations would reduce Monte Carlo variability and show more reliably whether the estimated tail probability is below the decision cutoff.

a) \(0.043\). b) Approximately \(0.00641\). c) The estimate is close to \(0.05\) relative to its simulation uncertainty, so more repetitions would make the decision more stable.

- Think about every number that rounds to the displayed hundredth. - Compare the possible unrounded values with the study’s decision boundary. - Do not assume the displayed value is exact.

1. Values that round to \(0.05\) to the nearest hundredth satisfy \(0.045\le p\text{-value}<0.055\). 2. Some values in that interval are below \(0.05\), while others are above \(0.05\). 3. Therefore, the decision cannot be determined without more digits from the software output.

a) \(0.045\le p\text{-value}<0.055\). b) No. The unrounded p-value could fall on either side of \(0.05\), so more precision is needed.

- Recall what probability a p-value must represent under a null model. - Ask what random quantity would have \(0.160\) as its tail probability. - Distinguish a numerical summary from a formally defined hypothesis test.

1. Each p-value is calculated from its own study’s data and null sampling distribution. 2. The arithmetic mean \(\frac{0.020+0.300}{2}=0.160\) is not the tail probability of a specified combined test statistic under a specified null model. 3. Combining evidence across studies requires a defined method with its own sampling distribution; simply averaging p-values does not provide that method.

The value \(0.160\) is only the arithmetic mean of two p-values. It is not a valid combined p-value because no combined test statistic or null distribution was defined.

- Locate the observed proportion relative to the null value. - Determine the sign of the standardized statistic without calculating its magnitude. - Visualize the area to the right of a negative point on a symmetric null distribution.

1. The observed proportion \(0.47\) is below the null value \(0.50\), so its standardized test statistic is negative. 2. In a right-tailed test, the p-value is the area to the right of that negative statistic. 3. More than half of a symmetric null distribution lies to the right of a negative statistic, so the p-value must be greater than \(0.50\). 4. Therefore, a right-tailed p-value of \(0.040\) cannot match the stated null value and observed sample proportion.

The report is inconsistent. Because \(\hat p<p_0\), a right-tailed p-value must exceed \(0.50\), not equal \(0.040\).

- Read both one-sided areas from the software display. - A probability cannot exceed \(1\), so use that fact to check the proposed result. - For a symmetric two-sided test, the two equal extreme regions are based on the smaller tail.

1. Reading the software output, the left-tail area is \(0.54\) and the right-tail area is \(0.46\). 2. For a symmetric null distribution, a two-sided p-value doubles the smaller one-sided tail area corresponding to the observed distance from the null center. 3. The smaller tail area is \(0.46\), so the two-sided p-value is \(2(0.46)=0.92\). 4. Doubling the larger tail would count more than the full probability distribution and can produce an impossible value above \(1\).

The correct two-sided p-value is \(0.92\). Double the smaller one-sided tail area, not the larger one.

- Translate each p-value into the tail area appropriate for that test’s alternative. - A two-sided p-value divides its total area between two tails. - Convert the resulting cumulative areas into standard normal cutoffs before comparing distances.

1. For the right-tailed test, the area to the right is \(0.080\), so the cumulative area to the left is \(0.920\). This gives \(z_A\approx 1.405\). 2. For the two-sided test, each tail has area \(0.040\), so the cumulative area to the left of the positive statistic is \(0.960\). This gives \(z_B\approx 1.751\). 3. Since \(1.751>1.405\), Test B’s observed statistic is farther from \(0\), even though the numerical p-values are equal.

Test A: \(z\approx 1.405\). Test B: \(z\approx 1.751\). The two-sided test has the more extreme standardized statistic.

- Identify which sample outcomes are at least as supportive of \(H_a\) as the observed result. - For a right-tailed test, calculate the probability of the observed count or a larger count. - Compare the p-value with \(\alpha\). - A p-value greater than \(\alpha\) leads to failing to reject \(H_0\).

1. Let \(X\) be the number of users in the sample who uninstall the app within one week. Under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.10)\). 2. Because \(H_a: p > 0.10\), the p-value is the probability of observing \(26\) or more uninstalls under \(H_0\): \(P(X \ge 26)\). 3. Using the cumulative binomial probability, \(P(X \le 25) \approx 0.8995\). Therefore, \(P(X \ge 26) = 1 - P(X \le 25) \approx 0.1005\). 4. Since \(0.1005 > 0.05\), fail to reject \(H_0\). The sample does not provide sufficient evidence that the one-week uninstall proportion exceeds \(10\%\).

The exact binomial p-value is \(P(X \ge 26) \approx 0.1005\). Because \(0.1005 > 0.05\), fail to reject \(H_0\). The result is not statistically significant at the \(0.05\) level.

- Standardize the same observed statistic separately under each null model. - Because both alternatives are two-sided, use extremeness in either direction. - A larger p-value means the data are more compatible with that null model.

1. For Test A, \(z=\frac{0.46-0.40}{\sqrt{0.40(0.60)/500}}\approx 2.74\), so the two-sided p-value is approximately \(0.0062\). 2. For Test B, \(z=\frac{0.46-0.50}{\sqrt{0.50(0.50)/500}}\approx -1.79\), so the two-sided p-value is approximately \(0.0736\). 3. The larger p-value for Test B indicates that \(p=0.50\) is more compatible with the observed sample proportion than \(p=0.40\).

a) Test A: approximately \(0.0062\). Test B: approximately \(0.0736\). b) \(p=0.50\) is more compatible. c) Test B has the larger p-value, so the observed result is less unusual under that null model.

- Read the count in the group that definitely exceeds the observed statistic. - Then consider the largest possible contribution from the group containing the observed value. - Determine what information was lost when individual simulated values were grouped.

1. The chart shows \(30\) simulated statistics at least \(2.5\). All are at least as large as the observed value, so the p-value is at least \(\frac{30}{5000}=0.006\). 2. It also shows \(90\) statistics between \(2.0\) and \(2.5\). At most, all \(90\) could be at least \(2.2\). Thus the p-value is at most \(\frac{30+90}{5000}=0.024\). 3. The grouped count does not show how many of the \(90\) statistics between \(2.0\) and \(2.5\) fall at or above \(2.2\).

a) \(0.006\). b) \(0.024\). c) The bin containing the observed statistic combines simulated values below and above \(2.2\), so the qualifying count within that bin is unknown.

- Measure extremeness by distance from the null sample proportion. - List the attainable sample proportions before combining probabilities from both sides. - Check whether the two tails of the discrete null distribution are mirror images.

1. The observed sample proportion is \(0\), which is \(0.10\) from the null value. Counts at least that far away are \(X=0\) and \(X\ge 4\), because \(\frac{4}{20}=0.20\). 2. Under \(X\sim\operatorname{Binomial}(20,0.10)\), \(P(X=0)=(0.90)^{20}\approx 0.12158\). 3. Also, \(P(X\ge 4)\approx 0.13295\). 4. The exact two-sided p-value is \(0.12158+0.13295\approx 0.25453\). 5. The shortcut gives \(2P(X\le 0)\approx 0.24315\), which differs because the discrete binomial null distribution is skewed and the equally distant opposite tail does not have the same probability.

a) \(X=0\) and \(X\ge 4\). b) The exact two-sided p-value is approximately \(0.25453\). c) \(2P(X\le 0)\approx 0.24315\), so doubling the smaller tail is not exact for this skewed discrete null distribution.