Free math worksheets

- Identify the boundary value in each claim. - Put the equality in the null hypothesis. - Use the direction of \(H_a\) to identify the tail. - Ask which sample results would contradict the original claim.

1. For part a), the boundary proportion is \(0.85\). A claim of “at least” is challenged by values below the boundary, so \(H_0: p = 0.85\) and \(H_a: p < 0.85\). This is a left-tailed test. 2. For part b), the boundary proportion is \(0.06\). A claim of “no more than” is challenged by values above the boundary, so \(H_0: p = 0.06\) and \(H_a: p > 0.06\). This is a right-tailed test.

a) \(H_0: p = 0.85\); \(H_a: p < 0.85\); left-tailed test. b) \(H_0: p = 0.06\); \(H_a: p > 0.06\); right-tailed test.

- Define \(p\) as the population proportion of installations that produce an error message. - Use the boundary value from “no more than \(2\%\)” in the null hypothesis. - Decide whether unusually small or unusually large sample proportions support the tester’s suspicion. - The inequality in \(H_a\) determines the tail.

1. Let \(p\) be the true proportion of installations that produce an error message. The boundary value in the developer’s claim is \(0.02\), so \(H_0: p = 0.02\). 2. The tester suspects a higher error rate, so \(H_a: p > 0.02\). 3. Because the alternative hypothesis concerns values greater than \(0.02\), unusually large error counts provide evidence against \(H_0\). Therefore, the test is right-tailed.

a) \(H_0: p = 0.02\); \(H_a: p > 0.02\). b) This is a right-tailed test because the alternative hypothesis states that the error proportion is greater than \(0.02\).

- Let \(p\) represent the current population proportion of students who use public transportation. - Put the benchmark value in the null hypothesis. - Translate the word “higher” into an inequality for the alternative.

1. The null hypothesis uses the reported benchmark: \(H_0:p=0.40\). 2. The suspicion is that the population proportion is higher, so \(H_a:p>0.40\). 3. An alternative of the form \(p>p_0\) gives a right-tailed test.

\(H_0:p=0.40\) and \(H_a:p>0.40\). The test is right-tailed.

- Look at which end of the possible count scale is included in the rejection region. - Identify the boundary proportion in the company’s claim. - Explain what it means when the observed count falls in \(R\).

1. The rejection region contains the larger possible values of the test statistic, so this is a right-tailed test. 2. The company’s boundary value is \(p = 0.12\), so the hypotheses are \(H_0: p = 0.12\) and \(H_a: p > 0.12\). The broader claim being tested is \(p \le 0.12\). 3. Reject \(H_0\) if at least \(15\) of the \(80\) sampled customers are dissatisfied. Otherwise, fail to reject \(H_0\).

This is a right-tailed test. \(H_0: p = 0.12\); \(H_a: p > 0.12\). Reject \(H_0\) if \(15\) or more sampled customers are dissatisfied; otherwise, fail to reject \(H_0\).

- Decide whether unusually small or unusually large germination counts would contradict “at least \(75\%\).” - Use the boundary proportion from the company’s claim in \(H_0\). - State exactly which sample counts lead to rejection.

1. The rejection region contains the smallest possible values of the test statistic, so this is a left-tailed test. 2. The company’s boundary value is \(p = 0.75\), so the hypotheses are \(H_0: p = 0.75\) and \(H_a: p < 0.75\). The broader claim being tested is \(p \ge 0.75\). 3. Reject \(H_0\) if \(81\) or fewer of the \(120\) seeds germinate within one week. Otherwise, fail to reject \(H_0\).

This is a left-tailed test. \(H_0: p = 0.75\); \(H_a: p < 0.75\). Reject \(H_0\) if \(81\) or fewer seeds germinate within one week; otherwise, fail to reject \(H_0\).

- Use the direction of \(H_a\) to identify the tail. - For a left-tailed test, the rejection region contains small values of \(X\). - The probability of the rejection region must be at most the significance level. - Find the largest table value of \(k\) whose cumulative probability is no greater than \(0.10\).

1. Because \(H_a: p < 0.60\), this is a left-tailed test. The rejection region has the form \(R = \{0, 1, \ldots, c\}\). 2. Choose the largest integer \(c\) for which \(P(X \le c) \le 0.10\). 3. From the table, \(P(X \le 41) = 0.0699 \le 0.10\), but \(P(X \le 42) = 0.1053 > 0.10\). 4. Therefore, \(c = 41\), so the rejection region is \(R = \{0, 1, \ldots, 41\}\).

The rejection region is \(R = \{0, 1, \ldots, 41\}\).

- In a right-tailed test, look at the largest possible values of \(X\). - Use \(P(X \ge k + 1) = 1 - P(X \le k)\). - Determine how large the cumulative probability must be so that the remaining upper-tail probability is at most \(0.05\). - Convert the cutoff into a complete decision rule.

1. For a right-tailed test, the rejection region has the form \(R = \{c + 1, c + 2, \ldots, 50\}\). 2. The requirement \(P(X \ge c + 1) \le 0.05\) is equivalent to \(P(X \le c) \ge 0.95\). 3. From the table, \(P(X \le 19) = 0.9152 < 0.95\), while \(P(X \le 20) = 0.9522 \ge 0.95\). 4. Thus, the smallest suitable value is \(c = 20\), and the rejection region begins at \(21\). 5. Reject \(H_0\) when \(X \ge 21\); otherwise, fail to reject \(H_0\).

Reject \(H_0\) when \(X \ge 21\). The rejection region is \(R = \{21, 22, \ldots, 50\}\).

- Use the direction of \(H_a\) to identify the upper tail. - Model the number of successes with a binomial distribution under \(H_0\). - Find the first cutoff whose upper-tail probability is at most \(0.05\). - Use the cumulative distribution to evaluate the upper tail.

1. Under \(H_0\), the number of successes is \(X \sim \operatorname{Bin}(60, 0.30)\). 2. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). Equivalently, find the smallest \(k\) such that \(P(X \le k - 1) \ge 0.95\). 3. The cumulative probabilities are \(P(X \le 23) \approx 0.9368\) and \(P(X \le 24) \approx 0.9638\). 4. Therefore, \(k = 25\), and the rejection region is \(R = \{25, 26, \ldots, 60\}\).

The rejection region is \(R = \{25, 26, \ldots, 60\}\).

- Use the direction of \(H_a\) to identify the lower tail. - Find the largest cutoff whose cumulative probability is no greater than \(0.10\). - State the cutoff as a complete reject-or-fail-to-reject rule.

1. Under \(H_0\), the number of successes is \(X \sim \operatorname{Bin}(45, 0.25)\). 2. Find the largest integer \(c\) such that \(P(X \le c) \le 0.10\). 3. The cumulative probabilities are \(P(X \le 6) \approx 0.0446\), \(P(X \le 7) \approx 0.0941\), and \(P(X \le 8) \approx 0.1725\). 4. Therefore, \(c = 7\). Reject \(H_0\) when \(X \le 7\); otherwise, fail to reject \(H_0\).

Reject \(H_0\) when \(X \le 7\). The rejection region is \(R = \{0, 1, \ldots, 7\}\).

- Convert the operational definition of “late” into a yes-or-no variable. - State the benchmark and its direction. - Include the flights and time period to which the conclusion applies.

1. The variable is whether a domestic flight arrives at least \(15\,\text{minutes}\) late. 2. The parameter is the population proportion of the airline's domestic flights during June 2026 that are late by that definition. 3. The direction is greater than \(15\%\).

One valid question is: “For the airline's domestic flights during June 2026, is the proportion that arrive at least \(15\,\text{minutes}\) late greater than \(15\%\)?”

- Identify the yes-or-no outcome recorded for each bag. - Translate the airport's claim into a statement about a population proportion. - Preserve the direction of the benchmark comparison.

1. The variable is whether a carry-on bag screened during July 2026 requires a manual recheck because of a false alarm. 2. The parameter is the population proportion of all carry-on bags screened at the airport during July 2026 that require such a recheck. 3. The direction is less than \(2\%\). 4. A complete question is: “For all carry-on bags screened at this airport during July 2026, is the proportion requiring a manual recheck because of a scanner false alarm less than \(2\%\)?”

For all carry-on bags screened at this airport during July 2026, is the proportion requiring a manual recheck because of a scanner false alarm less than \(2\%\)?

- Translate the verbal research question into the direction of the alternative. - Select the software row that matches that direction before comparing with the significance level. - State the conclusion about the population proportion, not about the sample alone.

1. A decrease corresponds to the left-tailed alternative \(H_a:p<0.70\), so the relevant p-value is \(0.0174\). 2. Because \(0.0174<0.05\), reject \(H_0\). 3. The sample provides convincing statistical evidence that the population proportion of customers who renew a membership is below \(0.70\).

a) Use \(0.0174\), the p-value for \(H_a:p<0.70\). b) Reject \(H_0\). c) There is sufficient evidence that fewer than \(70\%\) of customers renew a membership.

- Ask which proportion is assumed true while the test statistic is being evaluated. - Distinguish the standard error used for a confidence interval from the null-based spread used for a hypothesis test. - Match the numerator's comparison with the distribution assumed under the null hypothesis.

1. A hypothesis-test statistic measures how far the observed sample proportion is from the null value under the assumption that the null hypothesis is true. 2. Therefore, the null distribution uses \(p_0=0.30\) when calculating its standard deviation. 3. The correct standard error is \(\sqrt{\frac{0.30(0.70)}{200}}\).

Use \(\sqrt{\frac{0.30(0.70)}{200}}\). The test standard error is based on the null proportion because the test statistic is evaluated under \(H_0\).

- Check the sampling method separately from the numerical conditions. - For the independence check, compare the sample size with \(10\%\) of the population size. - For the large-count check in a test, use the null proportion rather than the observed sample proportion.

1. The sample is stated to be a simple random sample, so the random condition is satisfied. 2. For independence when sampling without replacement, the sample should be no more than \(10\%\) of the population. Here \(150>0.10(1200)=120\), so the \(10\%\) condition is not satisfied. 3. Under the null, \(np_0=150(0.40)=60\) and \(n(1-p_0)=150(0.60)=90\), so both large-count requirements are satisfied. 4. Because the \(10\%\) condition fails, the standard one-proportion \(z\)-test is not justified from the information given.

Random condition: satisfied. \(10\%\) condition: not satisfied because \(150>120\). Large-count condition: satisfied because \(np_0=60\) and \(n(1-p_0)=90\). Therefore, the standard one-proportion \(z\)-test is not justified from the information given.

- Match the two-sided \(5\%\) test with its corresponding confidence level. - Locate the null value relative to the interval. - Decide whether the interval rules out the null value or still includes it as plausible.

1. For a two-sided test at \(\alpha=0.05\), the corresponding \(95\%\) confidence interval can be used to assess the null value. 2. The interval \((0.47,0.55)\) contains \(0.50\). 3. Therefore, the sample does not provide sufficient evidence to reject \(H_0:p=0.50\) at the \(5\%\) significance level.

Fail to reject \(H_0\). The corresponding \(95\%\) confidence interval contains the null value \(0.50\), so a two-sided test at \(\alpha=0.05\) is not significant.

- Use the side of the null distribution that is shaded to identify the alternative direction. - The p-value is the null-model area at least as extreme as the observed test statistic in the alternative direction. - Compare the resulting tail area with the stated significance level.

1. The shaded region is to the right of the observed statistic, so the alternative is right-tailed. 2. The p-value is the standard normal area to the right of \(z=1.8\): \(P(Z\ge1.8)\approx0.0359\). 3. Because \(0.0359<0.05\), reject \(H_0\).

a) Right-tailed. b) The p-value is approximately \(0.0359\). c) Reject \(H_0\) at \(\alpha=0.05\).

- Use the number of shaded tails to determine the direction of the alternative. - Read the dashed boundary values from the horizontal axis before finding any probability. - For a symmetric two-sided test, combine the probabilities in both equally extreme tails.

1. The graph shades equally extreme outcomes in both tails, so the alternative is two-sided. The dashed boundaries are at \(-2.1\) and \(2.1\), so \(|z|=2.1\). 2. For \(|z|=2.1\), one tail has probability about \(0.0179\). Doubling gives a two-sided p-value of about \(0.0357\). 3. Since \(0.0357<0.05\), reject \(H_0\).

a) Two-sided, with \(|z|=2.1\). b) p-value \(\approx0.0357\). c) Reject \(H_0\) at \(\alpha=0.05\).

- Read the labeled positions of the gray and blue dashed lines before making a decision. - Ask what outcomes beyond critical values mean for a test performed at a fixed significance level. - For a two-sided p-value, use the observed magnitude rather than the critical-value magnitude to locate the tail boundaries.

1. The gray boundaries are at \(-1.96\) and \(1.96\), and the blue observed-statistic line is at \(z=1.40\). 2. The observed value lies between the critical values, so it is not in either shaded tail. Fail to reject \(H_0\). 3. The shaded tails are the rejection regions fixed by the significance level \(\alpha=0.05\). 4. For the observed magnitude \(|z|=1.40\), the two-sided p-value regions would begin at \(-1.40\) and \(1.40\). Those boundaries are closer to the center than \(\pm1.96\), so the p-value tails contain more total area than the displayed rejection regions. Therefore, the p-value is greater than \(0.05\).

a) Critical values \(-1.96\) and \(1.96\); observed \(z=1.40\). b) No. Fail to reject \(H_0\). c) The shaded tails are the rejection regions determined by \(\alpha=0.05\). d) The p-value regions would begin at \(z=-1.40\) and \(z=1.40\), and the p-value is greater than \(0.05\).

- Write hypotheses about the population proportion of defective screens. - Determine which tail corresponds to proportions greater than the claimed value. - Under the null hypothesis, model the defective-screen count with a binomial distribution. - Find the first count whose upper-tail probability is at most \(0.05\).

1. Let \(p\) be the true proportion of screens with defective pixels. The hypotheses are \(H_0: p = 0.04\) and \(H_a: p > 0.04\). 2. Under \(H_0\), the number of defective screens is \(X \sim \operatorname{Bin}(400, 0.04)\). 3. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). 4. The binomial probabilities give \(P(X \le 22) \approx 0.9455\) and \(P(X \le 23) \approx 0.9663\). Therefore, \(P(X \ge 23) \approx 0.0545 > 0.05\), while \(P(X \ge 24) \approx 0.0337 \le 0.05\). The rejection region is \(R = \{24, 25, \ldots, 400\}\). 5. The observed count \(22\) is not in the rejection region, so fail to reject \(H_0\). There is not sufficient evidence at the \(0.05\) level to conclude that the defective-screen proportion exceeds \(4\%\).

No. The rejection region is \(R = \{24, 25, \ldots, 400\}\), and the observed count \(22\) is not in that region. There is not sufficient evidence at the \(0.05\) significance level to conclude that more than \(4\%\) of the screens have defective pixels.

- Write hypotheses about the population germination proportion. - Decide whether unusually small or unusually large germination counts support the farmer’s suspicion. - Under the null hypothesis, model the germination count with a binomial distribution. - Find the greatest count whose lower-tail probability is at most \(0.05\).

1. Let \(p\) be the true proportion of wheat seeds that germinate. The hypotheses are \(H_0: p = 0.90\) and \(H_a: p < 0.90\). 2. Under \(H_0\), the number of seeds that germinate is \(X \sim \operatorname{Bin}(80, 0.90)\). 3. For a left-tailed test, find the largest integer \(k\) such that \(P(X \le k) \le 0.05\). 4. The binomial probabilities give \(P(X \le 66) \approx 0.0267\) and \(P(X \le 67) \approx 0.0538\). Therefore, the rejection region is \(R = \{0, 1, \ldots, 66\}\). 5. The observed count \(66\) is in the rejection region, so reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that the true germination rate is less than \(90\%\).

Hypotheses: \(H_0: p = 0.90\); \(H_a: p < 0.90\). Rejection region: \(R = \{0, 1, \ldots, 66\}\). Conclusion: Because \(66 \in R\), reject \(H_0\). There is sufficient evidence at the \(0.05\) significance level that the true germination rate is less than \(90\%\).

- In part 1, identify which parameter is being treated as known. - In part 2, locate the largest lower-tail cutoff with probability at most \(0.05\). - Compare the direction of reasoning in the two parts.

1. Let \(X\) be the number of bulbs in the random sample that last more than \(20{,}000\) hours. If \(p = 0.90\), then \(P(X \le 85) \approx 0.07257\), or about \(7.26\%\). 2. For the left-tailed test, find the largest integer \(k\) such that \(P_{p=0.90}(X \le k) \le 0.05\). Because \(P(X \le 84) \approx 0.03989\) and \(P(X \le 85) \approx 0.07257\), the rejection region is \(R = \{0, 1, \ldots, 84\}\). 3. In part 1, the population parameter is treated as known, and probability is used to predict possible sample outcomes. In part 2, sample evidence is used to assess a claim about an unknown population parameter.

1. \(P(X \le 85) \approx 0.07257\), or \(7.26\%\). 2. The rejection region is \(R = \{0, 1, \ldots, 84\}\). 3. Probability reasons from a specified population model to sample outcomes. Statistical inference uses sample outcomes to evaluate a claim about the population model.

- Define \(p\) as the population proportion of seeds that fail to germinate. - Decide whether small or large failure counts support the nursery’s suspicion. - Model the number of failures with a binomial distribution under \(H_0\). - For a right-tailed test, find the first count whose upper-tail probability is at most \(0.05\).

1. Let \(p\) be the true proportion of seeds that fail to germinate. The hypotheses are \(H_0: p = 0.08\) and \(H_a: p > 0.08\). Because larger failure proportions support \(H_a\), this is a right-tailed test. 2. Under \(H_0\), the number of seeds that fail to germinate is \(X \sim \operatorname{Bin}(150, 0.08)\). Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). 3. The cumulative probabilities are \(P(X \le 17) \approx 0.9449\) and \(P(X \le 18) \approx 0.9687\). Thus, \(P(X \ge 18) \approx 0.0551 > 0.05\), while \(P(X \ge 19) \approx 0.0313 \le 0.05\). 4. The rejection region is \(R = \{19, 20, \ldots, 150\}\). Reject \(H_0\) if at least \(19\) of the \(150\) seeds fail to germinate.

a) \(H_0: p = 0.08\); \(H_a: p > 0.08\). This is a right-tailed test. b) \(R = \{19, 20, \ldots, 150\}\). Reject \(H_0\) if \(19\) or more sampled seeds fail to germinate.

- Define the test statistic as the number of panels that still meet the performance standard. - Decide whether unusually small or unusually large counts would challenge the guarantee. - For a left-tailed test, find the largest lower-tail cutoff with probability at most \(0.05\). - Compare the observed count with the cutoff.

1. Let \(p\) be the true proportion of panels that still produce their rated power after \(20\) years. The hypotheses are \(H_0: p = 0.90\) and \(H_a: p < 0.90\). 2. Under \(H_0\), the number of panels that still produce their rated power is \(X \sim \operatorname{Bin}(250, 0.90)\). For this left-tailed test, find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). 3. The binomial probabilities are \(P(X \le 216) \approx 0.0410\) and \(P(X \le 217) \approx 0.0611\). Therefore, the rejection region is \(R = \{0, 1, \ldots, 216\}\). 4. The observed count \(218\) is not in the rejection region, so fail to reject \(H_0\). There is not sufficient evidence at the \(0.05\) level that fewer than \(90\%\) of the panels still produce their rated power after \(20\) years.

a) \(H_0: p = 0.90\); \(H_a: p < 0.90\). b) \(R = \{0, 1, \ldots, 216\}\). Reject \(H_0\) if \(216\) or fewer panels still produce their rated power. c) Because \(218 \notin R\), fail to reject \(H_0\). The data do not provide sufficient evidence at the \(0.05\) level to reject the manufacturer’s guarantee.

- Treat the two tests separately because they use different null values. - Team A rejects for large counts, while Team B rejects for small counts. - Find each exact binomial cutoff at the \(0.05\) level. - Compare the two rejection regions by finding their intersection.

1. Let \(X\) be the number of successful outcomes among the \(100\) patients. 2. For Team A, under \(H_0\), \(X \sim \operatorname{Bin}(100, 0.60)\). This is a right-tailed test. Since \(P(X \le 67) \approx 0.9385\) and \(P(X \le 68) \approx 0.9602\), the rejection region is \(R_A = \{69, 70, \ldots, 100\}\). 3. For Team B, under \(H_0\), \(X \sim \operatorname{Bin}(100, 0.80)\). This is a left-tailed test. Since \(P(X \le 72) \approx 0.0342\) and \(P(X \le 73) \approx 0.0558\), the rejection region is \(R_B = \{0, 1, \ldots, 72\}\). 4. Both teams reject when \(X\) belongs to the intersection of the two rejection regions. Thus, \(R_A \cap R_B = \{69, 70, 71, 72\}\).

a) Team A: \(R_A = \{69, 70, \ldots, 100\}\). Team B: \(R_B = \{0, 1, \ldots, 72\}\). b) Both teams reject their respective null hypotheses when \(X \in \{69, 70, 71, 72\}\).

- Model the count under the claimed population proportion. - Use the binomial probability formula for an exact count. - For the one-sided alternative, the p-value includes the observed count and all larger counts. - Compare consecutive upper-tail probabilities with \(\alpha=0.05\).

1. Under the null hypothesis, \(X\sim\operatorname{Bin}(20,0.15)\). Thus \(P(X=3)=\binom{20}{3}\cdot(0.15)^3\cdot(0.85)^{17}\approx 0.2428\). 2. For an observed count \(k\), the one-sided p-value is \(P(X\ge k)\). Seek the smallest \(k\) for which \(P(X\ge k)\le 0.05\). 3. For \(k=6\), \(P(X\ge 6)=1-P(X\le 5)\approx 0.0673>0.05\). 4. For \(k=7\), \(P(X\ge 7)=1-P(X\le 6)\approx 0.0219\le 0.05\). 5. Therefore, the smallest count that leads to rejection is \(7\).

1. \(P(X=3)\approx 0.2428\), or about \(24.28\%\) 2. Reject \(H_0\) when the sample contains at least \(7\) supporters.

- Define the parameter as this inspector’s probability of assigning a “no defects” rating for a comparable vehicle. - A stricter inspector would produce an unusually small count of “no defects” ratings. - Find the exact lower-tail cutoff at the \(0.05\) level. - Consider why comparable vehicle assignments are necessary for attributing the difference to the inspector.

1. Let \(p\) be the probability that this inspector gives a “no defects” rating to a vehicle drawn from the center’s usual population. The hypotheses are \(H_0: p = 0.72\) and \(H_a: p < 0.72\). 2. Let \(X\) be the number of “no defects” ratings among \(500\) inspections. Under \(H_0\), \(X \sim \operatorname{Bin}(500, 0.72)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). The exact probabilities are \(P(X \le 342) \approx 0.0419\) and \(P(X \le 343) \approx 0.0513\). Therefore, \(c = 342\). 4. The rejection region is \(R = \{0, 1, \ldots, 342\}\). 5. The observed count \(342\) is in the rejection region, so reject \(H_0\). Under the stated assumptions, there is sufficient evidence at the \(0.05\) level that this inspector gives fewer “no defects” ratings than the center average.

Hypotheses: \(H_0: p = 0.72\); \(H_a: p < 0.72\). Rejection region: \(R = \{0, 1, \ldots, 342\}\). Conclusion: Because \(342 \in R\), reject \(H_0\). Under the stated assumptions, the inspector gives significantly fewer “no defects” ratings than the center average.

- A claim that red occurs too often requires a right-tailed test. - Use the fair-wheel probability \(\frac{18}{37}\) under \(H_0\). - Find the first count whose upper-tail probability is at most \(0.05\). - The p-value is the probability of \(265\) or more red outcomes under \(H_0\).

1. Let \(p\) be the true probability of red. The hypotheses are \(H_0: p = \frac{18}{37}\) and \(H_a: p > \frac{18}{37}\). 2. Let \(X\) be the number of red outcomes in \(500\) spins. Under \(H_0\), \(X \sim \operatorname{Bin}\left(500, \frac{18}{37}\right)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). The exact probabilities are \(P(X \ge 262) \approx 0.0512\) and \(P(X \ge 263) \approx 0.0425\). Therefore, the rejection region is \(R = \{263, 264, \ldots, 500\}\). 4. The observed count \(265\) is in the rejection region. Its one-sided p-value is \(P(X \ge 265) \approx 0.0286\). 5. Since \(0.0286 < 0.05\), reject \(H_0\). The data provide sufficient evidence that red occurs more often than expected on a fair European roulette wheel.

Yes. The rejection region is \(R = \{263, 264, \ldots, 500\}\), and \(265\) is in that region. The one-sided p-value is approximately \(0.0286\), so reject \(H_0\) at the \(0.05\) significance level.

- A claim of “more than” requires a right-tailed test. - Calculate the mean and standard deviation under the null proportion. - Use the one-sided critical z-score and a continuity correction. - Compare the observed count with the approximate critical count.

1. Let \(p\) be the true satisfaction proportion. The hypotheses are \(H_0: p = 0.80\) and \(H_a: p > 0.80\). 2. Under \(H_0\), the count of satisfied users has mean \(\mu = 400 \cdot 0.80 = 320\) and standard deviation \(\sigma = \sqrt{400 \cdot 0.80 \cdot 0.20} = 8\). 3. For a right-tailed test with \(\alpha = 0.05\), use \(z_{0.95} \approx 1.645\). With continuity correction, the critical count \(k\) satisfies \(\frac{k - 0.5 - 320}{8} \ge 1.645\). This gives \(k - 0.5 \ge 333.16\), so the smallest integer cutoff is \(k = 334\). 4. The approximate rejection region is \(R = \{334, 335, \ldots, 400\}\). 5. The observed count \(335\) is in the rejection region, so reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that more than \(80\%\) of users are satisfied.

1. \(H_0: p = 0.80\); \(H_a: p > 0.80\). 2. The approximate rejection region is \(R = \{334, 335, \ldots, 400\}\). 3. Because \(335 \in R\), reject \(H_0\). The data support the company’s claim at the \(0.05\) significance level.

- A possible increase requires a right-tailed test. - Find the smallest count whose upper-tail probability is at most \(0.05\). - Compare the observed count with the rejection region. - Distinguish evidence of a change from evidence of what caused the change.

1. Let \(p\) be the provider’s current market-share proportion. The hypotheses are \(H_0: p = 0.12\) and \(H_a: p > 0.12\). 2. Let \(X\) be the number of sampled people who are customers. Under \(H_0\), \(X \sim \operatorname{Bin}(500, 0.12)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). The exact probabilities are \(P(X \le 71) \approx 0.94051\) and \(P(X \le 72) \approx 0.95445\). Thus, \(P(X \ge 73) \approx 0.04555\), and the rejection region is \(R = \{73, 74, \ldots, 500\}\). 4. Since \(75 \in R\), reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that the current market share exceeds \(12\%\). 5. This test compares the current proportion with last year’s proportion. It does not establish that the advertising campaign caused any increase.

The hypotheses are \(H_0: p = 0.12\) and \(H_a: p > 0.12\). The rejection region is \(R = \{73, 74, \ldots, 500\}\). Because \(75 \in R\), reject \(H_0\). The data provide sufficient evidence that the current market share is above \(12\%\), but they do not show that the advertising campaign caused the increase.

- Ask which population proportion is assumed when the null distribution is built. - Check sampling independence and expected outcome counts using the null claim. - Use the direction of the alternative to select the relevant tail after standardizing the sample result.

1. The sample proportion is \(\hat p=\frac{112}{250}=0.448\). Student A is correct because a hypothesis test calculates variability under the null value \(p_0=0.40\). 2. The sample is random, \(250\le 0.10(10{,}000)=1000\), and the null expected counts are \(250(0.40)=100\) and \(250(0.60)=150\), both at least \(10\). 3. The null standard error is \(\sqrt{\frac{0.40(0.60)}{250}}\approx 0.03098\). 4. The test statistic is \(z=\frac{0.448-0.40}{0.03098}\approx 1.549\). 5. For the right-tailed test, the p-value is \(P(Z\ge 1.549)\approx 0.0607\). 6. Because \(0.0607>0.05\), fail to reject \(H_0\). The sample does not provide convincing evidence that the proportion of visitors who stay at least two hours has increased above \(0.40\).

a) Student A, because the test standard error is calculated under \(H_0\). b) The randomization, \(10\%\), and null large-count conditions are satisfied. c) \(z\approx 1.549\) and p-value \(\approx 0.0607\). d) Fail to reject \(H_0\). There is not sufficient evidence at \(\alpha=0.05\) that more than \(40\%\) of visitors stay at least two hours.

- Translate “differs” into the appropriate pair of hypotheses. - Check the expected counts using the advertised proportion rather than the sample estimate. - Because departures in either direction matter, account for both tails of the null distribution.

1. Test \(H_0:p=0.92\) against \(H_a:p\ne 0.92\). 2. The random sample and \(10\%\) condition support independence. Under \(H_0\), \(np_0=368\) and \(n(1-p_0)=32\), so the large-count condition is met. 3. The sample proportion is \(\hat p=\frac{352}{400}=0.88\). 4. The null standard error is \(\sqrt{\frac{(0.92)(0.08)}{400}}\approx 0.01356\). 5. The test statistic is \(z=\frac{0.88-0.92}{0.01356}\approx -2.949\). 6. The two-sided p-value is approximately \(0.00319\). 7. Since \(0.00319<0.01\), reject \(H_0\). There is evidence that the true on-time proportion differs from \(92\%\).

\(z\approx -2.949\) and p-value \(\approx 0.00319\). Reject \(H_0\) at \(\alpha=0.01\); the data provide evidence that the true on-time proportion is not \(0.92\).

- Check the sampling method, population-to-sample size relationship, and expected counts under the null before calculating the statistic. - Compute the observed sample proportion, then compare it with the null value using the null-based standard error. - Match the tail probability to the direction of the alternative hypothesis before making the decision.

1. The sample is simple random, and \(250\) is no more than \(10\%\) of more than \(5000\) riders, so the random and \(10\%\) conditions are satisfied. 2. Under \(H_0\), \(np_0=250(0.50)=125\) and \(n(1-p_0)=125\), so the large-count condition is satisfied. 3. The sample proportion is \(\hat p=\frac{145}{250}=0.58\). 4. The null standard error is \(\sqrt{\frac{0.50(0.50)}{250}}\approx0.03162\), so \(z=\frac{0.58-0.50}{0.03162}\approx2.53\). 5. For a right-tailed test, the p-value is \(P(Z\ge2.53)\approx0.0057\). 6. Because \(0.0057<0.05\), reject \(H_0\). There is convincing evidence that more than half of Riverbend Transit riders use mobile tickets.

Conditions: satisfied. \(z\approx2.53\). p-value \(\approx0.0057\). Reject \(H_0\); there is convincing evidence that the proportion of Riverbend Transit riders who use mobile tickets is greater than \(0.50\).

- Identify the fixed number of trials and the probability of success under \(H_0\). - For a left-tailed test, find the largest cutoff whose cumulative probability does not exceed \(0.01\). - The nominal significance level and the attainable Type I error probability need not be exactly equal for a discrete distribution. - Compare the observed count directly with the rejection region.

1. Let \(X\) be the number of patients for whom the medication works within \(15\) minutes. Under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.90)\). 2. This is a left-tailed test. Find the largest integer \(k\) such that \(P(X \le k) \le 0.01\). The binomial probabilities give \(P(X \le 169) \approx 0.0095\) and \(P(X \le 170) \approx 0.0163\). Therefore, the rejection region is \(R = \{0, 1, \ldots, 169\}\). 3. At the null boundary \(p = 0.90\), the actual Type I error probability is \(P(X \le 169) \approx 0.0095\), or about \(0.95\%\). 4. The observed count \(172\) is not in the rejection region. Fail to reject \(H_0\). The study does not provide sufficient evidence at the \(0.01\) level that the medication works within \(15\) minutes for fewer than \(90\%\) of patients.

a) \(X\) is the number of patients for whom the medication works within \(15\) minutes, and \(X \sim \operatorname{Bin}(200, 0.90)\) under \(H_0\). b) \(R = \{0, 1, \ldots, 169\}\). c) \(P(X \le 169) \approx 0.0095\), or about \(0.95\%\). d) Because \(172 \notin R\), fail to reject \(H_0\). There is not sufficient evidence at the \(0.01\) level that the true proportion is less than \(0.90\).

- Define the population proportion and identify the direction of the organization’s suspicion. - For a left-tailed exact binomial test, look for a lower-tail cutoff. - Think about what a smaller significance level does to the chance of rejecting a true null hypothesis. - Decide whether the rejection region must expand or shrink when \(\alpha\) decreases.

1. Let \(p\) be the true proportion of patients for whom the medication produces the intended effect. The hypotheses are \(H_0: p = 0.70\) and \(H_a: p < 0.70\). This is a left-tailed test. 2. Under \(H_0\), the number of patients who experience the intended effect is \(X \sim \operatorname{Bin}(100, 0.70)\). Find the largest integer \(c\) such that \(P(X \le c) \le 0.10\). 3. The cumulative probabilities are \(P(X \le 62) \approx 0.0530\), \(P(X \le 63) \approx 0.0799\), and \(P(X \le 64) \approx 0.1161\). Therefore, \(c = 63\), and the rejection region is \(R = \{0, 1, \ldots, 63\}\). 4. Reducing \(\alpha\) makes the rejection criterion more stringent. The rejection region must have a smaller probability under \(H_0\), so its upper endpoint must decrease and the region becomes smaller.

a) \(H_0: p = 0.70\); \(H_a: p < 0.70\). b) \(R = \{0, 1, \ldots, 63\}\). Reject \(H_0\) if the medication works for \(63\) or fewer of the \(100\) patients. c) The rejection region would become smaller, so its upper endpoint would be less than \(63\). A smaller significance level requires more extreme evidence before rejecting \(H_0\).

- Analyze the right-tailed and left-tailed tests separately. - For the right-tailed test, use the complement of a cumulative probability. - For the left-tailed test, use the cumulative probability directly. - The requested middle range is outside both rejection regions.

1. Let \(X\) be the number of defective chips. Under \(H_0\), \(X \sim \operatorname{Bin}(500, 0.08)\). 2. For Test A, find the smallest integer \(k_A\) such that \(P(X \ge k_A) \le 0.05\). Since \(P(X \le 49) \approx 0.9378\) and \(P(X \le 50) \approx 0.9545\), the first qualifying cutoff is \(k_A = 51\). Thus, \(R_A = \{51, 52, \ldots, 500\}\). 3. For Test B, find the largest integer \(k_B\) such that \(P(X \le k_B) \le 0.05\). Since \(P(X \le 29) \approx 0.0372\) and \(P(X \le 30) \approx 0.0543\), \(k_B = 29\). Thus, \(R_B = \{0, 1, \ldots, 29\}\). 4. Neither test rejects \(H_0\) when the observed count is outside both rejection regions. Therefore, the shared nonrejection range is \(30\) through \(50\), inclusive.

1. Test A: \(R_A = \{51, 52, \ldots, 500\}\). Test B: \(R_B = \{0, 1, \ldots, 29\}\). 2. Neither separate test rejects \(H_0\) when the observed number of defective chips is from \(30\) through \(50\), inclusive.

- Put the previous commuter proportion in the null hypothesis and the claimed increase in the alternative. - For the rejection region, use the upper tail of a binomial distribution. - The actual Type I error probability is the probability of the chosen rejection region at the null value. - Recall how the standard deviation of a sample proportion depends on \(n\).

1. Let \(p\) be the true proportion of commuters who regularly use rail service. The hypotheses are \(H_0: p = 0.25\) and \(H_a: p > 0.25\). 2. Under \(H_0\), the number of regular rail users is \(X \sim \operatorname{Bin}(150, 0.25)\). Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). 3. The cumulative probabilities are \(P(X \le 45) \approx 0.9320\) and \(P(X \le 46) \approx 0.9527\). Therefore, \(k = 47\), and the rejection region is \(R = \{47, 48, \ldots, 150\}\). 4. The actual Type I error probability at \(p = 0.25\) is \(P(X \ge 47) = 1 - P(X \le 46) \approx 0.0473\), or about \(4.73\%\). 5. When the sample size is quadrupled, the standard deviation of the sample proportion is cut in half. At the same significance level, the critical sample proportion moves closer to \(0.25\) from above because smaller departures from the null value can be detected with the larger sample.

a) \(H_0: p = 0.25\); \(H_a: p > 0.25\). b) \(R = \{47, 48, \ldots, 150\}\). c) \(P(\text{Type I error}) \approx 0.0473\), or \(4.73\%\). d) The critical sample proportion would move closer to \(0.25\) from above because the larger sample has less sampling variability.

- A suspected decrease leads to a left-tailed test. - Use the exact binomial distribution at the null boundary value. - Find the largest lower-tail cutoff whose probability is at most \(0.01\). - Convert the critical count to a sample proportion by dividing by \(400\).

1. Let \(p\) be the true germination proportion for the new batch. The hypotheses are \(H_0: p = 0.90\) and \(H_a: p < 0.90\). 2. Let \(X\) be the number of seeds that germinate. Under \(H_0\), \(X \sim \operatorname{Bin}(400, 0.90)\). 3. For a left-tailed test, find the largest integer \(c\) such that \(P(X \le c) \le 0.01\). The exact probabilities are \(P(X \le 344) \approx 0.0066\) and \(P(X \le 345) \approx 0.01004 > 0.01\). Thus, \(c = 344\). 4. The rejection region is \(R = \{0, 1, \ldots, 344\}\). 5. The greatest sample proportion in the rejection region is \(\frac{344}{400} = 0.86\), or \(86\%\).

a) \(H_0: p = 0.90\); \(H_a: p < 0.90\). b) \(R = \{0, 1, \ldots, 344\}\). c) The greatest sample germination proportion that leads to rejection is \(\frac{344}{400} = 0.86\), or \(86\%\).

- A suspected increase requires a right-tailed test. - Use the cumulative binomial distribution to locate the upper-tail cutoff. - The greatest nonrejection count is one less than the first rejection count.

1. Let \(p\) be the true damage proportion. The hypotheses are \(H_0: p = 0.03\) and \(H_a: p > 0.03\). 2. Let \(X\) be the number of damaged packages. Under \(H_0\), \(X \sim \operatorname{Bin}(400, 0.03)\). 3. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). The cumulative probabilities are \(P(X \le 20) \approx 0.9895\) and \(P(X \le 21) \approx 0.9947\). Therefore, \(k = 22\). 4. Reject \(H_0\) if \(X \ge 22\). The rejection region is \(R = \{22, 23, \ldots, 400\}\). 5. The greatest count that does not lead to rejection is \(21\).

a) \(H_0: p = 0.03\); \(H_a: p > 0.03\). b) Reject \(H_0\) when \(X \ge 22\); \(R = \{22, 23, \ldots, 400\}\). c) The greatest number of damaged packages that does not lead to rejection is \(21\).

- Model the number of participants with the side effect using a binomial distribution. - For zero occurrences, use \(P(X = 0) = (1 - p_0)^n\). - Determine when the lower-tail probability of the observed result is greater than \(\alpha\).

1. Let \(X\) be the number of participants who experience the side effect. If \(p = 0.001\), then \(X \sim \operatorname{Bin}(2500, 0.001)\). 2. The probability of observing no side effects is \(P(X = 0) = (1 - 0.001)^{2500} = 0.999^{2500} \approx 0.08198\), or about \(8.20\%\). 3. For \(X = 0\) not to be in the rejection region of the left-tailed test, its lower-tail probability must satisfy \(P_{p_0}(X \le 0) > 0.01\). 4. Since \(P_{p_0}(X = 0) = (1 - p_0)^{2500}\), the boundary satisfies \((1 - p_0)^{2500} = 0.01\). Thus, \(p_0 = 1 - 0.01^{1/2500} \approx 0.001840\). 5. Therefore, among valid nondegenerate null values, \(X = 0\) is not in the rejection region for \(0 < p_0 < 0.001840\), or approximately \(0 < p_0 < 0.184\%\). At the boundary value, the lower-tail probability equals \(\alpha\), so \(X = 0\) is in the rejection region.

a) \(P(X = 0) = 0.999^{2500} \approx 0.08198\), or about \(8.20\%\). b) The boundary is \(p_0 = 1 - 0.01^{1/2500} \approx 0.001840\), or about \(0.184\%\). For valid nondegenerate null values, \(X = 0\) is not in the rejection region when \(0 < p_0 < 0.001840\).

- Translate the significance level into a cutoff on the null sampling distribution. - Convert the cutoff proportion into a count, paying attention to the strict inequality. - Check the first possible whole-number count beyond the cutoff by calculating its tail probability.

1. For a right-tailed test with \(\alpha=0.01\), the critical standard score is \(z^*\approx 2.326\). 2. Under \(H_0\), the standard error is \(\sqrt{\frac{(0.30)(0.70)}{150}}\approx 0.03742\). 3. Rejection requires \(\hat p>0.30+2.326(0.03742)\approx 0.38704\). 4. The corresponding count must satisfy \(x>150(0.38704)\approx 58.057\), so the smallest whole-number count is \(59\). 5. For \(x=59\), \(\hat p=\frac{59}{150}\approx 0.3933\), giving \(z\approx 2.494\) and a right-tail p-value of approximately \(0.00631\). 6. Because \(0.00631<0.01\), a count of \(59\) rejects \(H_0\), while \(58\) does not reach the critical cutoff.

The smallest rejecting count is \(59\) successes. For \(x=59\), the p-value is approximately \(0.00631\), which is below \(0.01\).

- Determine each rejection region from its own null distribution. - As a binomial parameter decreases, the distribution and its lower-tail cutoff shift left. - The overlap disappears when Marc’s largest rejection count drops below Julia’s smallest rejection count. - Use a continuity correction when replacing the binomial probability with a normal probability.

1. For Julia’s test, under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.25)\). Since \(P(X \le 59) \approx 0.9375\) and \(P(X \le 60) \approx 0.9546\), the right-tailed rejection region is \(R_J = \{61, 62, \ldots, 200\}\). 2. For Marc’s test with \(p_M = 0.40\), \(X \sim \operatorname{Bin}(200, 0.40)\) under \(H_0\). Since \(P(X \le 68) \approx 0.0475\) and \(P(X \le 69) \approx 0.0639\), the left-tailed rejection region is \(R_M = \{0, 1, \ldots, 68\}\). The overlap is \(R_J \cap R_M = \{61, 62, \ldots, 68\}\). 3. As \(p_M\) decreases, Marc’s null distribution shifts left, so the upper endpoint of his rejection region also moves left. The overlap shrinks and disappears once \(61\) is no longer in Marc’s rejection region. 4. At the approximate boundary, use \(P(X \le 61) \approx 0.05\). With continuity correction, \(\Phi\left(\frac{61.5 - 200p_M}{\sqrt{200p_M(1-p_M)}}\right) = 0.05\). Using \(z_{0.05} \approx -1.645\) gives \(p_M \approx 0.363\). Thus, for values below about \(0.363\), the two rejection regions no longer overlap.

a) \(R_J = \{61, 62, \ldots, 200\}\). b) \(R_M = \{0, 1, \ldots, 68\}\). Both tests reject for \(X \in \{61, 62, \ldots, 68\}\). c) Lowering \(p_M\) shifts Marc’s rejection region left and reduces the overlap. The overlap disappears at approximately \(p_M = 0.363\), so values below about \(36.3\%\) produce no common rejection count.

- Substitute the reported statistic, sample proportion, and sample size into the standardization equation. - The unknown appears in both the numerator and the null-based spread, so solve the resulting equation carefully. - Use the direction of the alternative when converting the standardized statistic to a tail probability.

1. The test statistic equation is \(\frac{0.46-p_0}{\sqrt{\frac{p_0(1-p_0)}{400}}}=-2.00\). 2. Solving for the valid proportion gives \(p_0\approx 0.50999\). 3. The left-tail probability for \(z=-2.00\) is approximately \(0.02275\). 4. Because \(0.02275<0.05\), reject \(H_0\). There is convincing evidence that the population proportion is less than approximately \(0.50999\).

a) \(p_0\approx 0.50999\). b) The p-value is approximately \(0.02275\). c) Reject \(H_0\); there is convincing evidence that the population proportion is below the null value.