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- 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 \(H_0\).

- 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 \(0.02\).

- 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.

- 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 \(R = \{21, 22, \ldots, 50\}\).

- Decide whether small or large values of the test statistic support \(H_a\). - For a right-tailed test, find the first count whose upper-tail probability is at most \(\alpha\). - Use the cumulative binomial distribution to evaluate the upper tail. - Compare the observed count with the rejection region.

1. Under \(H_0\), the number of successes is \(X \sim \operatorname{Bin}(80, 0.30)\). For a right-tailed test, find the smallest critical value \(k\) such that \(P(X \ge k) \le 0.05\). Equivalently, find the smallest \(k\) such that \(P(X \le k - 1) \ge 0.95\). 2. The cumulative probabilities are \(P(X \le 29) \approx 0.9084\), \(P(X \le 30) \approx 0.9413\), and \(P(X \le 31) \approx 0.9640\). Therefore, the critical value is \(k = 32\), and the rejection region is \(R = \{32, 33, \ldots, 80\}\). 3. Because the observed value \(x = 32\) is in the rejection region, reject \(H_0\) at the \(0.05\) significance level.

1. \(R = \{32, 33, \ldots, 80\}\). 2. Because \(32 \in R\), reject \(H_0\).

- Decide which values of \(X\) support the claim that \(p < 0.75\). - For a left-tailed test, use cumulative probabilities of the form \(P(X \le c)\). - Choose the largest cutoff whose probability does not exceed \(0.01\). - Compare the observed count with the rejection region.

1. Under \(H_0\), the number of successes is \(X \sim \operatorname{Bin}(120, 0.75)\). For a left-tailed test, choose the largest critical value \(c\) such that \(P(X \le c) \le 0.01\). 2. The cumulative probabilities are \(P(X \le 77) \approx 0.0054\), \(P(X \le 78) \approx 0.0093\), and \(P(X \le 79) \approx 0.0155\). Therefore, \(c = 78\), and the rejection region is \(R = \{0, 1, \ldots, 78\}\). 3. The observed count \(x = 78\) is in the rejection region, so reject \(H_0\).

1. \(R = \{0, 1, \ldots, 78\}\). 2. Because \(78 \in R\), reject \(H_0\).

- 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.

- Identify the distribution of the number of germinated seeds under \(H_0\). - A right-tailed test uses unusually large germination counts as evidence. - Express the upper-tail probability using the cumulative distribution if needed. - Choose the first count whose upper-tail probability is at most \(0.05\).

1. Let \(X\) be the number of treated seeds that germinate. Under \(H_0\), \(X \sim \operatorname{Bin}(100, 0.80)\). 2. For a right-tailed test, 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 \(P(X \le 85) \approx 0.9196\) and \(P(X \le 86) \approx 0.9531\). Therefore, \(k = 87\). 4. The rejection region is \(R = \{87, 88, \ldots, 100\}\). Reject \(H_0\) if at least \(87\) of the \(100\) treated seeds germinate.

The rejection region is \(R = \{87, 88, \ldots, 100\}\). Reject \(H_0\) if \(87\) or more treated seeds germinate.

- Identify which sample counts support a proportion below \(0.90\). - For a left-tailed test, use probabilities of the form \(P(X \le c)\). - Choose the largest cutoff whose cumulative probability does not exceed \(0.02\). - Check the probabilities on both sides of the proposed cutoff.

1. Let \(X\) be the number of batteries that retain full capacity for two years. Under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.90)\). 2. This is a left-tailed test, so find the largest integer \(c\) such that \(P(X \le c) \le 0.02\). 3. The cumulative probabilities are \(P(X \le 170) \approx 0.0163\) and \(P(X \le 171) \approx 0.0271\). 4. Therefore, \(c = 170\), and the rejection region is \(R = \{0, 1, \ldots, 170\}\).

The rejection region is \(R = \{0, 1, \ldots, 170\}\).

- A decrease in the defect rate is supported by unusually small defect counts. - Model the number of defective bulbs with a binomial distribution under \(H_0\). - Find the largest lower-tail cutoff whose probability is at most \(0.05\). - State the cutoff as a decision rule in context.

1. Let \(X\) be the number of defective bulbs in the sample. Under \(H_0\), \(X \sim \operatorname{Bin}(150, 0.06)\). 2. Because the alternative states that the defect rate has decreased, this is a left-tailed test. The rejection region has the form \(R = \{0, 1, \ldots, c\}\). 3. Choose the largest integer \(c\) such that \(P(X \le c) \le 0.05\). 4. The cumulative probabilities are \(P(X \le 3) \approx 0.0186\), \(P(X \le 4) \approx 0.0499\), and \(P(X \le 5) \approx 0.1083\). 5. Therefore, \(c = 4\). Reject \(H_0\) if \(4\) or fewer of the \(150\) bulbs are defective; otherwise, fail to reject \(H_0\).

Reject \(H_0\) if \(X \le 4\). The rejection region is \(R = \{0, 1, 2, 3, 4\}\).

- Identify which end of the distribution supports \(H_a\). - Express the upper-tail probability using a cumulative probability. - Find the first cutoff whose upper-tail probability is no greater than \(0.10\). - Write the result as a complete decision rule.

1. Under \(H_0\), the number of successes is \(X \sim \operatorname{Bin}(100, 0.30)\). 2. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.10\). Equivalently, find the smallest \(k\) such that \(P(X \le k - 1) \ge 0.90\). 3. The cumulative probabilities are \(P(X \le 35) \approx 0.8839\) and \(P(X \le 36) \approx 0.9201\). 4. Therefore, \(k = 37\). Reject \(H_0\) when \(X \ge 37\); otherwise, fail to reject \(H_0\).

Reject \(H_0\) when \(X \ge 37\). The rejection region is \(R = \{37, 38, \ldots, 100\}\).

- Decide whether unusually small or unusually large germination counts challenge the producer’s claim. - Use a binomial distribution with the null boundary proportion. - Find the largest lower-tail cutoff whose probability is at most \(0.05\). - Compare the observed count with the rejection region.

1. Let \(p\) be the true germination proportion. The hypotheses are \(H_0: p = 0.92\) and \(H_a: p < 0.92\). 2. Under \(H_0\), the number of seeds that germinate is \(X \sim \operatorname{Bin}(200, 0.92)\). For this left-tailed test, find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). 3. The cumulative probabilities are \(P(X \le 176) \approx 0.0306\) and \(P(X \le 177) \approx 0.0507\). Therefore, the rejection region is \(R = \{0, 1, \ldots, 176\}\). 4. The observed count \(178\) is not in the rejection region, so fail to reject \(H_0\). There is not sufficient evidence at the \(0.05\) level that the true germination rate is below \(92\%\).

a) \(H_0: p = 0.92\); \(H_a: p < 0.92\); \(R = \{0, 1, \ldots, 176\}\). b) Because \(178 \notin R\), fail to reject \(H_0\). The sample does not provide sufficient evidence that the germination rate is less than \(92\%\).

- Use the direction of \(H_a\) to identify the relevant tail. - For a right-tailed test, the rejection region contains large values of \(X\). - Relate \(P(X \ge k)\) to \(P(X \le k - 1)\). - Choose the first cutoff whose upper-tail probability is at most \(0.10\).

1. Under \(H_0\), the number of successes is \(X \sim \operatorname{Bin}(150, 0.40)\). 2. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.10\). Equivalently, find the smallest \(k\) such that \(P(X \le k - 1) \ge 0.90\). 3. The cumulative probabilities are \(P(X \le 67) \approx 0.8939\) and \(P(X \le 68) \approx 0.9210\). 4. Therefore, \(k = 69\), and the rejection region is \(R = \{69, 70, \ldots, 150\}\). Reject \(H_0\) if at least \(69\) successes are observed.

The rejection region is \(R = \{69, 70, \ldots, 150\}\). Reject \(H_0\) when \(X \ge 69\).

- Decide whether small or large values of the test statistic support \(H_a\). - For a left-tailed test, use cumulative probabilities directly. - Find the largest cutoff whose cumulative probability does not exceed \(0.05\).

1. Because \(H_a: p < 0.25\), this is a left-tailed test. 2. Under \(H_0\), the number of successes is \(X \sim \operatorname{Bin}(100, 0.25)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). 4. The cumulative probabilities are \(P(X \le 17) \approx 0.0376\) and \(P(X \le 18) \approx 0.0630\). 5. Therefore, \(c = 17\), and the rejection region is \(R = \{0, 1, \ldots, 17\}\).

The rejection region is \(R = \{0, 1, \ldots, 17\}\). Reject \(H_0\) if \(17\) or fewer successes are observed.

- Decide whether small or large germination counts challenge the company’s claim. - Model the number of germinated seeds with a binomial distribution under \(H_0\). - For a left-tailed test, find the largest cumulative probability cutoff that does not exceed \(0.05\). - Check the values immediately below and above the proposed cutoff.

1. Let \(X\) be the number of seeds that germinate. Under \(H_0\), \(X \sim \operatorname{Bin}(400, 0.75)\). 2. This is a left-tailed test, so find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). 3. The cumulative probabilities are \(P(X \le 285) \approx 0.0486\) and \(P(X \le 286) \approx 0.0610\). 4. Therefore, \(c = 285\), and the rejection region is \(R = \{0, 1, \ldots, 285\}\).

The rejection region is \(R = \{0, 1, \ldots, 285\}\).

- A challenge to an “at least” claim is supported by unusually small counts. - Model the number of supporters with a binomial distribution under the null boundary value. - Find the largest lower-tail cutoff whose probability is at most \(0.10\). - Verify the cumulative probabilities immediately around the cutoff.

1. Let \(X\) be the number of surveyed voters who support the candidate. Under \(H_0\), \(X \sim \operatorname{Bin}(400, 0.45)\). 2. This is a left-tailed test, so find the largest integer \(c\) such that \(P(X \le c) \le 0.10\). 3. The cumulative probabilities are \(P(X \le 166) \approx 0.0872\) and \(P(X \le 167) \approx 0.1043\). 4. Therefore, \(c = 166\), and the rejection region is \(R = \{0, 1, \ldots, 166\}\).

The rejection region is \(R = \{0, 1, \ldots, 166\}\).

- Define the parameter as the population proportion of late deliveries. - A higher-than-advertised rate is supported by unusually large late-delivery counts. - Express the upper-tail probability using the cumulative binomial distribution. - Choose the first cutoff whose upper-tail probability is at most \(0.05\).

1. Let \(X\) be the number of late orders in the sample. Under \(H_0\), \(X \sim \operatorname{Bin}(250, 0.10)\). 2. This is a right-tailed test. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\), or equivalently \(P(X \le k - 1) \ge 0.95\). 3. The cumulative probabilities are \(P(X \le 32) \approx 0.9389\) and \(P(X \le 33) \approx 0.9590\). 4. Therefore, \(k = 34\). Reject \(H_0\) if \(34\) or more of the \(250\) orders arrive late; otherwise, fail to reject \(H_0\).

Reject \(H_0\) when \(X \ge 34\). The rejection region is \(R = \{34, 35, \ldots, 250\}\).

- 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.

- 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 \(R = \{0, 1, \ldots, 7\}\).

- Define the parameter for the probability of a girl being born. - A lower-than-usual count requires a left-tailed test. - Find the largest exact lower-tail cutoff with probability at most \(0.05\). - Compare the observed count with the rejection region.

1. Let \(p\) be the probability that a randomly selected birth is a girl. The hypotheses are \(H_0: p = 0.485\) and \(H_a: p < 0.485\). 2. Let \(X\) be the number of girls among \(1200\) births. Under \(H_0\), \(X \sim \operatorname{Bin}(1200, 0.485)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). The exact probabilities are \(P(X \le 553) \approx 0.0498\) and \(P(X \le 554) \approx 0.0560\). Therefore, \(c = 553\). 4. The rejection region is \(R = \{0, 1, \ldots, 553\}\). 5. The observed count \(558\) is not in the rejection region, so fail to reject \(H_0\). The number of girls is not significantly low at the \(0.05\) level.

Hypotheses: \(H_0: p = 0.485\); \(H_a: p < 0.485\). Rejection region: \(R = \{0, 1, \ldots, 553\}\). Conclusion: Because \(558 \notin R\), fail to reject \(H_0\). The observed number of girls is not significantly low at the \(0.05\) level.

- 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 suspected increase requires a right-tailed test. - Use the cumulative binomial distribution to evaluate upper-tail probabilities. - Find the first count whose upper-tail probability is no greater than \(0.05\). - Be careful whether the boundary count belongs to the rejection region.

1. Let \(X\) be the number of sampled lights with a color variation. Under \(H_0\), \(X \sim \operatorname{Bin}(50, 0.15)\). 2. For a right-tailed test, 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 \(P(X \le 11) \approx 0.9372\) and \(P(X \le 12) \approx 0.9699\). 4. Therefore, \(k = 13\), and the rejection region is \(R = \{13, 14, \ldots, 50\}\).

The rejection region is \(R = \{13, 14, \ldots, 50\}\).

- For a fair die, the probability of a \(6\) is \(\frac{1}{6}\). - The player’s suspicion concerns an increase, so use the upper tail. - Find the first count whose probability of that many or more sixes is at most \(0.05\). - Compare the observed count with the rejection region.

1. Let \(p\) be the probability of rolling a \(6\). The hypotheses are \(H_0: p = \frac{1}{6}\) and \(H_a: p > \frac{1}{6}\). 2. Let \(X\) be the number of sixes in \(600\) rolls. Under \(H_0\), \(X \sim \operatorname{Bin}\left(600, \frac{1}{6}\right)\). 3. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). The cumulative probabilities are \(P(X \le 114) \approx 0.9419\) and \(P(X \le 115) \approx 0.9532\). Therefore, \(k = 116\). 4. The rejection region is \(R = \{116, 117, \ldots, 600\}\). 5. The observed count \(118\) is in the rejection region, so reject \(H_0\). There is sufficient evidence at the \(0.05\) level that this die rolls a \(6\) more often than a fair die.

Hypotheses: \(H_0: p = \frac{1}{6}\); \(H_a: p > \frac{1}{6}\). Rejection region: \(R = \{116, 117, \ldots, 600\}\). Conclusion: Because \(118 \in R\), reject \(H_0\). The data support the claim that the die rolls a \(6\) too often.

- Define the parameter as the long-term reduced-output proportion. - A suspected increase requires a right-tailed test. - Find the first upper-tail cutoff whose probability is at most \(0.05\). - Check whether the observed count equals the critical value.

1. Let \(p\) be the true proportion of panels with reduced output after \(10\) years. The hypotheses are \(H_0: p = 0.03\) and \(H_a: p > 0.03\). 2. Under \(H_0\), the number of panels with reduced output is \(X \sim \operatorname{Bin}(400, 0.03)\). 3. For the right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). The cumulative probabilities are \(P(X \le 17) \approx 0.9399\) and \(P(X \le 18) \approx 0.9649\). Therefore, \(k = 19\). 4. The rejection region is \(R = \{19, 20, \ldots, 400\}\). 5. The observed count \(19\) is in the rejection region, so reject \(H_0\). There is sufficient evidence at the \(0.05\) level that more than \(3\%\) of the panels have reduced output after \(10\) years.

a) \(H_0: p = 0.03\); \(H_a: p > 0.03\). b) \(R = \{19, 20, \ldots, 400\}\). c) Because \(19 \in R\), reject \(H_0\). The data support the advocacy group’s suspicion at the \(0.05\) significance level.

- A suspected decrease requires a left-tailed test. - Find the largest lower-tail cutoff whose probability is at most \(0.10\). - Check whether the observed count equals the critical value. - State the conclusion in terms of the population germination proportion.

1. Let \(p\) be the true germination proportion. 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 \(H_0\), \(X \sim \operatorname{Bin}(250, 0.90)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.10\). The cumulative probabilities are \(P(X \le 218) \approx 0.0886\) and \(P(X \le 219) \approx 0.1247\). Therefore, \(c = 218\). 4. The rejection region is \(R = \{0, 1, \ldots, 218\}\). 5. The observed count \(218\) is in the rejection region, so reject \(H_0\). The data provide sufficient evidence at the \(0.10\) level that the germination proportion is below \(90\%\).

Hypotheses: \(H_0: p = 0.90\); \(H_a: p < 0.90\). Rejection region: \(R = \{0, 1, \ldots, 218\}\). Conclusion: Because \(218 \in R\), reject \(H_0\). The gardener’s suspicion is supported at the \(0.10\) significance level.

- A suspected increase requires a right-tailed test. - Rewrite the upper-tail probability using the cumulative distribution. - Find the first cutoff whose upper-tail probability is at most \(0.01\).

1. Let \(X\) be the number of canceled orders in the sample. Under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.10)\). 2. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). Equivalently, find the smallest \(k\) such that \(P(X \le k - 1) \ge 0.99\). 3. The cumulative probabilities are \(P(X \le 29) \approx 0.9837\) and \(P(X \le 30) \approx 0.9905\). 4. Therefore, \(k = 31\), and the rejection region is \(R = \{31, 32, \ldots, 200\}\).

The rejection region is \(R = \{31, 32, \ldots, 200\}\).

- A fair die displays \(6\) with probability \(\frac{1}{6}\). - The concern involves an unusually large count of sixes. - Find the first upper-tail cutoff whose probability is at most \(0.05\). - Compare the observed count with the rejection region.

1. Let \(p\) be the probability that the digital die displays \(6\). The hypotheses are \(H_0: p = \frac{1}{6}\) and \(H_a: p > \frac{1}{6}\). 2. Let \(X\) be the number of sixes in \(1200\) rolls. Under \(H_0\), \(X \sim \operatorname{Bin}\left(1200, \frac{1}{6}\right)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). The exact cumulative probabilities are \(P(X \le 220) \approx 0.9424\) and \(P(X \le 221) \approx 0.9506\). Therefore, \(k = 222\). 4. The rejection region is \(R = \{222, 223, \ldots, 1200\}\). 5. The observed count \(230\) is in the rejection region, so reject \(H_0\). The data provide sufficient evidence at the \(0.05\) level that the digital die displays \(6\) too often.

a) \(H_0: p = \frac{1}{6}\); \(H_a: p > \frac{1}{6}\). b) \(R = \{222, 223, \ldots, 1200\}\). c) Because \(230 \in R\), reject \(H_0\). The concern is statistically supported at the \(0.05\) level.

- A suspected decrease requires a left-tailed test. - Model the number of users with a binomial distribution under the null value. - Find the largest lower-tail cutoff whose probability is at most \(0.05\).

1. Let \(X\) be the number of surveyed teenagers who use the streaming service. Under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.40)\). 2. For a left-tailed test, find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). 3. The cumulative probabilities are \(P(X \le 68) \approx 0.0475\) and \(P(X \le 69) \approx 0.0639\). 4. Therefore, \(c = 68\), and the rejection region is \(R = \{0, 1, \ldots, 68\}\).

The rejection region is \(R = \{0, 1, \ldots, 68\}\).

- A suspected increase requires a right-tailed test. - Use the cumulative binomial distribution to evaluate the upper tail. - Find the first count whose upper-tail probability is at most \(0.01\). - The requested answer is the first count in the rejection region.

1. Let \(p\) be the true defect proportion. The hypotheses are \(H_0: p = 0.05\) and \(H_a: p > 0.05\). 2. Let \(X\) be the number of defective cells. Under \(H_0\), \(X \sim \operatorname{Bin}(500, 0.05)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). The cumulative probabilities are \(P(X \le 36) \approx 0.9876\) and \(P(X \le 37) \approx 0.9923\). 4. Therefore, \(k = 38\). The rejection region is \(R = \{38, 39, \ldots, 500\}\).

The manufacturer’s claim is rejected when \(38\) or more defective cells are found. The rejection region is \(R = \{38, 39, \ldots, 500\}\).

- A suspected decrease requires a left-tailed test. - Use the exact binomial distribution at the claimed success proportion. - Find the largest lower-tail cutoff whose probability is at most \(0.05\). - Check whether the observed count equals the cutoff.

1. Let \(p\) be the true success proportion. The hypotheses are \(H_0: p = 0.85\) and \(H_a: p < 0.85\). 2. Let \(X\) be the number of patients for whom the medication works. Under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.85)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). The cumulative probabilities are \(P(X \le 161) \approx 0.0498\) and \(P(X \le 162) \approx 0.0720\). Therefore, \(c = 161\). 4. The rejection region is \(R = \{0, 1, \ldots, 161\}\). 5. The observed count \(161\) is in the rejection region, so reject \(H_0\). The data provide sufficient evidence at the \(0.05\) level that the true success proportion is below \(85\%\).

a) \(H_0: p = 0.85\); \(H_a: p < 0.85\). b) \(R = \{0, 1, \ldots, 161\}\). c) Because \(161 \in R\), reject \(H_0\). The data provide evidence against the company’s claim at the \(0.05\) level.

- A suspected increase requires a right-tailed test. - Express the upper-tail condition using a cumulative probability. - Find the first cutoff whose upper-tail probability is at most \(0.01\). - Apply the decision rule to the observed count.

1. Let \(p\) be the true proportion of packages damaged in transit. The hypotheses are \(H_0: p = 0.03\) and \(H_a: p > 0.03\). 2. Let \(X\) be the number of damaged packages. Under \(H_0\), \(X \sim \operatorname{Bin}(500, 0.03)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). The cumulative probabilities are \(P(X \le 24) \approx 0.9899\) and \(P(X \le 25) \approx 0.9945\). Therefore, \(k = 26\). 4. The rejection region is \(R = \{26, 27, \ldots, 500\}\). Reject \(H_0\) if \(26\) or more packages are damaged. 5. The observed count \(23\) is not in the rejection region, so fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(0.01\) level that the damage proportion exceeds \(3\%\).

a) Reject \(H_0\) when \(X \ge 26\); \(R = \{26, 27, \ldots, 500\}\). b) Because \(23 \notin R\), fail to reject \(H_0\). The consumer group’s suspicion is not statistically supported at the \(0.01\) level.

- A suspected increase requires a right-tailed test. - Model the defect count with a binomial distribution at the claimed proportion. - Find the first upper-tail cutoff whose probability is at most \(0.05\). - Check whether the observed count equals the cutoff.

1. Let \(p\) be the true proportion of bulbs with inaccurate color rendering. The hypotheses are \(H_0: p = 0.04\) and \(H_a: p > 0.04\). 2. Let \(X\) be the number of defective bulbs in the sample. Under \(H_0\), \(X \sim \operatorname{Bin}(500, 0.04)\). 3. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). The cumulative probabilities are \(P(X \le 26) \approx 0.9263\) and \(P(X \le 27) \approx 0.9511\). Therefore, \(k = 28\). 4. The rejection region is \(R = \{28, 29, \ldots, 500\}\). 5. The observed count \(28\) is in the rejection region, so reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that the defect proportion exceeds \(4\%\).

Hypotheses: \(H_0: p = 0.04\); \(H_a: p > 0.04\). Rejection region: \(R = \{28, 29, \ldots, 500\}\). Conclusion: Because \(28 \in R\), reject \(H_0\). The manufacturer’s claim is rejected at the \(0.05\) significance level.

- To challenge an “at least” claim, look for unusually small satisfaction counts. - Use the exact binomial distribution at the null boundary value. - Find the largest lower-tail cutoff whose probability is at most \(0.01\). - State the cutoff in the survey context.

1. Let \(p\) be the true proportion of users who are satisfied with the update. The hypotheses are \(H_0: p = 0.90\) and \(H_a: p < 0.90\). 2. Let \(X\) be the number of satisfied users in the sample. Under \(H_0\), \(X \sim \operatorname{Bin}(150, 0.90)\). 3. For a left-tailed test, find the largest integer \(c\) such that \(P(X \le c) \le 0.01\). The cumulative probabilities are \(P(X \le 125) \approx 0.0076\) and \(P(X \le 126) \approx 0.0143\). Therefore, \(c = 125\). 4. The rejection region is \(R = \{0, 1, \ldots, 125\}\). Reject \(H_0\) if \(125\) or fewer of the \(150\) surveyed users report being satisfied.

The rejection region is \(R = \{0, 1, \ldots, 125\}\). Reject the advertising claim if \(125\) or fewer surveyed users report being satisfied.

- Put the claimed increase in the alternative hypothesis. - Use the upper tail because large counts support the manager’s claim. - Find the first count whose upper-tail probability is at most \(0.05\). - Compare the observed count with the rejection region.

1. Let \(p\) be the true proportion of customers who bring reusable shopping bags. The hypotheses are \(H_0: p = 0.40\) and \(H_a: p > 0.40\). 2. Let \(X\) be the number of sampled customers who bring reusable bags. Under \(H_0\), \(X \sim \operatorname{Bin}(60, 0.40)\). 3. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.05\). Since \(P(X \le 29) \approx 0.9254\) and \(P(X \le 30) \approx 0.9555\), the rejection region is \(R = \{31, 32, \ldots, 60\}\). 4. The observed count \(31\) is in the rejection region, so reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that more than \(40\%\) of customers bring reusable bags.

Yes. The rejection region is \(R = \{31, 32, \ldots, 60\}\), and the observed count \(31\) is in that region. The sample supports the manager’s claim at the \(0.05\) significance level.

- 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 suspected decrease requires a left-tailed test. - Find the largest lower-tail cutoff whose probability is at most \(0.05\). - Check whether the observed count equals the critical value. - State the conclusion about the population germination proportion.

1. Let \(p\) be the true germination proportion. 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 \(H_0\), \(X \sim \operatorname{Bin}(200, 0.90)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). The cumulative probabilities are \(P(X \le 172) \approx 0.0434\) and \(P(X \le 173) \approx 0.0672\). Therefore, \(c = 172\). 4. The rejection region is \(R = \{0, 1, \ldots, 172\}\). 5. The observed count \(172\) is in the rejection region, so reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that the germination proportion is below \(90\%\).

a) \(H_0: p = 0.90\); \(H_a: p < 0.90\). b) \(R = \{0, 1, \ldots, 172\}\). c) Yes. Because \(172 \in R\), reject \(H_0\). The company’s claim is rejected at the \(0.05\) significance level.

- An improvement is supported by unusually large numbers of made shots. - Use the upper tail of the binomial distribution. - Find the first cutoff whose upper-tail probability is at most \(0.10\). - Compare the observed number of makes with the rejection region.

1. Let \(p\) be the player’s current probability of making a free throw. The hypotheses are \(H_0: p = 0.70\) and \(H_a: p > 0.70\). 2. Let \(X\) be the number of made free throws. Under \(H_0\), \(X \sim \operatorname{Bin}(80, 0.70)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.10\). The cumulative probabilities are \(P(X \le 60) \approx 0.8648\) and \(P(X \le 61) \approx 0.9127\). Therefore, \(k = 62\). 4. The rejection region is \(R = \{62, 63, \ldots, 80\}\). 5. The observed count \(62\) is in the rejection region, so reject \(H_0\). The results provide sufficient evidence at the \(0.10\) level that the player’s free-throw percentage has improved.

a) \(H_0: p = 0.70\); \(H_a: p > 0.70\). b) \(R = \{62, 63, \ldots, 80\}\). c) Because \(62 \in R\), reject \(H_0\). The data support the player’s improvement claim at the \(0.10\) 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 claim of “more than” requires a right-tailed test. - Use the null proportion to calculate the mean and standard deviation. - Apply the one-sided \(0.01\) critical z-score with continuity correction. - Compare the observed germination count with the approximate cutoff.

1. Let \(p\) be the true germination proportion. The hypotheses are \(H_0: p = 0.40\) and \(H_a: p > 0.40\). 2. Under \(H_0\), the germination count has mean \(\mu = 250 \cdot 0.40 = 100\) and standard deviation \(\sigma = \sqrt{250 \cdot 0.40 \cdot 0.60} = \sqrt{60} \approx 7.75\). 3. For a right-tailed test with \(\alpha = 0.01\), use \(z_{0.99} \approx 2.326\). With continuity correction, the critical count \(k\) satisfies \(\frac{k - 0.5 - 100}{\sqrt{60}} \ge 2.326\). This gives \(k - 0.5 \ge 118.02\), so the smallest integer cutoff is \(k = 119\). 4. The approximate rejection region is \(R = \{119, 120, \ldots, 250\}\). 5. The observed count \(115\) is not in the rejection region, so fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(0.01\) level that the germination proportion is greater than \(40\%\).

1. \(H_0: p = 0.40\); \(H_a: p > 0.40\). 2. \(\mu = 100\) and \(\sigma = \sqrt{60} \approx 7.75\). 3. The approximate rejection region is \(R = \{119, 120, \ldots, 250\}\). Because \(115 \notin R\), fail to reject \(H_0\); the advertising claim is not statistically supported at the \(0.01\) level.

- A suspected decrease requires a left-tailed test. - Use the claim’s boundary value as the null proportion. - Find the largest lower-tail cutoff whose cumulative probability is at most \(0.05\). - Compare the observed count with the rejection region.

1. Let \(p\) be the true proportion of eggs labeled XL that meet the XL weight standard. Use \(H_0: p = 0.95\) and \(H_a: p < 0.95\). The boundary value \(p = 0.95\) is used to test the producer’s claim that \(p \ge 0.95\). 2. Let \(X\) be the number of sampled eggs that meet the standard. Under \(H_0\), \(X \sim \operatorname{Bin}(400, 0.95)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). The exact probabilities are \(P(X \le 372) \approx 0.04799\) and \(P(X \le 373) \approx 0.07265\). Therefore, \(c = 372\). 4. The rejection region is \(R = \{0, 1, \ldots, 372\}\). 5. Since the observed count \(371\) is in \(R\), reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that fewer than \(95\%\) of the eggs labeled XL meet the weight standard.

a) \(H_0: p = 0.95\); \(H_a: p < 0.95\). b) \(R = \{0, 1, \ldots, 372\}\). c) Because \(371 \in R\), reject \(H_0\). There is sufficient evidence at the \(0.05\) level that the true proportion is below \(95\%\).

- The laboratory suspects a decrease, so use a left-tailed test. - Model the number of patients with a timely response using a binomial distribution. - Find the largest lower-tail cutoff whose probability is at most \(0.05\). - Compare the observed number of successes with the rejection region.

1. Let \(p\) be the true proportion of patients for whom the medication works within \(15\) minutes. Use \(H_0: p = 0.80\) and \(H_a: p < 0.80\). The boundary value \(p = 0.80\) is used to test the company’s claim that \(p \ge 0.80\). 2. Let \(X\) be the number of patients in the sample for whom the medication works within \(15\) minutes. Under \(H_0\), \(X \sim \operatorname{Bin}(250, 0.80)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.05\). The exact probabilities are \(P(X \le 188) \approx 0.03705\) and \(P(X \le 189) \approx 0.05097\). Therefore, \(c = 188\). 4. The rejection region is \(R = \{0, 1, \ldots, 188\}\). 5. The observed count \(188\) is in the rejection region, so reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that the success proportion is below \(80\%\).

Hypotheses: \(H_0: p = 0.80\); \(H_a: p < 0.80\). Rejection region: \(R = \{0, 1, \ldots, 188\}\). Decision: Because \(188 \in R\), reject \(H_0\). The data provide sufficient evidence at the \(0.05\) level that the medication’s success proportion is below \(80\%\).

- A suspected increase requires a right-tailed test. - Calculate the mean and standard deviation using the null proportion. - Use the one-sided critical z-score and a continuity correction. - Compare the observed count with the approximate rejection region.

1. Let \(p\) be the true proportion of city households that regularly buy organic food. The hypotheses are \(H_0: p = 0.35\) and \(H_a: p > 0.35\). 2. Under \(H_0\), the count has mean \(\mu = 500 \cdot 0.35 = 175\) and standard deviation \(\sigma = \sqrt{500 \cdot 0.35 \cdot 0.65} = \sqrt{113.75} \approx 10.67\). The expected counts of successes and failures are both at least \(10\), so the normal approximation is appropriate. 3. For a right-tailed test with \(\alpha = 0.05\), use \(z_{0.95} \approx 1.645\). With continuity correction, the smallest critical count \(k\) satisfies \(\frac{k - 0.5 - 175}{\sqrt{113.75}} \ge 1.645\). This gives \(k \ge 193.04\), so the smallest integer cutoff is \(k = 194\). 4. The approximate rejection region is \(R = \{194, 195, \ldots, 500\}\). 5. Since the observed count \(196\) is in \(R\), reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that the city proportion is greater than \(35\%\).

The hypotheses are \(H_0: p = 0.35\) and \(H_a: p > 0.35\). The approximate rejection region is \(R = \{194, 195, \ldots, 500\}\). Because \(196 \in R\), reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that more than \(35\%\) of city households regularly buy organic food.

- A suspected increase requires a right-tailed test. - Calculate the binomial mean and standard deviation using the null proportion. - Use the one-sided critical z-score with a continuity correction. - Compare the observed count with the approximate rejection region.

1. Let \(p\) be the true proportion of underweight bars. Use \(H_0: p = 0.10\) and \(H_a: p > 0.10\). The boundary value \(p = 0.10\) is used to test the manufacturer’s claim that \(p \le 0.10\). 2. Under \(H_0\), the number \(X\) of underweight bars has mean \(\mu = 500 \cdot 0.10 = 50\) and standard deviation \(\sigma = \sqrt{500 \cdot 0.10 \cdot 0.90} = \sqrt{45} \approx 6.71\). 3. The expected counts of underweight and acceptable-weight bars are both at least \(10\), so the normal approximation is appropriate. 4. For a right-tailed test with \(\alpha = 0.05\), use \(z_{0.95} \approx 1.645\). With continuity correction, the smallest critical count \(k\) satisfies \(\frac{k - 0.5 - 50}{\sqrt{45}} \ge 1.645\). This gives \(k \ge 61.53\), so \(k = 62\). The approximate rejection region is \(R = \{62, 63, \ldots, 500\}\). 5. Since \(65 \in R\), reject \(H_0\). The sample provides sufficient evidence at the \(0.05\) level that the underweight proportion exceeds \(10\%\).

Hypotheses: \(H_0: p = 0.10\); \(H_a: p > 0.10\). Mean and standard deviation: \(\mu = 50\); \(\sigma = \sqrt{45} \approx 6.71\). Approximate rejection region: \(R = \{62, 63, \ldots, 500\}\). Conclusion: Because \(65 \in R\), reject \(H_0\). There is sufficient evidence that more than \(10\%\) of the bars are underweight.

- A suspected decrease requires a left-tailed test. - Use the lower tail of the exact binomial distribution. - Find the largest cutoff whose cumulative probability is at most \(0.01\). - Failing to reject the null hypothesis does not prove that the claimed proportion is correct.

1. Let \(p\) be the true proportion of residents who approve of the policy. Use \(H_0: p = 0.70\) and \(H_a: p < 0.70\). The boundary value \(p = 0.70\) is used to test the campaign’s claim that \(p \ge 0.70\). 2. Let \(X\) be the number of surveyed residents who approve. Under \(H_0\), \(X \sim \operatorname{Bin}(400, 0.70)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.01\). The exact probabilities are \(P(X \le 257) \approx 0.00771\) and \(P(X \le 258) \approx 0.01026\). Therefore, \(c = 257\). 4. The rejection region is \(R = \{0, 1, \ldots, 257\}\). 5. The observed count \(265\) is not in \(R\), so fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(0.01\) level that the approval proportion is below \(70\%\).

Hypotheses: \(H_0: p = 0.70\); \(H_a: p < 0.70\). Rejection region: \(R = \{0, 1, \ldots, 257\}\). Conclusion: Because \(265 \notin R\), fail to reject \(H_0\). The data do not provide sufficient evidence at the \(0.01\) level that approval is below \(70\%\).

- 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.

- The consumer organization suspects that the success proportion is lower than claimed. - Use a left-tailed test and the claim’s boundary value. - Find the largest lower-tail cutoff whose probability is at most \(0.10\).

1. Let \(p\) be the true proportion of eligible users who reach B1 proficiency within three months. The hypotheses are \(H_0: p = 0.80\) and \(H_a: p < 0.80\). The boundary value \(p = 0.80\) is used to test the company’s claim that \(p \ge 0.80\). 2. Let \(X\) be the number of sampled users who reach B1 proficiency. Under \(H_0\), \(X \sim \operatorname{Bin}(200, 0.80)\). 3. Find the largest integer \(c\) such that \(P(X \le c) \le 0.10\). The exact probabilities are \(P(X \le 152) \approx 0.09440\) and \(P(X \le 153) \approx 0.12625\). Therefore, \(c = 152\). 4. The rejection region is \(R = \{0, 1, \ldots, 152\}\).

Hypotheses: \(H_0: p = 0.80\); \(H_a: p < 0.80\). Rejection region: \(R = \{0, 1, \ldots, 152\}\).

- 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 \(H_0\).

- 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.

- 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 decrease leads to a left-tailed test. - Use the binomial distribution with \(n = 500\) and the null proportion \(0.92\). - Choose the largest lower-tail cutoff whose probability is at most \(0.01\). - Check carefully whether the observed count equals or exceeds the cutoff.

1. Let \(p\) be the true germination proportion. The hypotheses are \(H_0: p = 0.92\) and \(H_a: p < 0.92\). 2. Under \(H_0\), the number of seeds that germinate is \(X \sim \operatorname{Bin}(500, 0.92)\). 3. For the left-tailed test, find the largest integer \(c\) such that \(P(X \le c) \le 0.01\). The exact probabilities are \(P(X \le 444) \approx 0.0072\) and \(P(X \le 445) \approx 0.0108\). Therefore, \(c = 444\). 4. The rejection region is \(R = \{0, 1, \ldots, 444\}\). 5. The observed count \(445\) is not in the rejection region. Fail to reject \(H_0\); the sample does not provide sufficient evidence at the \(0.01\) level that the germination rate is below \(92\%\).

a) \(R = \{0, 1, \ldots, 444\}\). b) No. Because \(445 \notin R\), fail to reject \(H_0\). The company’s claim cannot be rejected at the \(0.01\) significance level.

- A suspected increase leads to a right-tailed test. - Use the cumulative distribution to find an upper-tail cutoff. - Choose the first count whose upper-tail probability is at most \(0.01\). - Compare the observed damage count with the rejection region.

1. Let \(p\) be the true proportion of packages that arrive damaged. The hypotheses are \(H_0: p = 0.04\) and \(H_a: p > 0.04\). 2. Under \(H_0\), the number of damaged packages is \(X \sim \operatorname{Bin}(500, 0.04)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). The cumulative probabilities are \(P(X \le 30) \approx 0.9882\) and \(P(X \le 31) \approx 0.9931\). Therefore, \(k = 32\). 4. The rejection region is \(R = \{32, 33, \ldots, 500\}\). 5. The observed count \(29\) is not in the rejection region, so fail to reject \(H_0\). There is not sufficient evidence at the \(0.01\) level that the damage proportion exceeds \(4\%\).

a) \(R = \{32, 33, \ldots, 500\}\). b) Because \(29 \notin R\), fail to reject \(H_0\). The data do not provide sufficient evidence that the damage proportion is greater than \(4\%\).

- Define the parameter as the post-change rework proportion. - An increase is supported by unusually large rework counts. - Find the first upper-tail cutoff whose probability is at most \(0.01\). - Compare the observed count with the rejection region.

1. Let \(p\) be the true proportion of batches that require rework after the process change. The hypotheses are \(H_0: p = 0.06\) and \(H_a: p > 0.06\). 2. Let \(X\) be the number of sampled batches requiring rework. Under \(H_0\), \(X \sim \operatorname{Bin}(800, 0.06)\). 3. For a right-tailed test, find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). The cumulative probabilities are \(P(X \le 63) \approx 0.9870\) and \(P(X \le 64) \approx 0.9909\). Therefore, \(k = 65\). 4. The rejection region is \(R = \{65, 66, \ldots, 800\}\). 5. The observed count \(62\) is not in the rejection region, so fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(0.01\) level that the rework proportion has increased.

Hypotheses: \(H_0: p = 0.06\); \(H_a: p > 0.06\). Rejection region: \(R = \{65, 66, \ldots, 800\}\). Conclusion: Because \(62 \notin R\), fail to reject \(H_0\). The suspected increase is not statistically significant at the \(0.01\) level.

- A suspected decrease requires a left-tailed test. - Use the exact binomial distribution with the claimed germination rate. - Find the largest lower-tail cutoff whose probability is at most \(0.01\). - Compare the observed germination count with the cutoff.

1. Let \(p\) be the true germination proportion. The hypotheses are \(H_0: p = 0.95\) and \(H_a: p < 0.95\). 2. Let \(X\) be the number of seeds that germinate. Under \(H_0\), \(X \sim \operatorname{Bin}(400, 0.95)\). 3. For a left-tailed test, find the largest integer \(c\) such that \(P(X \le c) \le 0.01\). The exact probabilities are \(P(X \le 368) \approx 0.0067\) and \(P(X \le 369) \approx 0.0114\). Therefore, \(c = 368\). 4. The rejection region is \(R = \{0, 1, \ldots, 368\}\). 5. The observed count \(372\) is not in the rejection region, so fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(0.01\) level that the germination rate is below \(95\%\).

No. The rejection region is \(R = \{0, 1, \ldots, 368\}\), and \(372 \notin R\). Fail to reject \(H_0\); the company’s claim cannot be rejected at the \(0.01\) significance level.

- A suspected increase requires a right-tailed test. - Model the defect count with a binomial distribution at the claimed proportion. - Find the first upper-tail cutoff whose probability is at most \(0.01\). - Compare the observed count with the critical value.

1. Let \(p\) be the true defect proportion. The hypotheses are \(H_0: p = 0.02\) and \(H_a: p > 0.02\). 2. Let \(X\) be the number of defective chips in the sample. Under \(H_0\), \(X \sim \operatorname{Bin}(800, 0.02)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). The cumulative probabilities are \(P(X \le 25) \approx 0.9877\) and \(P(X \le 26) \approx 0.9931\). Therefore, \(k = 27\). 4. The rejection region is \(R = \{27, 28, \ldots, 800\}\). 5. The observed count \(25\) is not in the rejection region, so fail to reject \(H_0\). There is not sufficient evidence at the \(0.01\) level that the defect proportion exceeds \(2\%\).

Rejection region: \(R = \{27, 28, \ldots, 800\}\). Conclusion: Because \(25 \notin R\), fail to reject \(H_0\). The data do not provide sufficient evidence that the defect proportion is greater than \(2\%\).

- 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 \(21\).

- Sampling variability allows a sample to miss rare defects. - For a left-tailed test, determine when the lower-tail probability of \(X = 0\) exceeds \(\alpha\). - Use \(P(X = 0) = (1 - p_0)^n\). - Solve the resulting exponential inequality by taking an \(800\)th root.

1. A sample can contain no defective sensors even when the production process has a positive defect proportion. Defective sensors may simply be missed because of sampling variability. 2. For \(X = 0\) not to be in the rejection region of the left-tailed test, its lower-tail probability under \(H_0\) must satisfy \(P_{p_0}(X \le 0) > 0.05\). 3. Under \(H_0\), \(X \sim \operatorname{Bin}(800, p_0)\), so \(P_{p_0}(X = 0) = (1 - p_0)^{800}\). Therefore, solve \((1 - p_0)^{800} > 0.05\). 4. Taking the \(800\)th root gives \(1 - p_0 > 0.05^{1/800}\), so \(p_0 < 1 - 0.05^{1/800} \approx 0.003738\). 5. Thus, among valid nondegenerate null values, \(X = 0\) is not in the rejection region for \(0 < p_0 < 0.003738\), or approximately \(0 < p_0 < 0.374\%\).

a) A random sample can miss defective sensors even when the overall defect proportion is greater than \(0\). b) For valid nondegenerate null values, \(X = 0\) is not in the rejection region when \(0 < p_0 < 1 - 0.05^{1/800} \approx 0.003738\), or approximately \(0 < p_0 < 0.374\%\).

- 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 \(0 < p_0 < 0.001840\).

- A suspected increase requires a right-tailed test. - Find the smallest cutoff whose upper-tail probability is at most \(0.01\). - Use the relationship \(P(X \ge k) = 1 - P(X \le k - 1)\). - Compare the observed count with the critical cutoff.

1. Let \(p\) be the true proportion of users who experience drowsiness. Use \(H_0: p = 0.08\) and \(H_a: p > 0.08\). The boundary value \(p = 0.08\) is used to test the company’s claim that \(p \le 0.08\). 2. Let \(X\) be the number of users in the sample who report drowsiness. Under \(H_0\), \(X \sim \operatorname{Bin}(600, 0.08)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). The cumulative probabilities are \(P(X \le 63) \approx 0.98782\) and \(P(X \le 64) \approx 0.99155\). Therefore, \(P(X \ge 65) \approx 0.00845\), and the smallest cutoff is \(k = 65\). 4. The rejection region is \(R = \{65, 66, \ldots, 600\}\). 5. Since \(66 \in R\), reject \(H_0\). The sample provides sufficient evidence at the \(0.01\) level that the drowsiness proportion exceeds \(8\%\).

a) \(H_0: p = 0.08\); \(H_a: p > 0.08\). b) The rejection region begins at \(65\), so \(R = \{65, 66, \ldots, 600\}\). c) Because \(66 \in R\), reject \(H_0\). There is sufficient evidence at the \(0.01\) level that the drowsiness proportion is greater than \(8\%\).

- An increase requires a right-tailed test. - Find the smallest count whose upper-tail probability is at most \(0.01\). - Compare the observed count directly with the critical cutoff. - A before-and-after association does not by itself establish causation.

1. Let \(p\) be the current proportion of commuters who use public transportation. The hypotheses are \(H_0: p = 0.15\) and \(H_a: p > 0.15\). 2. Let \(X\) be the number of public-transit users in the sample. Under \(H_0\), \(X \sim \operatorname{Bin}(500, 0.15)\). 3. Find the smallest integer \(k\) such that \(P(X \ge k) \le 0.01\). The exact probabilities are \(P(X \le 93) \approx 0.98809\) and \(P(X \le 94) \approx 0.99129\). Thus, \(P(X \ge 95) \approx 0.00871\), and the rejection region is \(R = \{95, 96, \ldots, 500\}\). 4. Since \(92 \notin R\), fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(0.01\) level that the current public-transit-use proportion exceeds \(15\%\). 5. The test can identify evidence of an increase, but it does not by itself establish that the discounted pass caused any increase.

a) \(H_0: p = 0.15\); \(H_a: p > 0.15\). b) At least \(95\) public-transit users are required; \(R = \{95, 96, \ldots, 500\}\). c) No. Because \(92 \notin R\), fail to reject \(H_0\). The result does not provide sufficient evidence of an increase at the \(0.01\) level.

- 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.