Proportion Tests - Biostatistics

A quality control manager tests whether the defect rate has changed from the historical 8%. In 500 items, 50 are defective. If the calculated test statistic is $z = 1.47$ , what conclusion should be drawn at $\alpha = 0.05$ ?
Answer: Fail to reject $H_0$ because $|1.47| < 1.96$ When you encounter hypothesis testing problems involving proportions, you need to identify the test type and compare your test statistic to the appropriate critical value. This is a two-tailed test since we're checking whether the defect rate has "changed" (could be higher or lower) from the historical 8%. For a two-tailed z-test at $\alpha = 0.05$ , the critical values are $\pm 1.96$ . Since our calculated test statistic is $z = 1.47$ , we have $|1.47| = 1.47 < 1.96$ . Because the test statistic falls within the acceptance region (between -1.96 and +1.96), we fail to reject the null hypothesis. There isn't sufficient evidence that the defect rate has changed significantly from 8%. Option A incorrectly compares the test statistic directly to the significance level (1.47 > 0.05), which makes no statistical sense—you never compare z-values to alpha values directly. Option C treats this as a one-tailed test by using the critical value 1.645, but the phrase "has changed" indicates we're testing for difference in either direction, requiring a two-tailed approach. Option D mentions the sample proportion (50/500 = 0.10) but doesn't properly connect this to the hypothesis testing framework—the decision depends on the test statistic relative to critical values, not just the sample proportion value. Remember: "changed" or "different" signals a two-tailed test with critical values $\pm 1.96$ at $\alpha = 0.05$ , while "increased" or "decreased" would indicate one-tailed tests with $\pm 1.645$ .
A researcher claims that 65% of college students prefer online learning. In a sample of 120 students, 72 prefer online learning. What is the appropriate null hypothesis for testing this claim?
Answer: $H_0: p = 0.65$ When you encounter hypothesis testing questions, remember that the null hypothesis ( $H_0$ ) always represents the status quo or the claim being tested. It uses the population parameter, not sample statistics. In this problem, the researcher claims that 65% of college students prefer online learning. This claim becomes our null hypothesis, so we write $H_0: p = 0.65$ , where $p$ represents the true population proportion. The sample data (72 out of 120 students, or 60%) will be used to test whether this claim is reasonable. Let's examine why the other options are incorrect: Option A ( $H_0: \hat{p} = 0.65$ ) uses $\hat{p}$ , which represents the sample proportion. Null hypotheses are always about population parameters, never sample statistics. This is a fundamental error in hypothesis testing. Option B ( $H_0: p = 0.60$ ) uses the correct notation but the wrong value. While 0.60 is the observed sample proportion (72/120), the null hypothesis should reflect the researcher's claim (65%), not what we observed in our sample. Option D ( $H_0: \hat{p} = 0.60$ ) combines both errors: it uses sample notation ( $\hat{p}$ ) and bases the hypothesis on the sample result rather than the original claim. Study tip: Always distinguish between population parameters ( $p$ , $\mu$ , $\sigma$ ) and sample statistics ( $\hat{p}$ , $\bar{x}$ , $s$ ). The null hypothesis tests claims about populations, so it always uses parameter notation. The researcher's original claim, not your sample results, determines the null hypothesis value.
A researcher wants to test if the proportion of students who study abroad differs between two universities. University X: 45 out of 300 students; University Y: 38 out of 250 students. What is the pooled proportion for this test?
Answer: $\frac{45 + 38}{300 + 250} = 0.151$ When comparing proportions between two groups, you need to understand the concept of pooled proportion, which estimates the overall proportion assuming no difference between groups. The pooled proportion combines all successes from both groups and divides by the total sample size from both groups. For this study abroad comparison, you take the total number of students who studied abroad (45 + 38 = 83) and divide by the total number of students surveyed (300 + 250 = 550). This gives you $\frac{83}{550} = 0.151$ , which matches answer choice A. Let's examine why the other options are incorrect. Answer B adds the individual proportions ( $\frac{45}{300} + \frac{38}{250} = 0.302$ ), but this doesn't account for the different sample sizes between universities and isn't mathematically meaningful for hypothesis testing. Answer C calculates the difference in the number of students who studied abroad divided by total sample size ( $\frac{7}{550} = 0.013$ ), which has no statistical interpretation. Answer D finds the difference between the two individual proportions ( $0.15 - 0.152 = -0.002$ ), which would be useful for calculating the test statistic but isn't the pooled proportion. The pooled proportion serves as your best estimate of the true proportion under the null hypothesis that both universities have the same study abroad rate. Remember this pattern: for two-proportion tests, always pool by adding numerators and adding denominators separately—never add the calculated proportions themselves.
A political pollster wants to test if support for a candidate differs between urban and rural voters. Urban: 120 supporters out of 300 voters; Rural: 80 supporters out of 200 voters. Before conducting the test, which assumption should be verified first?
Answer: The samples must be independent and randomly selected When comparing proportions between two groups, you're typically performing a two-proportion z-test. The foundation of any valid statistical test lies in meeting its underlying assumptions before examining the data or calculating test statistics. The correct answer is C because independence and random sampling form the cornerstone assumptions for proportion tests. Your samples must be independent (one group's responses don't influence the other's) and randomly selected from their respective populations. Without these conditions, your statistical inference becomes meaningless regardless of sample sizes or proportions. Random sampling ensures your samples represent their populations, while independence prevents bias from creeping into your comparison. Option A is incorrect because equal population sizes aren't required for two-proportion tests. You can validly compare groups of different sizes - urban and rural populations naturally differ in real-world scenarios. Option B misunderstands the assumption about expected counts, not sample proportions. The actual requirement is that expected frequencies in each cell should be at least 5, which relates to the normal approximation validity. The specific range 0.3-0.7 has no statistical basis. Option D incorrectly focuses on total sample size. While larger samples generally improve power, there's no magic threshold of 500. The key is having adequate expected counts in each category (typically ≥5), which depends on both sample size and proportion values. Study tip: For any hypothesis test, always verify assumptions before diving into calculations. Independence and randomness are fundamental - without them, even perfect arithmetic won't yield valid conclusions.
A school district compares graduation rates between two high schools. School X: 180 graduates out of 200 seniors; School Y: 165 graduates out of 180 seniors. The calculated z-statistic is -1.45. What does this negative value indicate?
Answer: The sample proportion for School X is less than that for School Y When you encounter z-statistics in two-proportion hypothesis tests, remember that the sign of the z-value tells you about the direction of the difference between your sample proportions, not about statistical significance. Let's calculate the actual proportions: School X has 180/200 = 0.90 (90%) graduation rate, while School Y has 165/180 = 0.917 (91.7%) graduation rate. Since School X's proportion (0.90) is indeed lower than School Y's proportion (0.917), the z-statistic correctly reflects this relationship with its negative value. The z-statistic formula is $z = \frac{(\hat{p_1} - \hat{p_2}) - 0}{\text{standard error}}$ , where $\hat{p_1}$ represents School X's proportion. Since 0.90 - 0.917 = -0.017, the numerator is negative, producing a negative z-value. Choice A is incorrect because statistical significance depends on whether |z| exceeds the critical value (typically 1.96 for α = 0.05). Here, |-1.45| < 1.96, so there's no significant difference. Choice C reflects a fundamental misunderstanding—z-statistics can absolutely be negative and simply indicate which sample proportion is smaller. Choice D is nonsensical because pooled proportions in two-sample tests are always positive values between 0 and 1, calculated as the total successes divided by total sample size. Study tip: The sign of your z-statistic always matches the sign of (proportion₁ - proportion₂). Negative z-values are completely normal and simply tell you the direction of the difference—don't confuse the sign with significance testing.
A health inspector tests whether the contamination rate at a food processing plant exceeds 5%. In 300 samples, 21 show contamination. If $\alpha = 0.05$ and this is a one-tailed test, what is the critical value?
Answer: $1.645$ When you encounter hypothesis testing problems involving critical values, you need to identify three key elements: the significance level ( $\alpha$ ), whether it's a one-tailed or two-tailed test, and the appropriate distribution to use. This problem involves testing whether a contamination rate exceeds 5%, which is a one-tailed test since we're only concerned with rates greater than 5%. With a sample size of 300, we can use the standard normal distribution (z-distribution) because the sample is large enough for the Central Limit Theorem to apply. For a one-tailed test with $\alpha = 0.05$ , you need the z-value that leaves 5% in the upper tail of the standard normal distribution. This critical value is $z = 1.645$ , making B the correct answer. Let's examine why the other options are wrong: A) $1.96$ is the critical value for a two-tailed test with $\alpha = 0.05$ , where you split the significance level between both tails (2.5% each). C) $2.33$ would be the critical value for a one-tailed test with $\alpha = 0.01$ , which is more stringent than our given significance level. D) $-1.645$ would be correct for a left-tailed test (testing if the rate is significantly less than 5%), but this problem asks whether the rate exceeds 5%, making it a right-tailed test. Study tip: Remember that one-tailed tests use $1.645$ for $\alpha = 0.05$ and two-tailed tests use $1.96$ . The sign depends on which tail you're testing—positive for right-tailed, negative for left-tailed.
A manufacturer tests whether the defect rate has increased from the standard 3%. In 400 items, 18 are defective. The sample proportion is 0.045. What is the standard error for this one-proportion test?
Answer: $\sqrt{\frac{0.03 \times 0.97}{400}} = 0.0085$ When you encounter a hypothesis test for a single proportion, you need to carefully distinguish between the null hypothesis value and the sample proportion when calculating the standard error. This is a crucial concept that trips up many students. For hypothesis testing of proportions, the standard error uses the null hypothesis proportion, not the sample proportion. Since the manufacturer is testing whether the defect rate has increased from the standard 3%, your null hypothesis is that $p_0 = 0.03$ . The standard error formula is $SE = \sqrt{\frac{p_0(1-p_0)}{n}}$ , where $p_0$ is the hypothesized proportion under the null hypothesis. Choice A is correct: $\sqrt{\frac{0.03 \times 0.97}{400}} = 0.0085$ . This properly uses the null hypothesis value of 0.03 and its complement 0.97, giving you the standard error needed for the test statistic. Choice B uses the sample proportion (0.045) instead of the null hypothesis proportion. This is the standard error you'd use for a confidence interval, not a hypothesis test. Choice C incorrectly multiplies the null hypothesis proportion by the sample proportion, which has no statistical meaning. Choice D forgets to divide by the sample size entirely, giving you just $\sqrt{p_0(1-p_0)}$ without the $\sqrt{n}$ in the denominator. Key takeaway: For hypothesis tests about proportions, always use the null hypothesis proportion in your standard error calculation. Save the sample proportion for confidence intervals.
A medical researcher claims that less than 25% of patients experience side effects from a new medication. In a study of 80 patients, 16 experienced side effects. What type of test should be performed?
Answer: One-tailed test with $H_a: p < 0.25$ When you encounter a hypothesis testing question, start by carefully reading what the researcher claims and determine the direction of the test based on that claim. The researcher claims that "less than 25%" of patients experience side effects. This is a directional claim suggesting the true proportion is below 25%, not just different from 25%. In hypothesis testing, the alternative hypothesis ( $H_a$ ) represents what the researcher is trying to prove or support with evidence. Since the claim is specifically that the proportion is less than 25%, this calls for a one-tailed test with $H_a: p < 0.25$ . The null hypothesis would be $H_0: p \geq 0.25$ (the proportion is 25% or higher). The sample data (16 out of 80 patients = 20% experiencing side effects) would be used to test whether there's sufficient evidence to support the claim that the true proportion is less than 25%. Option B is incorrect because a two-tailed test ( $H_a: p \neq 0.25$ ) would be appropriate if the researcher claimed the proportion was simply "different from" 25%, without specifying a direction. Option C is wrong because $H_a: p > 0.25$ tests the opposite direction—whether the proportion is greater than 25%. Option D combines the wrong test type (two-tailed) with an impossible notation, since two-tailed tests use $\neq$ , not directional symbols. Study tip: Always match the alternative hypothesis to the researcher's specific claim. Words like "less than," "greater than," or "more than" signal one-tailed tests, while "different from" or "not equal to" indicate two-tailed tests.
A quality assurance manager tests whether the defect rate has changed from the historical 2%. In 600 items, 18 are defective. The sample proportion is 0.03. What is the test statistic?
Answer: $z = \frac{0.03 - 0.02}{\sqrt{\frac{0.02 \times 0.98}{600}}} = 1.75$ When testing a population proportion, you're comparing your sample proportion to a hypothesized value using a z-test. The key is identifying which proportion goes in the denominator of the standard error formula. For hypothesis tests about proportions, the test statistic formula is $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ , where $\hat{p}$ is your sample proportion, $p_0$ is the hypothesized proportion, and $n$ is sample size. Crucially, you use the hypothesized proportion ( $p_0 = 0.02$ ) in the standard error, not the sample proportion. Answer A correctly applies this formula: $z = \frac{0.03 - 0.02}{\sqrt{\frac{0.02 \times 0.98}{600}}} = 1.75$ . The denominator uses the historical rate of 0.02 because we're testing whether the true rate has changed from this value. Answer B makes a common error by using the sample proportion (0.03) in the standard error calculation instead of the hypothesized proportion. This is incorrect for hypothesis testing. Answer C uses the raw counts (18 defective items vs. expected 12) rather than proportions. While mathematically this can work, it doesn't match the standard hypothesis testing format and the calculation shown gives 1.73, not the correct 1.75. Answer D incorrectly omits the hypothesized proportion from the numerator, testing whether the sample proportion equals zero rather than whether it differs from 0.02. Remember: In proportion hypothesis tests, always use the null hypothesis value ( $p_0$ ) in your standard error calculation, never the sample proportion.
An educator tests whether online students have a different completion rate than in-person students. Online: 85 completions out of 120 students; In-person: 78 completions out of 100 students. The standard error for the difference in proportions is approximately 0.062. What is the test statistic?
Answer: $z = \frac{0.708 - 0.780}{0.062} = -1.16$ When testing for differences between two proportions, you're looking at whether two groups have significantly different rates of some outcome. The key is setting up your test statistic correctly using the formula $z = \frac{\hat{p_1} - \hat{p_2}}{SE}$ , where you compare the difference in sample proportions to the standard error. First, calculate the proportions: online students have $\hat{p_1} = \frac{85}{120} = 0.708$ and in-person students have $\hat{p_2} = \frac{78}{100} = 0.780$ . Using the given standard error of 0.062, the test statistic becomes $z = \frac{0.708 - 0.780}{0.062} = \frac{-0.072}{0.062} = -1.16$ . Answer A correctly applies this formula with the proper order (online minus in-person). Answer B makes the common error of switching the order to get a positive result, calculating $(0.780 - 0.708)/0.062 = 1.16$ . While the magnitude is correct, the sign matters for interpretation—it tells you which group has the higher rate. Answer C incorrectly uses raw counts (85 - 78 = 7) instead of proportions, yielding an unrealistically large test statistic of 112.90. Answer D nonsensically adds the proportions instead of finding their difference, giving 24.00. The negative z-score in the correct answer tells you that online students have a lower completion rate than in-person students. Remember: always use proportions (not raw counts) when comparing rates between groups, and pay attention to which group you subtract from which—the sign of your test statistic provides meaningful information about the direction of the difference.
A quality control manager tests whether the defect rate of a manufacturing process has increased from the acceptable level of 2%. In a random sample of 500 items, 15 defects are found. The calculated p-value is 0.032. If the company uses α = 0.05, what should the manager conclude, and what type of error might occur?
Answer: Reject H₀ and conclude the defect rate has increased; a Type I error could occur if the true rate is still 2% Since p-value = 0.032 < α = 0.05, we reject H₀ and conclude there is evidence that the defect rate has increased above 2%. When we reject H₀, a Type I error (rejecting a true H₀) could occur if the true defect rate is actually still 2%. Choice B gives the wrong decision based on the p-value. Choice C correctly rejects H₀ but incorrectly identifies the potential error type. Choice D avoids making a statistical decision, which is not appropriate when p-value < α.
Two laboratories are comparing error rates in blood tests. Lab A: 8 errors in 400 tests; Lab B: 15 errors in 500 tests. To perform a two-proportion z-test, what must be calculated first?
Answer: The pooled proportion $\hat{p} = \frac{8 + 15}{400 + 500}$ When comparing proportions between two groups, you're testing whether the difference between them is statistically significant. The two-proportion z-test requires a specific approach to calculate the test statistic properly. The correct answer is C because the two-proportion z-test uses a pooled proportion under the null hypothesis assumption that both populations have the same true proportion. This pooled proportion $\hat{p} = \frac{8 + 15}{400 + 500} = \frac{23}{900} = 0.0256$ combines all successes (errors) over all trials from both labs. This pooled value is then used to calculate the standard error for the test statistic. Option A is wrong because while you need both sample sizes for the calculation, simply finding their difference (100) serves no purpose in the two-proportion z-test formula. Option B represents an intermediate step - you do calculate the individual sample proportions (Lab A: 8/400 = 0.02; Lab B: 15/500 = 0.03), but these alone aren't sufficient to perform the test. The pooled proportion must be calculated first to proceed with the hypothesis test. Option D incorrectly suggests averaging the two sample proportions, which gives $\frac{0.02 + 0.03}{2} = 0.025$ . This average doesn't account for the different sample sizes and isn't the proper way to pool proportions in hypothesis testing. Remember: Two-proportion z-tests always require pooling under the null hypothesis. Look for the formula that combines total successes over total trials from both groups - this distinguishes it from confidence intervals, which use individual sample proportions.
A medical researcher is planning a study to detect a difference in cure rates between two treatments. Historical data suggests Treatment A has a 60% cure rate. The researcher wants 80% power to detect if Treatment B has a 75% cure rate, using α = 0.05 (two-tailed). Which factor would most effectively reduce the required sample size?
Answer: Increasing the target cure rate for Treatment B to 80%, since larger effect sizes require smaller samples Sample size for comparing two proportions is inversely related to the square of the effect size (|p₁ - p₂|). Increasing the difference from 15% (75% - 60%) to 20% (80% - 60%) would substantially reduce the required sample size. Choice A would reduce sample size but increases Type I error risk. Choice B would reduce sample size modestly but may not be scientifically justified. Choice D reduces sample size but increases Type II error risk, which may not be acceptable for medical research.
In a clinical trial, 45 out of 150 patients in the treatment group and 60 out of 150 patients in the control group experienced improvement. A statistician calculates a 95% confidence interval for the difference in proportions (treatment - control) as (-0.18, 0.08). What can be concluded about the hypothesis test H₀: p₁ = p₂ vs H₁: p₁ ≠ p₂?
Answer: Fail to reject H₀ since the confidence interval contains zero, indicating no significant difference between treatments When a confidence interval for the difference in proportions contains zero, it indicates that zero is a plausible value for the true difference, meaning we fail to reject H₀: p₁ = p₂. The observed difference is (45/150) - (60/150) = 0.30 - 0.40 = -0.10. Choice A incorrectly focuses on the interval containing 'mostly' negative values rather than containing zero. Choice C misinterprets what it means for the observed difference to fall within the CI. Choice D incorrectly suggests the test is inconclusive when the CI provides clear guidance.
A pharmaceutical company tests whether men and women respond differently to a new drug. Men: 45 positive responses out of 80 participants; Women: 38 positive responses out of 70 participants. What assumption is most critical for the validity of this test?
Answer: The participants in each group must be independent of each other When comparing proportions between two groups (like response rates between men and women), you're typically conducting a two-proportion z-test or chi-square test. The validity of these statistical tests hinges on several key assumptions about your data structure. The most fundamental assumption is that observations within each group must be independent of each other (C). This means one person's response cannot influence another person's response. If participants discussed the treatment, were family members, or were treated in groups where they could influence each other, this independence would be violated and your statistical test results would be unreliable. Independence ensures that each observation provides unique information about the treatment effect. Looking at why the other options are incorrect: (A) is wrong because equal sample sizes, while sometimes beneficial for power, are not required for valid statistical comparison between groups. You can legitimately compare groups of different sizes. (B) misunderstands the purpose of the test—you're actually testing whether the groups have different response rates, so assuming they're the same would defeat the point. (D) is incorrect because while larger samples generally provide more power, there's no universal minimum of 100 participants required for validity. Study tip: For any statistical test comparing groups, always ask yourself about the independence assumption first. Questions about "validity" in biostatistics often focus on whether the fundamental assumptions of independence, randomness, or appropriate sampling have been met, rather than technical details about sample sizes or calculation methods.
Two independent studies examined the effectiveness of the same intervention. Study 1 found 42 successes out of 60 trials (p̂₁ = 0.70), while Study 2 found 35 successes out of 50 trials (p̂₂ = 0.70). A researcher wants to test if there is a significant difference between the two studies. What is the most likely outcome and reasoning?
Answer: Fail to reject H₀ because the sample proportions are identical, making the test statistic exactly zero When testing H₀: p₁ = p₂, the test statistic is z = (p̂₁ - p̂₂)/SE. Since p̂₁ = p̂₂ = 0.70, the numerator is 0.70 - 0.70 = 0, making z = 0. Regardless of the standard error calculation, z = 0 leads to failing to reject H₀. Choice B incorrectly suggests that different sample sizes with identical proportions would yield a significant result. Choice C makes an unjustified leap about population proportions. Choice D incorrectly states that two-proportion tests require equal sample sizes.
A restaurant owner claims that more than 80% of customers are satisfied with service. A survey of 120 customers finds 102 are satisfied. To test this claim at $\alpha = 0.01$ , what decision should be made if the calculated test statistic is $z = 2.19$ ?
Answer: Fail to reject $H_0$ because $2.19 < 2.33$ When you encounter a hypothesis testing question, you need to systematically compare your test statistic to the critical value based on your significance level and test type. This is a one-tailed test since the claim is "more than 80%" (testing $H_0: p \leq 0.8$ vs $H_1: p > 0.8$ ). With $\alpha = 0.01$ for a right-tailed test, the critical value is $z_{0.01} = 2.33$ . Since your calculated test statistic $z = 2.19$ is less than 2.33, you fail to reject $H_0$ . The evidence isn't strong enough at the 1% significance level to support the owner's claim. Option A incorrectly states that $2.19 > 2.33$ , which is mathematically false. Option C uses the wrong critical value ( $z = 1.96$ corresponds to $\alpha = 0.05$ , not 0.01) and would lead to an incorrect rejection. Option D gives the right conclusion but for the wrong reason—being a one-tailed test doesn't automatically mean you fail to reject $H_0$ ; the decision depends on comparing your test statistic to the appropriate critical value. The correct answer is B because $2.19 < 2.33$ , meaning the test statistic falls in the non-rejection region. Study tip: Always identify three things before making your decision: (1) Is this one-tailed or two-tailed? (2) What's the correct critical value for your $\alpha$ ? (3) Does your test statistic exceed that critical value? Double-check your critical values—confusing 0.01 and 0.05 significance levels is a common mistake.
A pharmaceutical company claims that their new drug reduces the incidence of a particular side effect from the standard 15% to less than 10%. In a clinical trial of 200 patients, 16 patients experienced the side effect. If we test at the 0.05 significance level, what is the most appropriate conclusion?
Answer: Fail to reject the null hypothesis; there is insufficient evidence that the drug reduces the side effect rate to less than 10% This is a one-sample proportion test with H₀: p ≥ 0.10 vs H₁: p < 0.10. The sample proportion is p̂ = 16/200 = 0.08. The test statistic is z = (0.08 - 0.10)/√(0.10×0.90/200) = -0.94. The critical value for α = 0.05 (one-tailed) is -1.645. Since -0.94 > -1.645, we fail to reject H₀. Choice A incorrectly rejects H₀. Choice C misinterprets the hypothesis. Choice D uses incorrect language (we never 'accept' H₀).
A researcher claims that more than 60% of college students own laptops. In a sample of 150 students, 98 own laptops. What is the appropriate alternative hypothesis?
Answer: $H_a: p > 0.60$ When you encounter hypothesis testing problems, the key is to carefully read what the researcher is claiming and translate that claim into the appropriate alternative hypothesis. The alternative hypothesis ( $H_a$ ) always represents what the researcher is trying to prove or demonstrate. In this problem, the researcher claims that "more than 60% of college students own laptops." This directional claim translates directly to $H_a: p > 0.60$ , where $p$ represents the true population proportion of students who own laptops. The null hypothesis would be $H_0: p \leq 0.60$ , representing the status quo or opposing view. Option A ( $H_a: p > 0.60$ ) correctly captures the researcher's claim in proper hypothesis testing format. Option B ( $H_a: p < 0.60$ ) represents the opposite claim—that fewer than 60% own laptops—which contradicts what the researcher wants to prove. Option C ( $H_a: p \neq 0.60$ ) would be appropriate for a two-tailed test where the researcher claims the proportion is simply different from 60% (either higher or lower), but that's not what's stated here. Option D ( $H_a: \hat{p} > 0.653$ ) makes a fundamental error by using the sample proportion ( $\hat{p} = 98/150 = 0.653$ ) instead of the population parameter $p$ , and uses a value derived from the sample data rather than the researcher's original claim. Remember: alternative hypotheses are always about population parameters ( $p$ , $\mu$ ), never sample statistics ( $\hat{p}$ , $\bar{x}$ ), and they must directly reflect the researcher's specific claim.
An epidemiologist compares infection rates between two cities during an outbreak. City A: 65 infections out of 1000 residents tested; City B: 48 infections out of 800 residents tested. If the p-value is 0.12, what can be concluded at $\alpha = 0.05$ ?
Answer: There is insufficient evidence that infection rates differ between cities When you encounter a hypothesis testing question in biostatistics, focus on the relationship between the p-value and your significance level ( $\alpha$ ) to determine what the data tells you about the research question. Here, you're comparing infection rates: City A has 65/1000 = 6.5% and City B has 48/800 = 6.0%. The epidemiologist tested whether these rates are significantly different, yielding a p-value of 0.12. Since p = 0.12 > $\alpha$ = 0.05, you fail to reject the null hypothesis. In hypothesis testing, this means there's insufficient evidence to conclude the infection rates differ between cities. The observed difference could reasonably be due to random variation rather than a true difference between populations. Answer C correctly states this conclusion. Answer A is wrong because "significantly different" requires p ≤ $\alpha$ , which we don't have. Answer B makes two errors: it assumes significance when there isn't any, and it implies a one-tailed test focusing on City A being higher, when the question suggests a two-tailed comparison. Answer D commits a classic statistical error—failing to reject the null hypothesis never "proves" equality. We simply lack sufficient evidence to detect a difference; a true difference might exist but be too small for our sample sizes to detect. Study tip: Remember that p > $\alpha$ means "insufficient evidence," never "proof of no difference." Statistical tests can only provide evidence against the null hypothesis, not prove it true.

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