Setting Up Tests for Population Proportion
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AP Statistics › Setting Up Tests for Population Proportion
A restaurant chain reports that 70% of customers rate their experience as “satisfied” or “very satisfied.” A random sample of 150 customers is surveyed, and 96 give a satisfied rating ($\hat{p}=0.64$). Which hypotheses are appropriate for determining whether the reported proportion is accurate?
$H_0: p=0.64\quad H_a: p\ne0.64$
$H_0: p\ne0.70\quad H_a: p=0.70$
$H_0: p=0.70\quad H_a: p\ne0.70$
$H_0: p=0.70\quad H_a: p<0.70$
$H_0: \hat{p}=0.70\quad H_a: \hat{p}\ne0.70$
Explanation
The skill assessed is setting up hypotheses for testing a reported proportion of 0.70 in AP Statistics, to determine accuracy. The hypotheses H0: p = 0.70 and Ha: p ≠ 0.70 fit a two-sided test for verifying the exact value, as in choice A. A distractor is choice D, substituting ˆp for p. Choice C uses a one-sided alternative, unsuitable for accuracy checks. For a mini-lesson, use ≠ in Ha for assessing reported or exact values to allow rejection for any difference. Null includes equality to the reported figure, about p. This setup evaluates if the proportion matches without directional bias.
A political analyst claims that exactly 50% of adults in a state approve of the governors performance. A random sample of 1000 adults finds that 540 approve (so $
\hat{p}=540/1000$). Which hypotheses are appropriate to test whether approval differs from the analysts claim?
$H_0: p=0.50$; $H_a: p>0.50$
$H_0: p\ne 0.50$; $H_a: p=0.50$
$H_0: p=0.50$; $H_a: p\ne 0.50$
$H_0: \hat{p}=0.50$; $H_a: \hat{p}\ne 0.50$
$H_0: p=0.54$; $H_a: p\ne 0.54$
Explanation
This question tests setup for testing if approval differs from exactly 50%. Correct choice A: H0: p = 0.50 and Ha: p ≠ 0.50, for two-sided testing. Distractors use sample p-hat = 0.54 in H0, like option B, or p-hat in C. Option D reverses H0 and Ha. Mini-lesson: 'Differs from' calls for ≠ in Ha, checking both directions. Equality is always in H0, about p. Sample proportion is for the z-test, not hypotheses.
A school principal believes that the proportion of students who eat school breakfast is different from 55%. In a random sample of 200 students, 124 report eating school breakfast (so $
\hat{p}=124/200$). Which hypotheses are appropriate for testing the principals belief using a one-proportion $z$ test?
$H_0: p=0.62$; $H_a: p\ne 0.62$
$H_0: p\ne 0.55$; $H_a: p=0.55$
$H_0: p=0.55$; $H_a: p>0.55$
$H_0: \hat{p}=0.55$; $H_a: \hat{p}\ne 0.55$
$H_0: p=0.55$; $H_a: p\ne 0.55$
Explanation
This question tests hypothesis setup for a principal's belief that the proportion of students eating breakfast differs from 55%, requiring a two-sided test. The correct hypotheses are H0: p = 0.55 and Ha: p ≠ 0.55, as in choice B, matching the 'different from' claim. A distractor is using the sample p-hat = 0.62 in H0, like option A, which confuses sample with population. Option C reverses H0 and Ha, violating the rule that H0 includes equality. Mini-lesson: For two-sided tests, Ha uses ≠ to check for any difference, while one-sided tests use < or >. Hypotheses focus on p, the unknown population proportion. The sample data informs the test statistic but not the hypotheses statements.
A phone manufacturer claims that no more than 8% of its new phones are defective. A quality inspector wants to test whether the defect rate exceeds this value. In a random sample of 500 phones, 52 are found to be defective (so $
\hat{p}=52/500$). Which hypotheses are appropriate?
$H_0: p=0.08$; $H_a: p\ne 0.08$
$H_0: p=0.08$; $H_a: p>0.08$
$H_0: p\ge 0.08$; $H_a: p<0.08$
$H_0: \hat{p}=0.08$; $H_a: \hat{p}>0.08$
$H_0: p=0.104$; $H_a: p>0.104$
Explanation
This question involves hypotheses to test if defect rate exceeds 8%, challenging the 'no more than' claim. Choice B is correct: H0: p = 0.08 and Ha: p > 0.08, for a right-tailed test. Distractors include sample p-hat = 0.104 in H0, like option A, or p-hat in D. Option C reverses direction and structure. Mini-lesson: To test exceeding a maximum, use > in Ha, with H0 as equality. Hypotheses are about p; sample informs evidence against H0. This setup places the burden on showing a higher rate.
A nonprofit organization claims that more than $65%$ of adults in a county have volunteered in the past year. A random sample of $n=220$ adults finds $132$ have volunteered (so $\hat{p}=0.60$). Which hypotheses are appropriate to test the nonprofit’s claim?
$H_0: \hat{p}=0.65$; $H_a: \hat{p}>0.65$
$H_0: p=0.65$; $H_a: p>0.65$
$H_0: p=0.60$; $H_a: p>0.60$
$H_0: p=0.65$; $H_a: p<0.65$
$H_0: p>0.65$; $H_a: p=0.65$
Explanation
This question involves testing a claim about a proportion being greater than a specific value. The nonprofit claims MORE THAN 65% have volunteered, so we test if p > 0.65. The null hypothesis states H₀: p = 0.65 (the boundary value), and the alternative reflects the claim: Hₐ: p > 0.65. Choice B has the wrong direction. Choice C incorrectly uses the sample proportion (p̂). Choice D uses the sample proportion value (0.60) instead of the claimed threshold. Choice E reverses the null and alternative hypotheses. Note that even though the sample proportion (0.60) is less than 65%, we still set up hypotheses to test the original claim, not based on what the sample shows.
A streaming service claims that 60% of its subscribers watch at least one documentary each month. A market analyst believes the true proportion is less than 60%. In a random sample of 150 subscribers, 81 report watching at least one documentary (so $
\hat{p}=81/150$). Which hypotheses are appropriate?
$H_0: p=0.60$; $H_a: p<0.60$
$H_0: \hat{p}=0.60$; $H_a: \hat{p}<0.60$
$H_0: p<0.60$; $H_a: p=0.60$
$H_0: p=0.60$; $H_a: p>0.60$
$H_0: p=0.54$; $H_a: p<0.54$
Explanation
This question tests setting up hypotheses for believing a documentary viewing proportion is less than 60%. Choice B is correct: H0: p = 0.60 and Ha: p < 0.60, matching the 'less than' belief. Distractors include using sample p-hat = 0.54 in H0, like option A, or p-hat in option D. Option C reverses H0 and Ha incorrectly. Mini-lesson: For 'less than' claims, Ha uses <, making it left-tailed, with H0 as equality. Hypotheses concern p, not p-hat. The sample proportion helps compute the z-score but doesn't appear in the hypotheses.
A principal suspects that the proportion of all students at a high school who arrive late at least once per week is different from $10%$. A random sample of $n=120$ students finds $9$ who arrive late at least once per week (so $\hat{p}=0.075$). Which hypotheses are appropriate?
$H_0: p=0.10$; $H_a: p\ne 0.10$
$H_0: \hat{p}=0.10$; $H_a: \hat{p}\ne 0.10$
$H_0: p=0.10$; $H_a: p<0.10$
$H_0: p\ne 0.10$; $H_a: p=0.10$
$H_0: p=0.075$; $H_a: p\ne 0.075$
Explanation
This question tests setting up hypotheses for a two-sided test about a population proportion. The principal suspects the proportion is DIFFERENT FROM 10%, indicating a two-sided alternative. The null hypothesis is H₀: p = 0.10 (the value being tested against), and the alternative is Hₐ: p ≠ 0.10 (different from 10%). Choice B uses a one-sided alternative when the claim is about any difference. Choice C incorrectly uses the sample proportion (p̂) in the hypotheses. Choice D uses the sample proportion value (0.075) instead of the stated comparison value. Choice E reverses the null and alternative hypotheses. The phrase "different from" always indicates a two-sided test, regardless of what the sample data shows.
A restaurant chain claims that the proportion of customers who would recommend the restaurant is $0.85$. A random sample of $n=100$ customers finds $80$ would recommend it (sample proportion $\hat{p}=0.800$). Which hypotheses are appropriate to test whether the true proportion is different from $0.85$?
$H_0: p=0.85$; $H_a: p<0.85$
$H_0: \hat{p}=0.85$; $H_a: \hat{p}\ne 0.85$
$H_0: p=0.80$; $H_a: p\ne 0.80$
$H_0: p=0.85$; $H_a: p\ne 0.85$
$H_0: p\ne 0.85$; $H_a: p=0.85$
Explanation
This question involves a two-tailed test for a specific claimed proportion. The restaurant chain claims the proportion is exactly 0.85, and we want to test if the true proportion is "different from" this value. This language indicates a two-tailed test with H₀: p = 0.85 and H_a: p ≠ 0.85. Choice A correctly represents these hypotheses. The sample proportion of 0.800 might suggest the true value is lower, but since we're testing for any difference, we use the two-sided alternative. Never reverse the null and alternative hypotheses (as in choice D) or use sample statistics (p̂) in place of population parameters. The phrase "different from" always signals a two-tailed test with ≠ in the alternative hypothesis.
A school district reports that $30%$ of all households in the district have no home internet access. A community group suspects the true proportion is lower. In a random sample of $n=160$ households, $38$ report no home internet access (so $\hat{p}=38/160$). Which hypotheses are appropriate?
$H_0: p=0.30$; $H_a: p>0.30$
$H_0: p=38/160$; $H_a: p<38/160$
$H_0: p=0.30$; $H_a: p<0.30$
$H_0: \hat{p}=0.30$; $H_a: \hat{p}<0.30$
$H_0: p<0.30$; $H_a: p=0.30$
Explanation
This question involves setting up hypotheses to test if a proportion is lower than a reported value. The district reports 30% have no internet access, but the community group suspects it's LOWER. The null hypothesis states the reported value: H₀: p = 0.30, and the alternative hypothesis reflects the suspicion: Hₐ: p < 0.30. Choice B has the wrong direction (greater than). Choice C incorrectly uses the sample proportion (p̂) in the hypotheses. Choice D reverses the null and alternative hypotheses. Choice E uses the sample proportion value instead of the reported population value. The key principle is that we test the claim or reported value in the null hypothesis, and the alternative reflects what we're trying to show evidence for.
A gym owner believes that more than 35% of members attend at least 3 times per week. To check, a random sample of 160 members is selected, and 49 report attending at least 3 times per week (so $
\hat{p}=49/160$). Which hypotheses are appropriate for a one-proportion $z$ test of the owners belief?
$H_0: \hat{p}=0.35$; $H_a: \hat{p}>0.35$
$H_0: p>0.35$; $H_a: p=0.35$
$H_0: p=0.306$; $H_a: p>0.306$
$H_0: p=0.35$; $H_a: p>0.35$
$H_0: p=0.35$; $H_a: p<0.35$
Explanation
This question assesses hypotheses for believing gym attendance exceeds 35%. Choice D is correct: H0: p = 0.35 and Ha: p > 0.35, aligning with 'more than' for right-tailed. Distractors use sample p-hat ≈ 0.306 in B, or p-hat in C. Option A has wrong direction. Mini-lesson: 'More than' uses > in Ha, H0 with equality on p. Sample data tests H0 but isn't in statements. This tests if evidence supports higher attendance.