Sampling Distributions for Sample Proportions - AP Statistics

A school reports that 10% of students are left-handed. A teacher repeatedly takes many random samples of $n=300$ students and calculates $\hat p$ . Which statement about the sampling distribution is correct?
Answer: The standard deviation of $\hat p$ is $\sqrt{0.10(0.90)/300}$ . Question on standard deviation of $\hat p$ in AP Statistics. SD = $\sqrt{0.10(0.90)/300}$ for n=300, confirming A. Choice B distracts by multiplying by n instead of dividing. Mini-lesson: Mean of $\hat p$ distribution is p, spread $\sqrt{p(1-p)/n}$ ; not zero as in C, variability decreases with n.
In a large city, 52% of registered voters support Candidate A. A pollster repeatedly selects random samples of $n=50$ registered voters and computes the sample proportion $\hat p$ who support Candidate A. Which statement about the sampling distribution of $\hat p$ is correct?
Answer: The sampling distribution of $\hat p$ is approximately Normal and centered at $0.52$ . This question evaluates understanding of sampling distributions for sample proportions in AP Statistics. The center of the sampling distribution of (\hat{p}) is the population proportion p = 0.52, and with n = 50, np = 26 and n(1-p) = 24 both exceeding 10, it's approximately normal, supporting choice A. The spread is quantified by the standard deviation ( $\sqrt{\frac{p(1-p)}{n}}$ ), which reflects variability around the center. Distractor choice D suggests centering at 0.50 due to 'balancing,' but the distribution is unbiased and centered at p, not 0.50. Choice E wrongly asserts no variability, overlooking that samples fluctuate around p. Mini-lesson: sampling distributions describe how statistics like (\hat{p}) vary across many samples; they're centered at the parameter, have spread inversely related to sqrt(n), and approximate normality when expected successes and failures are at least 10.
A school has 1,200 students, and 45% participate in at least one club. A counselor repeatedly takes many SRSs of $n=80$ students (without replacement) and computes $\hat p$ . Which statement about the sampling distribution is correct?
Answer: The independence assumption is reasonable because $80$ is less than 10% of $1{,}200$ . This question tests the independence assumption via 10% condition in AP Statistics. 80 is 6.67% of 1200 (<10%), reasonable, confirming A. Distractor B says fails without replacement, but condition allows it. Mini-lesson: For without replacement, independence approx if n<10% N; mean p, spread adjusted but approx sqrt{p(1-p)/n}.
A school nurse estimates that 28% of students have seasonal allergies. Many random samples of $n=75$ students are taken and $\hat p$ is computed each time. Which statement about the sampling distribution is correct?
Answer: The standard deviation of $\hat p$ is $\sqrt{0.28(0.72)/75}$ . This question tests the standard deviation formula for sampling distributions of sample proportions in AP Statistics. The sd is sqrt(p(1-p)/n) = sqrt(0.28*0.72/75), measuring spread around the center p=0.28. Choice B omits the division by n, incorrectly calculating variance instead. A distractor is choice E, claiming sd=0 because p is fixed, but sampling variability always exists. In a mini-lesson, the sampling distribution of î p is centered at p with sd sqrt(p(1-p)/n), and approximates normal when n p and n(1-p) ≥ 10. Here, 75*0.28≈21 and 75*0.72≈54, both fine. Larger n reduces spread, improving precision.
A survey organization knows that 30% of households in a region have a pet cat. Many random samples of $n=20$ households are selected, and for each sample the statistic $\hat p$ (the sample proportion with a cat) is recorded. Which statement about the sampling distribution of $\hat p$ is correct?
Answer: The sampling distribution of $\hat p$ is centered at $0.30$ , but an approximately Normal model may not be justified because the sample size is small. This question tests sampling distributions for sample proportions in AP Statistics. The distribution is centered at p = 0.30, but with n = 20, np = 6 < 10, the normal model may not be justified due to potential skewness, as in choice B. Spread is given by ( $\sqrt{\frac{p(1-p)}{n}}$ ), which is larger for smaller n. Distractor choice A assumes normality for any proportion, but conditions must be checked. Choice C centers at 0.70, mistakenly using the complement. Mini-lesson: sampling distributions illustrate variability of (\hat{p}) around p; center is unbiased, spread depends on n and p, and normality is approximate only if np and n(1-p) ≥ 10—small samples or extreme p can lead to non-normal shapes requiring caution.
A company says that 33% of its emails are opened within 1 hour. A data analyst repeatedly takes many random samples of $n=150$ emails and computes $\hat p$ . Which statement about the sampling distribution is correct?
Answer: The mean of $\hat p$ is 0.33, and the standard deviation is $\sqrt{0.33(0.67)/150}$ . This question examines the mean and sd of sampling distributions for sample proportions in AP Statistics. The mean is p=0.33, sd sqrt(0.33*0.67/150), describing center and spread. Choice B swaps mean and sd, a common error. A distractor is choice E, claiming sd=0 due to fixed p, ignoring sampling variation. In a mini-lesson, sampling distributions have mean p, sd sqrt(p(1-p)/n), normal if conditions met: here, 150*0.33=49.5 ≥ 10, 150*0.67=100.5 ≥ 10. This enables probability calculations for î p.
A manufacturer states that 2% of batteries fail within the first month. Many random samples of $n=100$ batteries are taken and $\hat p$ is recorded each time. Which statement about the sampling distribution is correct?
Answer: The Normal approximation may be questionable because $np$ may be less than 10. This question assesses normality conditions for sampling distributions of sample proportions in AP Statistics. The normal approximation is questionable since n p=100*0.02=2<10, failing success-failure. Center is p=0.02, spread sqrt(p(1-p)/n), but shape may be skewed. Choice B is a distractor, claiming normality just because n=100, ignoring other conditions. In a mini-lesson, require n p ≥ 10 and n(1-p) ≥ 10 for normality, plus random sampling. Here, n(1-p)=98 ≥ 10 but n p=2<10, so not reliable. Use caution in such cases for inferences.
A state reports that 57% of drivers have used a hands-free device while driving. Many random samples of $n=1000$ drivers are taken and $\hat p$ is computed each time. Which statement about the sampling distribution is correct?
Answer: The sampling distribution of $\hat p$ is likely approximately Normal and has relatively small variability. This question evaluates understanding of sampling distributions for sample proportions in AP Statistics, emphasizing normality and variability with large n. The sampling distribution of (\hat{p}) is centered at p = 0.57, with a small standard deviation ( $\sqrt{0.57 \times 0.43 / 1000} \approx 0.016$ ), indicating low variability. Given np = 570 ≥ 10 and n(1-p) = 430 ≥ 10, the distribution is approximately normal. A frequent distractor is choice E, which incorrectly suggests that large n prevents normality, when actually larger n improves the normal approximation. Mini-lesson: Sampling distributions describe how statistics like (\hat{p}) vary across many random samples; for proportions, the mean is p, spread decreases with larger n, and normality holds when success and failure counts are sufficient. This setup allows us to model probabilities about sample results.
A company knows that 40% of its shipped orders arrive late. A quality analyst repeatedly takes many random samples of $n=100$ orders from all shipments and records the sample proportion late, $\hat p$ , each time. Which statement about the sampling distribution of $\hat p$ is correct?
Answer: The sampling distribution of $\hat p$ is approximately Normal and centered at $0.40$ . This question tests knowledge of sampling distributions for sample proportions in AP Statistics. The sampling distribution of the sample proportion (\hat{p}) is centered at the true population proportion p = 0.40, and for large enough samples, it is approximately normal. Here, with n = 100, np = 40 and n(1-p) = 60, both greater than 10, the normal approximation is appropriate, making choice B correct. A common distractor is choice A, which incorrectly states that the sample proportion must equal the population proportion, ignoring sampling variability. Another distractor, choice D, claims no variability, but the distribution does have spread given by the standard deviation ( $\sqrt{\frac{p(1-p)}{n}}$ ). In a mini-lesson on sampling distributions: when repeatedly sampling from a population, the sample proportions form a distribution centered at p, with variability decreasing as n increases, and it's approximately normal under the conditions np ≥ 10 and n(1-p) ≥ 10.
A survey suggests that 19% of adults have run a marathon. Many random samples of $n=500$ adults are taken and $\hat p$ is computed each time. Which statement about the sampling distribution is correct?
Answer: The sampling distribution of $\hat p$ is centered at 0.19. This question probes the center of sampling distributions for sample proportions in AP Statistics. The sampling distribution of (\hat{p}) is centered at p = 0.19, not at n=500 or the complement 0.81. Its spread is ( $\sqrt{0.19 \times 0.81 / 500} \approx 0.018$ ), showing some variability. Distractor E mistakenly identifies the center as the standard deviation, a mix-up of concepts. Mini-lesson: Sampling distributions for proportions are centered at the true p, with spread decreasing as n increases; they're approximately normal if np and n(1-p) ≥ 10, here both are satisfied (95 and 405). This framework underlies confidence intervals and hypothesis tests.
A large university states that 16% of its students are first-generation college students. A researcher repeatedly takes many random samples of $n=250$ students and records $\hat p$ . Which statement about the sampling distribution is correct?
Answer: The standard deviation of $\hat p$ is $\sqrt{0.16(0.84)/250}$ . This question examines the standard deviation of sampling distributions for sample proportions in AP Statistics. Sd is sqrt(0.16*0.84/250), measuring spread around p=0.16. Choice B multiplies by n incorrectly. A distractor is choice E, claiming sd=0. In a mini-lesson, sd formula captures variability, smaller with larger n; center fixed. Here, n=250, 250*0.16=40 ≥ 10, 250*0.84=210 ≥ 10, normal ok. Used in margin of error calculations.
A county reports that 72% of homes have high-speed internet. Many random samples of $n=64$ homes are taken and $\hat p$ is computed each time. Which statement about the sampling distribution is correct?
Answer: The standard deviation of $\hat p$ would be smaller if the sample size were increased. This question focuses on how n affects spread in sampling distributions for sample proportions in AP Statistics. Increasing n decreases sd sqrt(p(1-p)/n), smaller variability around p=0.72. Choice B says opposite. A distractor is choice D, wrongly saying center changes with n. In a mini-lesson, center fixed, spread shrinks with n; here n=64, 64*0.72≈46, 64*0.28≈18, normal ok. Larger n improves accuracy.
A website knows that the true proportion of visitors who accept cookies is $p=0.90$ . The site repeatedly takes many random samples of size $n=60$ and computes $\hat p$ each time. Which statement about the sampling distribution of $\hat p$ is correct?
Answer: The sampling distribution of $\hat p$ is centered at $0.90$ , but it may not be approximately normal because $n(1-p)$ may be too small. This question tests understanding of normality conditions when p is very large. With p = 0.90 and n = 60, we check: np = 60(0.90) = 54 ≥ 10 (satisfied) but n(1-p) = 60(0.10) = 6 < 10 (not satisfied). The failure condition is not met, so the sampling distribution may be noticeably skewed rather than approximately normal. Choice A incorrectly applies the n > 30 rule for means to proportions. Choice C incorrectly centers at 0.10 and misinterprets why. Choice D incorrectly restricts the formula to non-random samples. Choice E incorrectly claims no variability. When p is very close to 0 or 1, larger sample sizes are needed to achieve approximate normality.
A community health report states that 40% of adults have had a flu shot this year. A student repeatedly takes many random samples of $n=30$ adults and records $\hat p$ . Which statement about the sampling distribution is correct?
Answer: The sampling distribution of $\hat p$ may be approximately Normal because $np$ and $n(1-p)$ are likely at least 10. This evaluates normality conditions in AP Statistics for hat p. np=12, n(1-p)=18 >=10, so approx Normal possible, making A correct. Distractor C says only if p=0.50, but works for other p if conditions hold. Mini-lesson: Distribution approx Normal with mean p, spread sqrt{p(1-p)/n} under rules; not exact as in E.
A company reports that 66% of its employees participate in a retirement plan. Many random samples of $n=120$ employees are taken and $\hat p$ is recorded. Which statement about the sampling distribution is correct?
Answer: The sampling distribution of $\hat p$ is centered at 0.66 and its standard deviation is $\sqrt{0.66(0.34)/120}$ . This question assesses the center and spread of sampling distributions for sample proportions in AP Statistics. The distribution of $\hat{p}$ is centered at the population proportion p = 0.66, with standard deviation $\sqrt{0.66 \times 0.34 / 120} \approx 0.043$ . Since np = 79.2 ≥ 10 and n(1-p) = 40.8 ≥ 10, it's approximately normal. Choice C is a common distractor, confusing the center with n=120 and spread with p. Mini-lesson: In sampling distributions, repeated random samples yield $\hat{p}$ values that cluster around p, with variability quantified by the standard deviation formula; this helps predict how close $\hat{p}$ will be to p. Understanding this is key for inference in statistics.
A survey reports that 49% of students prefer online homework submission. Many random samples of $n=25$ students are taken and $\hat p$ is computed each time. Which statement about the sampling distribution is correct?
Answer: The sampling distribution of $\hat p$ may be approximately Normal because $np$ and $n(1-p)$ are both likely at least 10. This question evaluates normality conditions for sampling distributions of sample proportions in AP Statistics. May be normal: n=25, p=0.49, n p=12.25 ≥ 10, n(1-p)=12.75 ≥ 10 barely. Center 0.49, spread sqrt(p(1-p)/n). Choice E says normal for any n=25, too absolute. In a mini-lesson, check conditions carefully; marginal here, shape close to normal. Proceed with caution for small n.
A researcher estimates that 64% of households own a pet. Many random samples of $n=10$ households are taken and $\hat p$ is recorded. Which statement about the sampling distribution is correct?
Answer: The sampling distribution of $\hat p$ may not be approximately Normal because the success-failure condition may fail. This question evaluates normality conditions for sampling distributions of sample proportions in AP Statistics. May not be normal because n=10, n p=10*0.64=6.4<10, failing condition. Center p=0.64, but shape skewed. Choice B is distractor, wrongly tying to p>0.50. In a mini-lesson, need n p ≥ 10, n(1-p) ≥ 10 for normality; here n(1-p)=3.6<10 too. Small n limits approximation. Use exact methods instead.
A factory has 2,000 bolts in a bin, and 6% are slightly too long. An inspector repeatedly selects many SRSs of $n=150$ bolts without replacement and computes $\hat p$ . Which statement about the sampling distribution is correct?
Answer: The 10% condition is satisfied because $150$ is less than 10% of $2{,}000$ . Focus on 10% condition for finite population sampling in AP Statistics. 150 is 7.5% of 2000 (<10%), satisfied, so A correct. Distractor B says not because p small, but unrelated. Mini-lesson: Condition justifies ignoring finite correction; mean p, hat p unbiased unlike E suggests.
A company knows that the true proportion of its customers who prefer online support is $p=0.62$ . The company repeatedly takes many random samples of size $n=200$ from this population and computes the sample proportion $\hat p$ each time. Which statement about the sampling distribution of $\hat p$ is correct?
Answer: The sampling distribution of $\hat p$ is centered at $0.62$ , and its standard deviation is $\sqrt{\dfrac{0.62(0.38)}{200}}$ . This question tests understanding of the center and spread of the sampling distribution of sample proportions. The sampling distribution of p̂ is centered at the true population proportion p = 0.62, and its standard deviation (standard error) is calculated using the formula √[p(1-p)/n] = √[0.62(0.38)/200]. Choice B incorrectly claims there's no variability - sample proportions will vary from sample to sample even though they center around 0.62. Choice C uses an incorrect denominator of 0.62 instead of n = 200. Choice D incorrectly states the distribution is exactly normal for any n - normality is approximate and requires checking conditions. Choice E incorrectly uses n-1 in the denominator, which is for sample standard deviations, not sampling distributions of proportions.
A hospital reports that 80% of patients rate their experience as "satisfied." A researcher repeatedly takes random samples of $n=60$ patients and computes $\hat p$ , the sample proportion satisfied. Which statement about the sampling distribution of $\hat p$ is correct?
Answer: The sampling distribution of $\hat p$ is centered at $0.80$ and is approximately Normal. This question explores sampling distributions for sample proportions in AP Statistics. With p = 0.80 and n = 60, np = 48 and n(1-p) = 12 both ≥ 10, the distribution is centered at 0.80 and approximately normal, confirming choice A. The standard deviation ( $\sqrt{\frac{p(1-p)}{n}}$ ) measures spread around this center. Distractor choice B centers at 0.20, confusing satisfied with unsatisfied proportions. Choice C claims no variability, but samples inherently vary. Mini-lesson: repeated sampling produces a distribution of (\hat{p}) centered at p with decreasing spread as n increases; normality approximation is valid when expected counts of success and failure are at least 10, facilitating inference like confidence intervals.

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