Sampling Distributions for Sample Proportions
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AP Statistics › Sampling Distributions for Sample Proportions
A retailer estimates that 53% of customers use a coupon. Many random samples of $n=15$ customers are taken and $\hat p$ is computed each time. Which statement about the sampling distribution is correct?
The sampling distribution of $\hat p$ is centered at 15.
The sampling distribution of $\hat p$ is exactly Normal because sampling is random.
The sampling distribution of $\hat p$ has no spread because $p$ is fixed.
The sampling distribution of $\hat p$ is approximately Normal because $p$ is close to 0.50.
The sampling distribution of $\hat p$ may not be approximately Normal because the success-failure condition may fail.
Explanation
This question examines conditions for normality in sampling distributions for sample proportions in AP Statistics. With p=0.53 and n=15, np=7.95 <10 and n(1-p)=7.05 <10, so the distribution may not be approximately normal. The center is at 0.53, and spread is ($\sqrt{0.53 \times 0.47 / 15}$ \approx 0.129), but normality isn't guaranteed. Choice E is a distractor, wrongly centering at n=15 instead of p. Mini-lesson: For the sampling distribution of (\hat{p}) to be useful, it should be nearly normal, requiring at least 10 expected successes and failures; without this, shape can be skewed, affecting probability calculations. Always check these conditions before applying normal models.
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?
If sampling is without replacement, $\hat p$ is biased and not centered at 0.06.
The sampling distribution of $\hat p$ has no variability when sampling from a finite population.
The 10% condition is not satisfied because $n$ must be less than 10% of $p$.
The 10% condition is not satisfied because $p=0.06$ is small.
The 10% condition is satisfied because $150$ is less than 10% of $2{,}000$.
Explanation
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 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?
The sampling distribution of $\hat p$ is centered at 0.81.
The sampling distribution of $\hat p$ has no variability because 19% is fixed.
The sampling distribution of $\hat p$ is centered at 0.19.
The sampling distribution of $\hat p$ is centered at $\sqrt{0.19(0.81)/500}$.
The sampling distribution of $\hat p$ is centered at 500.
Explanation
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 researcher knows that 40% of seeds germinate under certain conditions. She repeatedly takes many random samples of $n=100$ seeds and computes $\hat p$. Which statement about the sampling distribution is correct?
The sampling distribution of $\hat p$ has mean 0.40 and standard deviation $\sqrt{0.40(0.60)\cdot 100}$.
The sampling distribution of $\hat p$ has mean 0 and standard deviation 0 because $p$ is fixed.
The sampling distribution of $\hat p$ has mean 0.40 and standard deviation $\sqrt{0.40(0.60)/100}$.
The sampling distribution of $\hat p$ has mean 100 and standard deviation 0.40.
The sampling distribution of $\hat p$ has mean $\sqrt{0.40(0.60)/100}$ and standard deviation 0.40.
Explanation
This question focuses on the mean and standard deviation formulas for sampling distributions of sample proportions in AP Statistics. The distribution is centered at $p=0.40$ with standard deviation $\sqrt{0.40 \times 0.60 / 100}$, capturing the center and spread accurately. Choice B incorrectly multiplies by n instead of dividing, inflating the spread. A common distractor is choice E, suggesting mean and sd of zero because p is fixed, but statistics vary while parameters do not. In a mini-lesson, the sampling distribution of $\hat p$ has mean $p$ and sd $\sqrt{p(1-p)/n}$, and is approximately normal if $n p \geq 10$ and $n(1-p) \geq 10$. Here, $100 \times 0.40 = 40$ and $100 \times 0.60 = 60$, both sufficient. This allows probabilistic statements about $\hat p$'s range.
A company states that 50% of its products are shipped from Warehouse A. Many random samples of $n=100$ products are selected and $\hat p$ is computed each time. Which statement about the sampling distribution is correct?
The distribution of $\hat p$ is more concentrated around 0.50 than it would be if $n=25$.
The distribution of $\hat p$ is centered at $\sqrt{0.50(0.50)/100}$.
The distribution of $\hat p$ is more concentrated around 0.50 than it would be if $n=400$.
The distribution of $\hat p$ has no spread because $p=0.50$.
The distribution of $\hat p$ is centered at 100 because $n=100$.
Explanation
This question examines how sample size affects the spread of sampling distributions for sample proportions in AP Statistics. The sampling distribution is centered at p=0.50, and with larger n=100 compared to n=25, the spread sqrt(p(1-p)/n) is smaller, making it more concentrated. For n=400, it would be even more concentrated, so choice B is incorrect. A distractor like choice C claims no spread because p=0.50, ignoring that variability exists regardless of p's value. In a mini-lesson, sampling distributions show that as n increases, the standard deviation decreases, leading to more precise estimates of p. Here, conditions for normality are met since n p=50 ≥ 10 and n(1-p)=50 ≥ 10. This principle underscores why larger samples are preferred in statistical inference.
A store estimates that 75% of customers use a reusable bag. Many random samples of $n=20$ customers are taken and $\hat p$ is recorded each time. Which statement about the sampling distribution is correct?
The sampling distribution of $\hat p$ cannot be studied because $\hat p$ changes each sample.
The sampling distribution of $\hat p$ has no spread because $p=0.75$.
The sampling distribution of $\hat p$ is exactly Normal whenever $n=20$.
The sampling distribution of $\hat p$ may not be approximately Normal because $n(1-p)$ may be less than 10.
The sampling distribution of $\hat p$ is approximately Normal because $p$ is close to 1.
Explanation
This question evaluates conditions for normality in sampling distributions for sample proportions in AP Statistics. The distribution may not be approximately normal because n(1-p)=20*0.25=5 <10, failing the success-failure condition. The center is p=0.75, with spread sqrt(p(1-p)/n), but shape is skewed when conditions fail. Choice B is a distractor, incorrectly linking normality to p near 1 rather than the conditions. In a mini-lesson, for î p's distribution to be normal, need random sample, n p ≥ 10, n(1-p) ≥ 10, and for finite pops, n<10% N. Here, n p=15 ≥ 10 but n(1-p)=5<10, so questionable. This affects reliability of normal-based inferences.
A charity reports that 5% of mailed donation requests receive a response. Many random samples of $n=500$ requests are considered, and $\hat p$ is the sample proportion that receive a response. Which statement about the sampling distribution is correct?
The sampling distribution of $\hat p$ is centered at 500 because $n=500$.
The sampling distribution of $\hat p$ is centered at 0.05 and typically varies from sample to sample.
The sampling distribution of $\hat p$ is centered at $\sqrt{0.05(0.95)/500}$.
The sampling distribution of $\hat p$ is centered at 0.05 and has no variation because $p$ is fixed.
The sampling distribution of $\hat p$ is centered at 0.95 because most do not respond.
Explanation
This question assesses the center and variability of sampling distributions for sample proportions in AP Statistics. The distribution is centered at $p=0.05$ and varies from sample to sample, with spread $\sqrt{p(1-p)/n}$. Choice B wrongly claims no variation, confusing the fixed $p$ with the variable $\hat p$. A distractor like choice C centers it at $n=500$, mistaking count for proportion. In a mini-lesson, sampling distributions describe the long-run behavior of $\hat p$: mean $p$, sd decreasing with larger $n$, and normality if success-failure conditions hold. Here, $n p=500 \times 0.05=25 \geq 10$, $n(1-p)=475 \geq 10$, so normal approximation is good. This is crucial for confidence intervals and hypothesis tests.
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?
The sampling distribution of $\hat p$ has mean $\sqrt{0.45(0.55)/80}$.
The sampling distribution of $\hat p$ has mean 0.55, the proportion not in clubs.
The sampling distribution of $\hat p$ will be centered at 80.
The independence assumption fails because sampling is without replacement.
The independence assumption is reasonable because $80$ is less than 10% of $1{,}200$.
Explanation
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 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?
The sampling distribution of $\hat p$ is likely approximately Normal but has no variability.
The sampling distribution of $\hat p$ is centered at $\sqrt{0.57(0.43)/1000}$.
The sampling distribution of $\hat p$ is centered at 1000.
The sampling distribution of $\hat p$ is likely approximately Normal and has relatively small variability.
The sampling distribution of $\hat p$ is not approximately Normal because $n$ is too large.
Explanation
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 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?
The mean of $\hat p$ is 0.67, and the standard deviation is $\sqrt{0.33(0.67)/150}$.
The mean of $\hat p$ is 0.33, and the standard deviation is $\sqrt{0.33(0.67)/150}$.
The standard deviation of $\hat p$ is 0 because the population proportion is fixed.
The mean of $\hat p$ is $\sqrt{0.33(0.67)/150}$, and the standard deviation is 0.33.
The mean of $\hat p$ is 150, and the standard deviation is 0.33.
Explanation
This question examines the mean and sd of sampling distributions for sample proportions in AP Statistics. The mean is p=0.33, sd sqrt(0.330.67/150), 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, 1500.33=49.5 ≥ 10, 150*0.67=100.5 ≥ 10. This enables probability calculations for î p.