Justifying Claims: Confidence Interval, Population Proportion
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AP Statistics › Justifying Claims: Confidence Interval, Population Proportion
A streaming service randomly samples 2,000 subscribers to estimate the proportion who use the service daily. The company wants to claim that between 45% and 55% of all subscribers use the service daily. A 95% confidence interval for the population proportion is $(0.46, 0.52)$. Is the claim supported by the confidence interval?
No, because the interval includes more than one value, so no range claim can be supported.
No, because 95% confidence means 95% of subscribers use the service daily.
Yes, because the entire confidence interval lies within $[0.45,0.55]$, making the claim plausible.
No, because the interval does not include $0.55$.
Yes, because there is a 95% probability that $p$ is exactly $0.49$.
Explanation
This question examines whether a confidence interval supports a range claim about a population proportion. The claim is that between 45% and 55% use the service daily (0.45 ≤ p ≤ 0.55), and the 95% confidence interval is (0.46, 0.52). Since the entire confidence interval (0.46, 0.52) lies within the claimed range [0.45, 0.55], all plausible values for the true proportion satisfy the claim. This provides support because every value in our confidence interval falls within the proposed range. The correct answer properly identifies that when a confidence interval is entirely contained within a claimed range, it supports that range claim. This is a special case where we're checking if the CI is a subset of the claimed interval.
A school principal claims that at least 80% of students at the school eat breakfast before arriving. A random sample was taken and a 90% confidence interval for the true proportion $p$ of students who eat breakfast was computed as $(0.78,\ 0.85)$. Is the claim supported by the confidence interval?
No, because the interval includes values below 0.80, so it does not support $p\ge 0.80$.
Yes, because the upper end of the interval is above 0.80.
Yes, because 0.80 is in the interval, so $p\ge 0.80$ must be true.
No, because 90% of students are between 0.78 and 0.85.
Yes, because 90% confidence means there is a 90% chance $p\ge 0.80$.
Explanation
This question asks whether a confidence interval supports a claim about a minimum proportion. The claim is that at least 80% eat breakfast (p ≥ 0.80), but the interval (0.78, 0.85) includes values below 0.80. Since some plausible values of p are less than 0.80, we cannot conclude that p ≥ 0.80. Choice A incorrectly thinks containing 0.80 proves the claim. Choice C misinterprets the confidence level. Choice D confuses the confidence interval with a range of student percentages. To support a claim that p ≥ some value, the entire confidence interval must be at or above that value.
A technology blog randomly samples 700 readers to estimate the proportion who prefer Android over iOS. The blog wants to claim that fewer than 35% of its readers prefer Android. A 92% confidence interval for the population proportion is $(0.33, 0.38)$. Is the claim supported by the confidence interval?
Yes, because 92% confidence means 92% of readers prefer Android.
No, because a confidence interval cannot be used to assess a claim about being less than a value.
Yes, because the lower endpoint is below $0.35$.
No, because $0.35$ is in the interval, so the data do not provide convincing evidence that $p<0.35$.
Yes, because there is a 92% probability that the true proportion is between $0.33$ and $0.38$, and $0.35$ is closer to $0.33$ than to $0.38$.
Explanation
This question examines whether a confidence interval can justify a claim about a proportion being less than a specific value. The claim is that fewer than 35% prefer Android (p < 0.35), and the 92% confidence interval is (0.33, 0.38). Since 0.35 is contained within the confidence interval, it represents a plausible value for the true population proportion. When a confidence interval contains a hypothesized value, we cannot reject that value, which means we cannot claim the proportion is definitively less than it. The correct answer recognizes that containing 0.35 means we lack convincing evidence that the true proportion is less than 0.35. Remember: to support p < k, the entire interval must be below k; if k is in the interval, p = k is plausible.
A city council member claims that more than 60% of city residents support a proposed recycling ordinance. A random sample of residents was used to construct a 95% confidence interval for the true population proportion $p$ of residents who support the ordinance: $(0.56,\ 0.64)$. Is the claim supported by the confidence interval?
No, because the interval includes values less than 0.60, so it does not support $p>0.60$.
No, because 95% confidence means there is a 95% chance that $p$ is less than 0.60.
Yes, because 0.60 is in the interval, so $p>0.60$.
Yes, because the sample proportion must be 0.60 exactly.
Yes, because 95% of residents are in the interval.
Explanation
This question tests whether you can use a confidence interval to evaluate a claim about a population proportion. The claim states that more than 60% support the ordinance (p > 0.60), but the confidence interval (0.56, 0.64) includes values both above and below 0.60. Since the interval contains values less than 0.60, we cannot conclude that p > 0.60. Choice A incorrectly interprets the confidence level as referring to the percentage of residents. Choice B wrongly assumes that containing 0.60 supports p > 0.60. Choice D misinterprets the confidence level as a probability about p. When using confidence intervals to test claims, if the interval includes values that contradict the claim, the claim is not supported.
A public health agency randomly samples 1,100 residents to estimate the proportion who received a flu shot this season. The agency wants to claim that at least 65% of residents received a flu shot (that is, $p\ge 0.65$). A 95% confidence interval for the population proportion is $(0.62, 0.66)$. Is the claim supported by the confidence interval?
No, because $0.65$ is not the upper endpoint of the interval.
Yes, because the midpoint of the interval is close to $0.65$.
No, because the interval includes values below $0.65$, so it does not support the claim $p\ge 0.65$.
Yes, because 95% confidence means there is a 95% chance that $p\ge 0.65$.
Yes, because the interval includes values above $0.65$, so it proves $p\ge 0.65$.
Explanation
This question tests understanding of using confidence intervals to support "at least" claims. The claim is that at least 65% received a flu shot (p ≥ 0.65), and the 95% confidence interval is (0.62, 0.66). Since the interval includes values below 0.65 (specifically 0.62 through 0.64), we cannot conclude that the true proportion is definitely at least 0.65. The interval tells us the true proportion could plausibly be anywhere from 62% to 66%, which includes possibilities where p < 0.65. The correct answer recognizes this principle: to support p ≥ k, the entire interval must be at or above k. When the interval contains values below k, the claim is not supported by the data.
An environmental group claims that more than 35% of households in a county compost food scraps. A survey produced a 95% confidence interval for the true proportion $p$ of households that compost: $(0.31,\ 0.36)$. Is the claim supported by the confidence interval?
Yes, because the upper bound is greater than 0.35.
Yes, because 0.35 is in the interval.
No, because the interval includes values less than or equal to 0.35, so it does not support $p>0.35$.
Yes, because 95% confidence means there is a 95% chance $p>0.35$.
No, because 95% of households compost.
Explanation
This question asks whether a confidence interval supports a claim about exceeding a proportion. The claim states that more than 35% compost (p > 0.35), but the interval (0.31, 0.36) includes values both above and below 0.35, including 0.35 itself. Since the interval contains values less than or equal to 0.35, the claim is not supported. Choice A incorrectly thinks containing 0.35 supports p > 0.35. Choice B misinterprets confidence levels. Choice E wrongly focuses only on the upper bound. To support a strict inequality claim (p > value), the entire interval must exceed that value.
A museum randomly samples 650 visitors to estimate the proportion who would recommend the museum to a friend. The museum wants to claim that at most 70% of all visitors would recommend it (that is, $p\le 0.70$). A 90% confidence interval for the population proportion is $(0.68, 0.74)$. Is the claim supported by the confidence interval?
Yes, because 90% confidence means 90% of visitors would recommend the museum.
Yes, because the interval contains $0.70$, so it proves $p\le 0.70$.
No, because $0.70$ is not the midpoint of the interval.
No, because the interval includes values greater than $0.70$, so it does not support the claim $p\le 0.70$.
Yes, because the lower bound is below $0.70$.
Explanation
This question examines whether a confidence interval supports a claim about a proportion being at most a certain value. The claim is that at most 70% would recommend (p ≤ 0.70), and the 90% confidence interval is (0.68, 0.74). Since the interval includes values greater than 0.70 (specifically 0.71 through 0.74), we cannot conclude that the true proportion is definitely at most 0.70. The interval suggests the true proportion could plausibly be anywhere from 68% to 74%, which includes possibilities where p > 0.70. The correct answer recognizes this principle: to support p ≤ k, the entire interval must be at or below k. When the interval extends above k, the claim is not supported by the data.
A museum director claims that more than 35% of visitors are from out of town. Based on a random sample of visitors, an 88% confidence interval for the population proportion of out-of-town visitors is $(0.31, 0.36)$. Is the claim supported by the confidence interval?
No, because 88% confidence means there is an 88% probability the true proportion is less than 0.35.
Yes, because 0.35 is inside the interval, so the true proportion is more than 0.35.
Yes, because an 88% confidence interval is wide enough to confirm the claim.
Yes, because the interval’s upper bound is above 0.35, so the true proportion must be more than 0.35.
No, because the interval includes values less than or equal to 0.35, so it does not support that the true proportion is more than 0.35.
Explanation
In AP Statistics, this assesses justifying directional claims with confidence intervals for proportions. The director claims more than 0.35 are out-of-town visitors, but the 88% confidence interval (0.31, 0.36) includes values <= 0.35, so the true proportion could be 0.35 or less, not supporting the claim. Distractor B wrongly claims that including 0.35 means it's more than 0.35, confusing containment with directionality. Mini-lesson: 'More than p' requires the entire interval > p; values <= p in the interval undermine support. With 0.31 < 0.35, the claim lacks backing. Choice C is correct, highlighting precise claim-interval alignment.
A bookstore owner claims that between 35% and 45% of customers buy a magazine at checkout. A random sample of customers is observed, and a 95% confidence interval for the true proportion $p$ is $(0.37,\ 0.44)$. Is the claim supported by the confidence interval?
Yes, because 95% of customers in the sample bought a magazine.
Yes, because the entire confidence interval lies within $(0.35, 0.45)$, so the claim is supported.
No, because 95% confidence means there is a 95% probability that $p$ is between 0.35 and 0.45.
No, because the interval does not include exactly $0.35$ and $0.45$ as endpoints.
No, because the interval contains 0.40, which proves $p=0.40$ exactly rather than a range.
Explanation
This question asks whether a claim that between 35% and 45% buy magazines (0.35 < p < 0.45) is supported by a 95% CI of (0.37, 0.44). Since the entire confidence interval (0.37, 0.44) lies within the claimed range (0.35, 0.45), all plausible values for p satisfy the claim. Choice B correctly identifies this - when the CI is entirely contained within the claimed range, the claim is supported. Choice A misinterprets what confidence means, Choice C incorrectly requires exact endpoint matching, Choice D confuses sample statistics with confidence levels, and Choice E wrongly interprets a contained value as proving exact equality. For range claims, check if the entire CI falls within the specified range.
A streaming service claims that at least 70% of its subscribers watch content weekly. A random sample is used to compute a 92% confidence interval for the true proportion $p$: $(0.68,\ 0.74)$. Is the claim supported by the confidence interval?
Yes, because 92% confidence means 92% of subscribers watch weekly.
Yes, because there is a 92% probability that $p$ is at least $0.70$.
No, because $0.70$ is not the midpoint of the interval, so the claim is false.
Yes, because the interval contains values above $0.70$, so $p\ge 0.70$ is guaranteed.
No, because the interval includes values less than $0.70$, so the data do not support the claim $p\ge 0.70$.
Explanation
This question tests whether a claim that at least 70% watch weekly (p ≥ 0.70) is supported by a 92% CI of (0.68, 0.74). The confidence interval includes values less than 0.70 (specifically 0.68 to 0.70), meaning some plausible values for p do not satisfy p ≥ 0.70. Choice B correctly identifies this - when the CI contains values that violate the claim, we cannot support it. Choice A wrongly focuses only on values above 0.70, Choice C confuses the confidence level with the proportion, Choice D misinterprets confidence as probability about p, and Choice E irrelevantly discusses the midpoint. Remember: for claims with inequalities, check whether ALL values in the CI satisfy the inequality.