Aimathic
Login | English | Deutsch

Free math worksheets

Build your own math worksheets from 28,000 problems for grades 3 to 12, from fractions to calculus. Every problem includes step-by-step solutions.

Confidence interval for a proportion

Click problems to add them to your worksheet.

54837212
A random sample produces a \(95\%\) confidence interval of \((0.41, 0.49)\) for the proportion of all district households that compost food scraps. For each statement, decide whether it is a valid interpretation and explain why. a) “There is a \(95\%\) probability that the true household proportion is between \(0.41\) and \(0.49\).” b) “We are \(95\%\) confident that between \(41\%\) and \(49\%\) of all district households compost food scraps.” c) “About \(95\%\) of random samples taken by this method would produce intervals that contain the true household proportion.”

Hints

- Distinguish the fixed population parameter from intervals that would vary across samples. - A contextual interval interpretation should name the population and the response. - Think about what repeats in the long-run meaning of a confidence level.

Solution

1. Statement a) is not valid because, after the interval is calculated, the fixed population proportion is either in the interval or it is not; the \(95\%\) refers to the long-run success rate of the method. 2. Statement b) is valid because it states confidence about the population proportion in context. 3. Statement c) is valid as a description of the confidence level under repeated random sampling with the same method.

Answer

a) Invalid. The \(95\%\) describes the method’s long-run capture rate, not a probability assigned to the fixed parameter after the interval is computed. b) Valid. c) Valid.
54838312
The same random sample produces the two confidence intervals shown. a) Which interval most likely has the higher confidence level? b) What point estimate was used for both intervals? c) Explain why the intervals have the same center but different widths.
Figure for problem 548383

Hints

- Compare how far each displayed interval extends from its center. - Read the endpoints and average them to recover the point estimate. - Changing the confidence level affects the distance from the center, not the observed sample proportion.

Solution

1. Interval B is wider, so it most likely uses the higher confidence level. 2. The midpoint of Interval A is \(\frac{0.284+0.356}{2}=0.320\), and the midpoint of Interval B is \(\frac{0.276+0.364}{2}=0.320\). 3. Both intervals use the same sample proportion as the point estimate. The higher confidence level requires a larger critical value, increasing the margin of error while leaving the center unchanged.

Answer

a) Interval B. b) \(0.320\). c) Both use the same sample estimate, but Interval B uses a larger critical value and therefore has a larger margin of error.
54838612
A random survey yields the \(90\%\) confidence interval shown for the proportion of all residents who support a local bond measure. a) Is there convincing evidence that a majority of residents support the measure? Explain. b) Is there convincing evidence that support is exactly \(60\%\)? Explain. c) State an appropriate interval interpretation.
Figure for problem 548386

Hints

- Compare the full displayed interval with the labeled value \(0.50\). - A labeled value inside an interval is plausible, not proven to be the exact parameter. - Keep the interpretation focused on the population rather than the sampled residents.

Solution

1. The entire displayed interval is greater than \(0.50\), so it provides convincing evidence that a majority of residents support the measure. 2. The value \(0.60\) lies inside the interval, so it is a plausible value for the population proportion; the interval does not show that support is exactly \(60\%\). 3. We are \(90\%\) confident that the interval from \(0.57\) to \(0.65\) contains the true proportion of all residents who support the bond measure.

Answer

a) Yes, because the entire interval is above \(0.50\). b) No. The interval includes \(0.60\), but it cannot establish exact equality. c) We are \(90\%\) confident that \(57\%\) to \(65\%\) of all residents support the measure.
54840112
Three random samples of size \(100\) produce these success counts: Sample A: \(9\) Sample B: \(10\) Sample C: \(90\) a) For which samples is the observed-count condition met for a one-sample \(z\)-interval? b) Explain the decision for each sample. c) Why is the condition symmetric with respect to “success” and “failure”?

Hints

- Convert each success count into the corresponding failure count. - Check both categories against the same threshold. - Imagine reversing which category is called a success.

Solution

1. Sample A has \(9\) successes and \(91\) failures, so the condition is not met. 2. Sample B has \(10\) successes and \(90\) failures, so the condition is met. 3. Sample C has \(90\) successes and \(10\) failures, so the condition is met. 4. The normal approximation requires both response categories to have enough observations; relabeling the categories should not change whether the procedure is appropriate.

Answer

a) Samples B and C. b) A fails because one count is \(9\); B and C each have at least \(10\) in both categories. c) Both outcomes contribute to the shape of the sample-proportion distribution.
54840312
A class simulates \(100\) random samples and constructs a \(95\%\) confidence interval from each sample. Exactly \(94\) of the intervals contain the true population proportion. a) Is this result inconsistent with a \(95\%\) confidence procedure? b) Does it mean the procedure should now be called a \(94\%\) confidence procedure? c) Explain the expected long-run behavior.

Hints

- Distinguish a theoretical long-run rate from one finite simulation result. - Random variation affects the number of successful intervals in any batch. - Think about what would happen over many more repetitions.

Solution

1. The result is not inconsistent. In a finite set of \(100\) repetitions, the capture count can vary randomly around \(95\). 2. The procedure remains a \(95\%\) confidence procedure; one simulation run does not redefine its theoretical long-run capture rate. 3. Over many repeated random samples, approximately \(95\%\) of intervals constructed by the method are expected to contain the true population proportion.

Answer

a) No. b) No. c) The long-run capture proportion should approach \(95\%\), although any set of \(100\) intervals can capture slightly more or fewer than \(95\).
54840812
A \(90\%\) confidence interval for the proportion of all commuters who work from home at least one day per week is \((0.31, 0.37)\). A manager says, “\(90\%\) of commuters have a work-from-home rate between \(31\%\) and \(37\%\).” a) Explain why this statement is incorrect. b) Give a correct interpretation of the interval. c) State what quantity the interval estimates.

Hints

- Identify whether the interval is about individuals or a single population parameter. - Keep the response variable binary in the interpretation. - Name the population and the characteristic measured.

Solution

1. The interval estimates one population proportion; it does not describe a separate work-from-home percentage for each commuter. 2. A correct interpretation is: We are \(90\%\) confident that between \(31\%\) and \(37\%\) of all commuters work from home at least one day per week. 3. The estimated quantity is \(p\), the population proportion of commuters with the stated characteristic.

Answer

a) The interval concerns one population proportion, not a distribution of individual percentages. b) We are \(90\%\) confident that \(31\%\) to \(37\%\) of all commuters work from home at least one day per week. c) The population proportion \(p\).
54841412
A random sample has \(\hat p=0.64\). Three proposed two-sided confidence intervals are shown. a) Which intervals could have been formed as \(\hat p\pm\text{margin of error}\)? b) Explain why the remaining interval cannot have that form.
Figure for problem 548414

Hints

- Compare the given sample proportion with the midpoint of each displayed interval. - A standard two-sided interval extends the same distance on both sides of the point estimate. - Read the endpoints when a midpoint calculation is needed.

Solution

1. Interval A has midpoint \(0.64\), and Interval C also has midpoint \(0.64\), so both could have the usual symmetric form. 2. Interval B has midpoint \(\frac{0.56+0.69}{2}=0.625\), not \(0.64\). 3. Therefore, B cannot be a standard interval centered at the given sample proportion.

Answer

a) A and C. b) B is centered at \(0.625\), not at \(\hat p=0.64\).
54842812
A random-sample study reports a \(95\%\) confidence interval of \((0.28, 0.36)\) for the proportion of households in a county that use a curbside food-scrap program. a) Find the corresponding \(95\%\) confidence interval for the proportion of county households that do not use the program. b) Interpret the new interval in context. c) Explain why the confidence level stays the same after taking the complement.

Hints

- The two household categories are complements and must add to one. - Apply the complement to both endpoints and then place them in increasing order. - Think about whether transforming every successful interval changes which samples capture the parameter.

Solution

1. If \(p\) is the proportion that use the program, then the proportion that do not use it is \(1-p\). 2. Complementing reverses the endpoints: \((1-0.36, 1-0.28)=(0.64, 0.72)\). 3. The transformed interval contains \(1-p\) exactly when the original interval contains \(p\), so the long-run capture rate is unchanged.

Answer

a) \((0.64, 0.72)\). b) We are \(95\%\) confident that between \(64\%\) and \(72\%\) of county households do not use the curbside food-scrap program. c) Complementing is a one-to-one transformation, so an interval that captures \(p\) produces an interval that captures \(1-p\) on the same samples.
54843212
A regional airport surveys a simple random sample of \(350\) passengers from \(5000\) departing passengers during a month. Of those surveyed, \(91\) say they used public transportation to reach the airport. Statistical software reports the following \(93\%\) confidence interval output. <table> <tr><th>Estimate</th><th>Standard error</th><th>Lower endpoint</th><th>Upper endpoint</th></tr> <tr><td>\(0.2600\)</td><td>\(0.02345\)</td><td>\(0.2175\)</td><td>\(0.3025\)</td></tr> </table> a) State the population parameter estimated by the interval. b) Interpret the interval in context. c) Does the interval provide convincing evidence that fewer than \(30\%\) of all departing passengers use public transportation to reach the airport? Explain.

Hints

- Describe the parameter using the full population and the measured response. - Interpret the endpoints as plausible values for that parameter. - Compare the entire interval with the benchmark in the claim.

Solution

1. The parameter is \(p\), the proportion of all \(5000\) departing passengers during the month who used public transportation to reach the airport. 2. The sample is random, \(350\le0.10(5000)=500\), and the observed counts are \(91\) and \(259\), both at least \(10\). 3. The reported interval is \((0.2175, 0.3025)\). 4. A valid interpretation is that we are \(93\%\) confident the true airport-wide proportion for that month is between \(0.2175\) and \(0.3025\). 5. The interval includes values greater than \(0.30\), so it does not provide convincing evidence that the population proportion is below \(0.30\).

Answer

a) The proportion of all departing passengers during the month who used public transportation to reach the airport. b) We are \(93\%\) confident that the true proportion is between \(21.75\%\) and \(30.25\%\). c) No. The interval includes values at and above \(30\%\).
54843612
A valid \(95\%\) confidence interval for the proportion of registered voters in a city who plan to vote early is \((0.31, 0.39)\). The city has \(25{,}000\) registered voters. a) Use the interval to find a corresponding range of plausible numbers of registered voters who plan to vote early. b) Interpret the range in context. c) Explain why the calculation does not mean that exactly \(95\%\) of individual voters have been classified correctly.

Hints

- Translate each plausible population proportion into a population count. - Keep the same population and response definition in the interpretation. - Distinguish uncertainty about a total from certainty about individual members.

Solution

1. Multiply each endpoint by the population size: \(0.31(25{,}000)=7750\) and \(0.39(25{,}000)=9750\). 2. The transformed interval gives a plausible range from \(7750\) to \(9750\) registered voters who plan to vote early. 3. The confidence level describes the long-run success rate of the interval procedure for estimating the population parameter. It does not assign a confidence label to individual voters.

Answer

a) From \(7750\) to \(9750\) voters. b) We are \(95\%\) confident that the number of registered city voters who plan to vote early is between \(7750\) and \(9750\). c) The confidence level applies to the interval procedure for the population total, not to classification of individual voters.
54844312
A \(90\%\) confidence interval for a population proportion is \((0.48, 0.56)\). A student says, “Because \(0.50\) is inside the interval, \(0.50\) is the most likely value of the population proportion.” a) Explain why the student's conclusion does not follow from the confidence interval. b) State what the interval does allow the student to say about \(0.50\). c) Identify the point estimate used to construct the interval.

Hints

- Distinguish a set of plausible values from a probability distribution over those values. - Ask what inclusion of a benchmark permits and what it does not prove. - Locate the interval's center to recover the sample estimate.

Solution

1. A standard confidence interval identifies a range of plausible parameter values but does not assign probabilities or likelihood rankings to individual values after the interval is computed. 2. Because \(0.50\) lies inside the interval, the sample data are compatible with \(p=0.50\) at the confidence level used; the interval does not establish that \(0.50\) is the most likely value. 3. The point estimate is the midpoint: \(\hat p=\frac{0.48+0.56}{2}=0.52\).

Answer

a) A confidence interval does not rank the values inside it by probability or likelihood. b) The value \(0.50\) is a plausible value for the population proportion based on this interval. c) \(\hat p=0.52\).
54846112
A museum randomly selects \(400\) names from its membership list for a survey. After data collection, \(20\) selected people are found to be ineligible because their memberships had expired before the survey date. Among the \(380\) eligible sampled members, \(228\) favor extending weekend hours. Construct and interpret a \(95\%\) confidence interval for the proportion of eligible current members who favor the extension. Assume the remaining interval conditions are satisfied.

Hints

- Identify which selected observations belong to the population named in the question. - Use the eligible-response count as the denominator for the sample proportion. - Interpret the completed interval for current eligible members only.

Solution

1. The relevant sample size is the number of eligible current members, \(n=380\), and the sample proportion is \(\hat p=\frac{228}{380}=0.60\). 2. The estimated standard error is \(\sqrt{\frac{0.60(0.40)}{380}}\approx 0.02513\). 3. The margin of error is \(1.96(0.02513)\approx 0.04926\). 4. The interval is \(0.60\pm 0.04926\approx(0.5507, 0.6493)\).

Answer

The \(95\%\) confidence interval is approximately \((0.551, 0.649)\). We are \(95\%\) confident that between \(55.1\%\) and \(64.9\%\) of eligible current museum members favor extending weekend hours.
54847012
A valid \(95\%\) confidence interval is shown for the proportion of shipped orders that arrive by the promised date. a) Find the point estimate and margin of error. b) Does the interval provide convincing evidence that more than \(80\%\) of all shipped orders arrive by the promised date? Explain. c) Explain why the interval does not guarantee that at least \(80\%\) of orders will arrive on time in every future week.
Figure for problem 548470

Hints

- Read the interval’s endpoints and recover its center and half-width. - Compare the full displayed interval with the labeled benchmark \(0.80\). - Separate a population estimate from the behavior of each future sample or time period.

Solution

1. Reading the endpoints, the point estimate is the midpoint: \(\hat p=\frac{0.81+0.87}{2}=0.84\). 2. The margin of error is half the width: \(\frac{0.87-0.81}{2}=0.03\). 3. Because the entire displayed interval lies above \(0.80\), it provides convincing evidence that the population proportion exceeds \(0.80\). 4. The interval estimates a long-run population proportion. Week-to-week sample proportions can vary and are not guaranteed to exceed \(0.80\).

Answer

a) Point estimate \(0.84\); margin of error \(0.03\). b) Yes. Every value in the interval is greater than \(0.80\). c) The interval concerns the overall population proportion, while individual future weeks can vary.
54847612
Two confidence intervals from the same sample are shown. Both use the same one-proportion \(z\)-interval method. Interval A was reported as an \(80\%\) interval, and Interval B was reported as a \(95\%\) interval. a) Explain why the two reports cannot both be correct. b) Which interval should be wider? c) If the endpoints are otherwise accurate, identify the most likely reporting error.
Figure for problem 548476

Hints

- Compare the displayed widths while holding the sample estimate and estimated standard error fixed. - Consider how confidence level changes the critical multiplier. - Check whether the nesting of the two intervals matches that relationship.

Solution

1. For the same sample and method, both intervals have the same center and estimated standard error. 2. The \(95\%\) interval uses a larger critical value than the \(80\%\) interval, so it must have a larger margin of error and be wider. 3. The display shows that reported Interval B is narrower and lies inside reported Interval A, which reverses the required relationship. 4. The most likely error is that the confidence-level labels were swapped.

Answer

a) Higher confidence cannot produce the narrower interval from the same sample and method. b) The \(95\%\) interval. c) The \(80\%\) and \(95\%\) labels were likely reversed.
54848212
In a simple random sample of \(250\) recreation-center members, \(130\) say they use the center's mobile reservation system. A student reports the interval \((0.468, 0.572)\) and labels it a \(95\%\) confidence interval. The student used \(z^*=1.645\). Assume the confidence-interval conditions are satisfied. a) Identify the error. b) Construct the correct \(95\%\) confidence interval. c) Interpret the corrected interval.

Hints

- Match the confidence label to the standard normal multiplier used. - Keep the sample estimate fixed while correcting the critical value. - Recalculate the endpoint distance before giving a population interpretation.

Solution

1. The sample proportion is \(\hat p=\frac{130}{250}=0.52\). The multiplier \(1.645\) corresponds to \(90\%\), not \(95\%\), confidence. 2. The estimated standard error is \(\sqrt{\frac{0.52(0.48)}{250}}\approx 0.03160\). 3. For \(95\%\) confidence, the margin of error is \(1.96(0.03160)\approx 0.06193\). 4. The corrected interval is \(0.52\pm 0.06193\approx(0.4581, 0.5819)\).

Answer

a) The student used the \(90\%\) critical value while labeling the interval \(95\%\). b) Approximately \((0.458, 0.582)\). c) We are \(95\%\) confident that between \(45.8\%\) and \(58.2\%\) of all recreation-center members use the mobile reservation system.
54850912
In a random sample of \(625\) commuters, \(203\) say they usually take public transportation to work. Assume the confidence-interval conditions are satisfied. Construct a \(95\%\) confidence interval for the population proportion. Then report the interval as whole percentages by rounding the lower endpoint down and the upper endpoint up.

Hints

- Estimate the proportion from the success count and sample size. - Keep the unrounded interval until the final reporting step. - “Rounding outward” means rounding the lower endpoint down and the upper endpoint up, even when ordinary rounding would move them differently.

Solution

1. The sample proportion is \(\hat p=\frac{203}{625}=0.3248\). 2. The standard error is \(\sqrt{\frac{(0.3248)(0.6752)}{625}}\approx 0.01873\). 3. The margin of error is \(1.96(0.01873)\approx 0.03671\). 4. The interval is approximately \((0.2881, 0.3615)\). 5. In percentages, the endpoints are about \(28.81\%\) and \(36.15\%\). Rounding outward gives \(28\%\) to \(37\%\).

Answer

The \(95\%\) confidence interval is approximately \((0.2881, 0.3615)\). Rounded outward to whole percentages, it is \(28\%\) to \(37\%\).
54851312
The exact, unrounded \(95\%\) confidence interval for the proportion of residents who support a proposal is shown. Does this interval support the claim that a strict majority of residents favor the proposal? Explain the role of the lower endpoint.
Figure for problem 548513

Hints

- Translate “strict majority” into an inequality for the population proportion. - Read the exact interval and the boundary value from the display. - Pay attention to whether equality is allowed in the claim.

Solution

1. A strict majority requires \(p>0.50\). 2. Reading the display, the exact interval is \((0.500, 0.580)\), so it includes \(p=0.50\) as its lower endpoint. 3. Because the interval does not place every plausible value strictly above \(0.50\), it does not support the strict-majority claim at the \(95\%\) confidence level.

Answer

No. The exact interval includes \(0.50\), so it does not establish that the population proportion is strictly greater than \(0.50\).
54851612
A standard symmetric \(95\%\) one-proportion \(z\)-interval calculation produces interval a). An analyst changes the result and reports interval b), while still labeling it a standard \(95\%\) confidence interval. Explain why that label is not justified.
Figure for problem 548516

Hints

- Compare the two displayed intervals and identify exactly what changed. - Ask whether changing an endpoint leaves the original repeated-sampling method unchanged. - Consider what an endpoint outside the parameter range says about the approximation being used.

Solution

1. Reading the display, the standard procedure produced \((-0.012, 0.068)\), while the analyst reported \((0, 0.068)\). 2. Replacing the negative endpoint by \(0\) changes the procedure after the calculation and changes its repeated-sampling coverage behavior. 3. The negative endpoint also signals that the symmetric normal interval may be unsuitable for data this close to the boundary. 4. The altered interval cannot automatically inherit the \(95\%\) confidence level of the original formula.

Answer

Interval b) is not the standard \(95\%\) one-proportion \(z\)-interval. Changing an endpoint alters the procedure and its coverage; a method designed for proportions near a boundary should be used instead.
54852912
A \(95\%\) confidence interval for a population proportion is \((0.42, 0.48)\). A manager says, “In the next random sample of \(100\) people, between \(42\) and \(48\) people must have the characteristic.” Explain the error in this interpretation.

Hints

- Identify the parameter estimated by the interval. - Distinguish uncertainty about a population value from randomness in a future sample. - Ask whether confidence intervals make guarantees about individual repeated samples.

Solution

1. The interval estimates the fixed population proportion \(p\); it is not a prediction interval for a future sample count. 2. Even if \(p\) lies between \(0.42\) and \(0.48\), a future sample proportion varies randomly around \(p\). 3. Therefore, a future sample of \(100\) can contain fewer than \(42\) or more than \(48\) successes.

Answer

The interval describes plausible values of the population proportion, not guaranteed counts in a future sample. Sampling variability can place a future count outside \(42\) through \(48\).
54855712
A \(95\%\) confidence interval for the proportion of current seniors at one high school who plan to take a gap year is \((0.12, 0.19)\). A student writes: “There is a \(95\%\) chance that between \(12\%\) and \(19\%\) of all high school students next year will take a gap year.” Identify the errors and write a correct interpretation.

Hints

- Check whether the population named in the interpretation matches the population sampled. - Distinguish the confidence level of a method from a probability about one fixed parameter. - Keep the time frame exactly aligned with the data collection.

Solution

1. After the interval is calculated, the fixed population proportion is either inside the interval or outside it; the \(95\%\) describes the long-run success rate of the procedure, not a probability assigned to this fixed parameter. 2. The interval concerns current seniors at the sampled high school, not all high school students and not a future cohort. 3. A correct interpretation states confidence that the interval contains the true proportion of current seniors at that school who plan to take a gap year.

Answer

The statement incorrectly assigns a \(95\%\) probability to the fixed parameter and changes both the population and time period. A correct interpretation is: We are \(95\%\) confident that between \(12\%\) and \(19\%\) of current seniors at this high school plan to take a gap year.
54836712
A public library system has \(6000\) active cardholders. A simple random sample of \(250\) cardholders finds that \(142\) borrowed at least one e-book during the past year. Construct and interpret a \(95\%\) confidence interval for the proportion of all active cardholders who borrowed at least one e-book during the past year. Verify the conditions for the procedure.

Hints

- Identify the population and the binary response before naming the parameter. - Confirm that the sample design and the two observed outcome counts support the procedure. - Build the interval around the sample proportion and interpret it as a range for the population proportion.

Solution

1. The parameter is \(p\), the proportion of all active cardholders who borrowed at least one e-book during the past year, and \(\hat p=\frac{142}{250}=0.568\). 2. The sample is random, \(250\le 0.10(6000)=600\), and the observed counts are \(142\) successes and \(108\) failures, both at least \(10\). 3. The standard error is \(\sqrt{\frac{0.568(0.432)}{250}}\approx 0.0313\). 4. Using \(z^*=1.96\), the interval is \(0.568\pm 1.96(0.0313)\), or approximately \((0.507, 0.629)\). 5. We are \(95\%\) confident that the interval from \(0.507\) to \(0.629\) contains the true proportion of all active cardholders who borrowed at least one e-book during the past year.

Answer

The \(95\%\) confidence interval is approximately \((0.507, 0.629)\). We are \(95\%\) confident that this interval contains the true proportion of all active cardholders who borrowed at least one e-book during the past year.
54837912
The \(95\%\) confidence interval shown was calculated from a random sample of \(400\) observations. a) Estimate the sample proportion used to construct the interval. b) Determine the corresponding number of successes in the sample. c) Does the interval provide convincing evidence that the population proportion differs from \(0.50\)? Explain.
Figure for problem 548379

Hints

- Read the endpoints and recover the estimate from the center of the displayed interval. - Convert the estimated proportion back to a count using the sample size. - Compare \(0.50\) with the full interval rather than only with its midpoint.

Solution

1. Reading the displayed rounded endpoints, the sample proportion is approximately the midpoint: \(\hat p\approx \frac{0.39+0.48}{2}=0.435\). 2. The success count is \(400(0.435)=174\). 3. The value \(0.50\) is outside the displayed interval because the upper endpoint is approximately \(0.48\). 4. Therefore, the interval provides convincing evidence at the corresponding two-sided \(5\%\) level that the population proportion differs from \(0.50\).

Answer

a) \(\hat p\approx 0.435\). b) \(174\) successes. c) Yes. The interval does not contain \(0.50\).
54839312
A random sample of \(200\) voters from a population of \(5000\) voters includes \(84\) who support a proposal. A student calculates a \(95\%\) confidence interval using the standard error \(\sqrt{\frac{0.50(0.50)}{200}}\) because \(0.50\) is the value in a related null hypothesis. a) Explain the error. b) Construct the correct \(95\%\) confidence interval. c) Compare the correct margin of error with the student’s margin of error.

Hints

- Distinguish the standard error used for estimation from the one used under a null model. - Use the observed data to estimate the variability of the estimator. - Compare the two products inside the square roots to anticipate which margin is larger.

Solution

1. A confidence interval estimates variability using the observed sample proportion, not a null hypothesized value. Here, \(\hat p=\frac{84}{200}=0.42\). 2. The sample is random, \(200\le0.10(5000)=500\), and the observed counts are \(84\) and \(116\), both at least \(10\). 3. The correct standard error is \(\sqrt{\frac{0.42(0.58)}{200}}\approx 0.0349\). 4. The correct interval is \(0.42\pm1.96(0.0349)\), or approximately \((0.352, 0.488)\). 5. The correct margin of error is approximately \(0.0684\), while the student’s is approximately \(0.0693\).

Answer

a) The interval standard error must use \(\hat p=0.42\), not the null value \(0.50\). b) Approximately \((0.352, 0.488)\). c) Correct margin of error: \(0.0684\); student’s margin of error: \(0.0693\).
54840612
A confidence interval for a population proportion is shown. The estimated standard error is \(0.025\). a) Find the point estimate and margin of error. b) Estimate the critical value. c) Identify the most likely common confidence level.
Figure for problem 548406

Hints

- Read the endpoints from the display and recover the center and half-width. - Compare the half-width with the estimated standard error. - Match the resulting multiplier to a familiar confidence level.

Solution

1. Reading the endpoints from the display, the point estimate is the midpoint: \(\frac{0.491+0.589}{2}=0.540\). 2. The margin of error is \(\frac{0.589-0.491}{2}=0.049\). 3. The critical value is \(z^*=\frac{0.049}{0.025}=1.96\). 4. A critical value near \(1.96\) corresponds to \(95\%\) confidence.

Answer

a) Point estimate \(0.540\); margin of error \(0.049\). b) \(z^*\approx 1.96\). c) Approximately \(95\%\) confidence.
54841912
An arts center has \(4000\) season-ticket holders. It selects a simple random sample of \(240\) holders and asks how likely they are to renew next season. The responses are shown. Treat “very likely” or “somewhat likely” as a success. Construct and interpret a \(95\%\) confidence interval for the proportion of all season-ticket holders who are likely to renew. Verify the conditions for the interval.
Figure for problem 548419

Hints

- Read the category counts from the chart and combine the two categories defined as success. - Check how the sample was selected, its size relative to the population, and the two observed outcome counts. - Build the interval around the combined sample proportion and interpret the population parameter in context.

Solution

1. Reading the chart, the number of successes is \(74+62=136\), so \(\hat p=\frac{136}{240}=0.5667\). 2. The sample was randomly selected. Also, \(240\le 0.10(4000)=400\), and the observed counts are \(136\) successes and \(104\) failures, both at least \(10\). 3. The estimated standard error is \(\sqrt{\frac{0.5667(0.4333)}{240}}\approx 0.03199\). 4. The margin of error is \(1.96(0.03199)\approx 0.06269\). 5. The interval is \(0.5667\pm 0.06269\approx(0.5040, 0.6294)\).

Answer

The \(95\%\) confidence interval is approximately \((0.504, 0.629)\). We are \(95\%\) confident that between \(50.4\%\) and \(62.9\%\) of all season-ticket holders are very likely or somewhat likely to renew. The randomization, \(10\%\), and observed-count conditions are satisfied.
54842312
In a random sample of \(120\) transit-card users from a population of \(3000\) users, \(84\) say they check arrival times before leaving home. Two analysts propose the rounded \(95\%\) confidence intervals shown. One interval is centered at the current sample proportion, and the other is centered at a historical value of \(0.65\). a) Which interval is the appropriate one-sample confidence interval for the current population proportion? b) Explain why the other interval is not appropriate. c) Interpret the appropriate interval in context.
Figure for problem 548423

Hints

- Calculate the statistic that estimates the current population proportion. - Compare that statistic with the centers of the displayed rounded intervals. - A confidence interval should extend outward from the estimate supplied by the sample.

Solution

1. The sample proportion is \(\hat p=\frac{84}{120}=0.700\). 2. The sample is random, \(120\le0.10(3000)=300\), and the observed counts are \(84\) and \(36\), both at least \(10\). 3. A confidence interval for a population proportion must be centered at the point estimate \(\hat p\), not at a historical null value. The display shows that Analyst A’s interval is centered at \(0.700\). 4. The estimated standard error is \(\sqrt{\frac{0.70(0.30)}{120}}\approx 0.04183\), and the margin of error is \(1.96(0.04183)\approx 0.08199\). 5. The resulting interval is \(0.700\pm 0.08199\approx(0.618, 0.782)\), confirming the rounded interval shown for Analyst A.

Answer

a) Analyst A’s interval, \((0.618, 0.782)\). b) Analyst B incorrectly centered the interval at the historical value \(0.65\) instead of the sample proportion. c) We are \(95\%\) confident that between \(61.8\%\) and \(78.2\%\) of all current transit-card users check arrival times before leaving home.
54845112
A random sample of \(100\) manufactured parts contains no defective parts. A student uses the standard one-proportion \(z\)-interval formula and reports the \(95\%\) confidence interval \((0, 0)\) for the population defect proportion. a) Show why the formula produces this degenerate interval. b) Explain why the standard \(z\)-interval is not appropriate here. c) Explain why observing no defects does not prove that the population defect proportion is zero.

Hints

- Substitute the observed sample proportion into the usual estimated-variability calculation. - Check the two observed category counts before trusting the normal-based interval. - Distinguish “not observed in this sample” from “impossible in the population.”

Solution

1. The sample proportion is \(\hat p=0\), so the estimated standard error from the usual formula is \(\sqrt{\frac{0(1-0)}{100}}=0\). This produces \(0\pm 1.96(0)=(0, 0)\). 2. The observed success count is \(0\), which is below \(10\), so the normality condition for the standard one-proportion \(z\)-interval fails. 3. A positive but small population defect proportion can still produce a sample with no defects. The sample result alone does not establish that the population proportion is exactly zero.

Answer

a) \(\hat p=0\) gives estimated standard error \(0\), so the formula returns \((0, 0)\). b) The observed-count condition fails because there are \(0\) defective parts. c) A rare defect can be missed in a finite random sample, so the result does not prove \(p=0\).
54845612
A city has more than \(20{,}000\) households. In January, a simple random sample of \(100\) households finds that \(61\) use a city alert app. In February, a simple random sample of \(150\) additional households is selected from the same complete household list, with no household included in both samples; \(84\) of the February households use the app. Assume the population proportion did not change between the two months. Pool the two samples to construct and interpret a \(95\%\) confidence interval for the household app-use proportion. Verify the relevant conditions.

Hints

- Confirm that the two samples estimate the same population parameter and contain distinct households before combining them. - Pool both the success counts and sample sizes. - Check the combined sample against the population-size and observed-count conditions before building the interval.

Solution

1. The combined sample has \(61+84=145\) successes out of \(100+150=250\) distinct households, so \(\hat p=\frac{145}{250}=0.58\). 2. Both samples were randomly selected from the same household list, no household appears twice, and the population proportion is assumed unchanged. The combined sample size satisfies \(250\le 0.10(20{,}000)=2000\), and the combined observed counts are \(145\) and \(105\), both at least \(10\). 3. The estimated standard error is \(\sqrt{\frac{0.58(0.42)}{250}}\approx 0.03122\). 4. The margin of error is \(1.96(0.03122)\approx 0.06118\). 5. The interval is \(0.58\pm 0.06118\approx(0.5188, 0.6412)\).

Answer

The \(95\%\) confidence interval is approximately \((0.519, 0.641)\). We are \(95\%\) confident that between \(51.9\%\) and \(64.1\%\) of city households use the alert app. The randomization, common-population, no-overlap, \(10\%\), and observed-count conditions are satisfied under the stated assumption that the proportion was unchanged.
54846512
A database contains \(500\) active service accounts. A researcher selects \(200\) accounts by sampling with replacement, and \(126\) of the selections have enabled two-factor authentication. Treat repeated selections as independent observations. Construct and interpret a \(95\%\) confidence interval for the probability that a randomly selected active account has two-factor authentication enabled. Explain why the usual \(10\%\) condition is not required here.

Hints

- Determine how the sampling method affects independence between selections. - Check the two observed outcome counts before using the interval. - Build the interval around the sample proportion and interpret the account-selection probability.

Solution

1. The sample proportion is \(\hat p=\frac{126}{200}=0.63\). 2. Sampling with replacement makes the selections independent, so the \(10\%\) condition used for sampling without replacement is not needed. 3. The observed counts are \(126\) and \(74\), both at least \(10\). 4. The estimated standard error is \(\sqrt{\frac{0.63(0.37)}{200}}\approx 0.03414\), and the margin of error is \(1.96(0.03414)\approx 0.06691\). 5. The interval is \(0.63\pm 0.06691\approx(0.5631, 0.6969)\).

Answer

The \(95\%\) confidence interval is approximately \((0.563, 0.697)\). We are \(95\%\) confident that the probability a randomly selected active account has two-factor authentication enabled is between \(56.3\%\) and \(69.7\%\). The \(10\%\) condition is unnecessary because the selections were made with replacement.
54847812
A quality office randomly inspects \(600\) records, and \(10\) contain a particular documentation error. Construct and interpret a \(90\%\) confidence interval for the population error proportion. Assume the remaining interval conditions are satisfied, verify the observed-count condition, and use \(z^*=1.645\).

Hints

- Check whether equality with the minimum observed-count threshold is sufficient. - Use the error count over the full inspected sample to find the estimate. - Keep enough precision when working with a small sample proportion.

Solution

1. The sample proportion is \(\hat p=\frac{10}{600}=0.01667\). 2. The observed counts are \(10\) errors and \(590\) nonerrors. Both are at least \(10\), so the observed-count condition is met at its boundary. 3. The estimated standard error is \(\sqrt{\frac{0.01667(0.98333)}{600}}\approx 0.005226\). 4. The margin of error is \(1.645(0.005226)\approx 0.008597\). 5. The interval is \(0.01667\pm 0.008597\approx(0.00807, 0.02526)\).

Answer

The \(90\%\) confidence interval is approximately \((0.0081, 0.0253)\). We are \(90\%\) confident that between \(0.81\%\) and \(2.53\%\) of all records contain the documentation error. The observed-count condition is satisfied because the counts are \(10\) and \(590\).
54848912
A local news graphic summarizes a random sample of \(600\) registered voters. Of those surveyed, \(330\) support a proposed transit measure. Assume the confidence-interval conditions are satisfied. The graphic gives a \(95\%\) confidence summary. Compute the one-proportion \(z\)-interval and decide whether the graphic is a reasonable rounded summary.
Figure for problem 548489

Hints

- Begin with the observed number of supporters out of the total sample. - Distinguish the estimated standard error from the final margin of error. - Read the graphic’s center and endpoint distance, then compare them with the calculated values.

Solution

1. The sample proportion is \(\hat p=\frac{330}{600}=0.55\). 2. The standard error is \(\sqrt{\frac{(0.55)(0.45)}{600}}\approx 0.02031\). 3. The \(95\%\) margin of error is \(1.96(0.02031)\approx 0.03981\). 4. The interval is \(0.55\pm 0.03981\), or approximately \((0.5102, 0.5898)\). 5. The displayed margin is \(4\) percentage points around \(55\%\), which closely matches the calculated interval after rounding.

Answer

The \(95\%\) confidence interval is approximately \((0.5102, 0.5898)\), or \(51.02\%\) to \(58.98\%\). The displayed summary of \(55\%\pm 4\) percentage points is reasonable after rounding.
54849312
Five years ago, \(50\%\) of residents in a county used curbside recycling. In a new random sample of \(300\) residents, \(162\) say they currently use it. Assume the confidence-interval conditions are satisfied. a) Construct a \(90\%\) confidence interval for the current proportion \(p\). b) Convert that interval into a confidence interval for the change \(p-0.50\). c) Does the interval provide clear evidence that use has increased?

Hints

- First estimate the present-day proportion from the new sample. - A shift in the parameter produces the same shift in both confidence-interval endpoints. - Decide what value represents “no change” on the transformed scale.

Solution

1. The sample proportion is \(\hat p=\frac{162}{300}=0.54\). 2. The standard error is \(\sqrt{\frac{(0.54)(0.46)}{300}}\approx 0.02877\). 3. With \(z^*=1.645\), the margin of error is \(1.645(0.02877)\approx 0.04733\). 4. The interval for \(p\) is \((0.4927, 0.5873)\). 5. Subtracting \(0.50\) from both endpoints gives an interval for \(p-0.50\) of \((-0.0073, 0.0873)\). 6. Because the change interval includes \(0\), it does not give clear evidence of an increase at the \(90\%\) confidence level.

Answer

a) Approximately \((0.4927, 0.5873)\). b) Approximately \((-0.0073, 0.0873)\). c) No. A change of \(0\) remains plausible because it lies in the interval.
54850412
A random sample contains \(200\) households. Assume the confidence-interval conditions are satisfied. In \(128\) households, the household’s selected respondent supports a zoning proposal. An analyst copies each household response twice and then treats the resulting \(400\) entries as independent observations. a) Compute the correct \(95\%\) confidence interval using the \(200\) independent households. b) Compute the interval produced by the analyst’s duplicated data. c) Explain why the second interval is misleading.

Hints

- Identify the actual independently sampled units. - Check whether copying a response changes the estimate or adds information. - Compare how the denominator in the standard error affects interval width.

Solution

1. The sample proportion is \(\hat p=\frac{128}{200}=0.64\). 2. Using \(n=200\), the margin of error is \(1.96\sqrt{\frac{(0.64)(0.36)}{200}}\approx 0.06652\), giving \((0.5735, 0.7065)\). 3. Duplicating every entry leaves \(\hat p=0.64\) but incorrectly uses \(n=400\). The resulting margin of error is \(1.96\sqrt{\frac{(0.64)(0.36)}{400}}=0.04704\), giving \((0.5930, 0.6870)\). 4. The copied entries add no independent information, so the narrower second interval falsely overstates precision.

Answer

a) Approximately \((0.5735, 0.7065)\). b) Approximately \((0.5930, 0.6870)\). c) The duplicated records are not new independent observations, so using \(n=400\) understates uncertainty.
54852012
A city randomly invites \(500\) residents to answer a transportation survey, and \(280\) of them support a proposal. Assume the confidence-interval conditions are satisfied. The city also posts an open link; \(7600\) of \(9500\) voluntary respondents support the proposal. Construct a \(95\%\) confidence interval that can validly estimate support among city residents, and explain why the open-link responses should not be combined with the random sample for this purpose.

Hints

- Separate observations obtained through random selection from those obtained through an open invitation. - Use the data source that supports generalization to the target population. - A much larger response count does not automatically make a biased design representative.

Solution

1. Use the random sample only, giving \(\hat p=\frac{280}{500}=0.56\). 2. The standard error is \(\sqrt{\frac{(0.56)(0.44)}{500}}\approx 0.02220\). 3. The margin of error is \(1.96(0.02220)\approx 0.04351\). 4. The confidence interval is approximately \((0.5165, 0.6035)\). 5. The open-link respondents self-selected, so combining them with the random sample would replace a representative design with a largely voluntary-response design.

Answer

The valid \(95\%\) confidence interval is approximately \((0.5165, 0.6035)\), based only on the \(500\) randomly invited residents. The voluntary responses should not be pooled with it because self-selection can introduce bias.
54853412
A utility company randomly selects \(300\) business days from its records. On \(72\) of those days, at least one service outage occurred. Assume the confidence-interval conditions are satisfied. Construct and interpret a \(95\%\) confidence interval for the proportion of business days with at least one outage. Explain why the interval does not estimate the proportion of minutes with an outage.

Hints

- Identify exactly what one observation represents in the sample. - Define the binary outcome before forming the sample proportion. - Match the interval interpretation to the sampled unit rather than to a different time scale.

Solution

1. The observational unit is a business day, and a success is a day with at least one outage. Thus, \(\hat p=\frac{72}{300}=0.24\). 2. The standard error is \(\sqrt{\frac{(0.24)(0.76)}{300}}\approx 0.02466\). 3. The margin of error is \(1.96(0.02466)\approx 0.04833\). 4. The confidence interval is approximately \((0.1917, 0.2883)\). 5. Each observation records whether a day had any outage, not how many minutes the outage lasted, so the parameter concerns days rather than minutes.

Answer

The \(95\%\) confidence interval is approximately \((0.1917, 0.2883)\). It estimates that about \(19.17\%\) to \(28.83\%\) of business days have at least one outage; it does not estimate outage time as a proportion of all minutes.
54853712
In a random sample of \(480\) event attendees, respondents may select more than one reason for attending. Exactly \(210\) respondents select “professional networking.” Assume the confidence-interval conditions are satisfied. Construct a \(95\%\) confidence interval for the proportion of attendees who would select professional networking. Explain why allowing multiple selections does not prevent this one-proportion analysis.

Hints

- Focus on one named response option rather than on whether all category percentages sum to \(100\%\). - Recode each respondent into two groups for the parameter being estimated. - Use the full respondent count as the denominator.

Solution

1. For this question, classify each respondent as either selecting professional networking or not selecting it. 2. The sample proportion is \(\hat p=\frac{210}{480}=0.4375\). 3. The standard error is \(\sqrt{\frac{(0.4375)(0.5625)}{480}}\approx 0.02264\). 4. The margin of error is \(1.96(0.02264)\approx 0.04438\), giving approximately \((0.3931, 0.4819)\). 5. Other response choices do not change this respondent-level binary classification.

Answer

The \(95\%\) confidence interval is approximately \((0.3931, 0.4819)\). Multiple selections are allowed because each respondent still has a clear yes-or-no outcome for the specific networking option.
54854612
A sample of \(100\) observations has \(\hat p=0.40\). An analyst reports the unrounded \(95\%\) one-proportion \(z\)-interval shown. Could this interval have been produced by the standard formula? Verify your conclusion.
Figure for problem 548546

Hints

- Read the reported endpoints, center, and margin of error from the display. - Independently calculate the margin of error from the sample proportion and sample size. - Compare the two margins before deciding whether the report is consistent.

Solution

1. Reading the display, the reported interval is \((0.31, 0.49)\). It is centered at \(0.40\) and has margin of error \(0.09\). 2. The estimated standard error is \(\sqrt{\frac{(0.40)(0.60)}{100}}\approx 0.04899\). 3. The standard \(95\%\) margin of error is \(1.96(0.04899)\approx 0.09602\). 4. The corresponding interval is approximately \((0.3040, 0.4960)\), not the displayed interval.

Answer

No. The standard formula gives a margin of error of approximately \(0.09602\) and an interval of about \((0.3040, 0.4960)\).
54849712
A random sample of \(500\) adults gives the responses shown to a ballot question. Assume the confidence-interval conditions are satisfied. a) Construct a \(95\%\) confidence interval for the proportion of all adults who support the question. b) Construct a \(95\%\) confidence interval for the proportion who support it among adults who express either support or opposition. c) Explain why the intervals estimate different parameters.
Figure for problem 548497

Hints

- Read the three response counts from the chart before choosing a denominator. - Treat the two requested parameters as separate estimation problems. - In the interpretation, state clearly who is included in each population base.

Solution

1. Reading the chart, \(230\) adults support the question, \(215\) oppose it, and \(55\) have no opinion. 2. For all adults, \(\hat p=\frac{230}{500}=0.46\). The margin of error is \(1.96\sqrt{\frac{(0.46)(0.54)}{500}}\approx 0.04369\), giving \((0.4163, 0.5037)\). 3. Among adults expressing an opinion, the denominator is \(230+215=445\), so \(\hat p=\frac{230}{445}\approx 0.5169\). 4. The second margin of error is \(1.96\sqrt{\frac{(0.5169)(0.4831)}{445}}\approx 0.04643\), giving \((0.4704, 0.5633)\). 5. The first parameter includes adults with no opinion in the population base; the second conditions on expressing support or opposition.

Answer

a) Approximately \((0.4163, 0.5037)\). b) Approximately \((0.4704, 0.5633)\). c) The first interval estimates support among all adults, while the second estimates support only among adults who express a side.
54850012
A sample proportion is \(\hat p=0.56\), and its estimated standard error is \(0.025\). Consider symmetric one-proportion \(z\)-intervals based on these values. At what central confidence level would the lower endpoint be exactly \(0.50\)? Explain what happens to the lower endpoint at lower confidence levels.

Hints

- Express the lower endpoint in terms of the point estimate, critical value, and standard error. - Find the cutoff that makes the endpoint equal the stated benchmark. - Translate that standard normal cutoff into the central area between its symmetric values.

Solution

1. Set the lower endpoint equal to \(0.50\): \(0.56-z^*(0.025)=0.50\). 2. Solving gives \(z^*=\frac{0.06}{0.025}=2.40\). 3. The central area between \(-2.40\) and \(2.40\) is \(2\Phi(2.40)-1\approx 0.9836\). 4. The required confidence level is approximately \(98.36\%\). At any lower confidence level, the critical value and margin of error are smaller, so the lower endpoint is above \(0.50\).

Answer

The lower endpoint equals \(0.50\) at a confidence level of approximately \(98.36\%\). Lower confidence levels produce a lower endpoint greater than \(0.50\).
54852412
A report says that \(43\%\) of a random sample of \(350\) people answered “yes,” with the percentage rounded to the nearest whole percent. The original success count is unavailable. Assume the confidence-interval conditions are satisfied. a) List the possible success counts. b) For the corresponding \(95\%\) one-proportion \(z\)-intervals, give the range of possible lower endpoints and the range of possible upper endpoints.

Hints

- Translate rounding to the nearest percent into an interval of possible unrounded proportions. - Convert that interval into whole-number success counts. - Recalculate the confidence interval for each feasible count before summarizing the endpoint ranges.

Solution

1. Percentages that round to \(43\%\) satisfy \(0.425\le \hat p<0.435\). 2. Multiplying by \(350\) gives \(148.75\le x<152.25\), so \(x\) can be \(149\), \(150\), \(151\), or \(152\). 3. For \(x=149\), the \(95\%\) interval is approximately \((0.3739, 0.4775)\). 4. For \(x=150\), it is approximately \((0.3767, 0.4804)\); for \(x=151\), approximately \((0.3795, 0.4833)\); and for \(x=152\), approximately \((0.3824, 0.4862)\). 5. Thus, the possible lower endpoints range from about \(0.3739\) to \(0.3824\), and the possible upper endpoints range from about \(0.4775\) to \(0.4862\).

Answer

a) The possible success counts are \(149\), \(150\), \(151\), and \(152\). b) Lower endpoints range from approximately \(0.3739\) to \(0.3824\); upper endpoints range from approximately \(0.4775\) to \(0.4862\).
54854812
A simple random sample of \(300\) customers asks whether they would recommend a service. Assume the confidence-interval conditions are satisfied. The recorded responses are shown; some electronic files cannot be read. a) Construct a \(95\%\) confidence interval if all unreadable responses are treated as “no.” b) Construct a \(95\%\) confidence interval if all unreadable responses are treated as “yes.” c) Explain what the two intervals show about the effect of the unreadable responses.
Figure for problem 548548

Hints

- Read all three response counts from the chart. - Treat each extreme scenario as a separate binary data set with the same total sample size. - Compare the locations of the two intervals, not only their widths.

Solution

1. Reading the chart gives \(174\) recorded “yes” responses, \(108\) recorded “no” responses, and \(18\) unreadable responses. 2. If all unreadable responses are treated as “no,” then \(\hat p=\frac{174}{300}=0.58\). The interval is \(0.58\pm 1.96\sqrt{\frac{(0.58)(0.42)}{300}}\approx (0.5241, 0.6359)\). 3. If all unreadable responses are treated as “yes,” then \(\hat p=\frac{192}{300}=0.64\). The interval is \(0.64\pm 1.96\sqrt{\frac{(0.64)(0.36)}{300}}\approx (0.5857, 0.6943)\). 4. The unreadable responses shift both the point estimate and the confidence interval. Without knowing how those responses should be classified, a single interval is not justified by the available data.

Answer

a) Approximately \((0.5241, 0.6359)\). b) Approximately \((0.5857, 0.6943)\). c) The possible coding of the unreadable responses materially changes the interval, so the data do not support one definitive confidence interval without an additional assumption.
54855212
An analyst claims that the exact, unrounded endpoints of a standard one-proportion \(z\)-interval from \(600\) binary observations are \((0.443, 0.523)\). Determine whether any possible whole-number success count could produce this exact interval. Justify your answer without needing to know the confidence level.

Hints

- Use the structural relationship between a confidence interval and its point estimate. - Recover the implied sample proportion from the two endpoints. - Check whether that proportion can come from a whole-number count with the stated sample size.

Solution

1. A standard one-proportion \(z\)-interval is centered at the sample proportion. 2. The claimed center is \(\frac{0.443+0.523}{2}=0.483\). 3. A sample proportion from \(600\) binary observations must equal \(\frac{x}{600}\) for a whole-number success count \(x\). 4. The implied count is \(600(0.483)=289.8\), which is not a whole number. 5. Therefore, no possible data set of \(600\) binary observations can have this exact interval as a standard one-proportion \(z\)-interval.

Answer

No. The interval’s center is \(0.483\), which would require \(289.8\) successes out of \(600\). Because a success count must be a whole number, the claimed exact interval is impossible.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Building free math worksheets