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- Identify which group was actually measured and which larger group is of interest. - A sample mean is used to estimate the corresponding population mean.

1. The value \(11\) minutes is a statistic calculated from the sampled riders. 2. It is used to estimate the mean wait time for all riders on that bus route.

It estimates the population mean wait time for all riders on the bus route.

- Form the sample's part-to-whole ratio first. - Apply that same proportion to the larger group. - Distinguish using a sample rate from observing the larger group directly.

1. The sample relative frequency is \(\frac{18}{60}=0.30=30\%\). 2. Applying the same rate to \(300\) visitors gives \(0.30\times300=90\). 3. The result is an estimate because a new group's observed relative frequency may differ from the sample rate.

Relative frequency: \(0.30=30\%\) Estimated users among \(300\): about \(90\)

- First use the sample proportion to make a population prediction. - Then distinguish a sample-based estimate from a count obtained by surveying every student. - Consider whether a second random sample of \(80\) students would have to contain exactly \(30\) students who bring lunch.

1. The sample proportion is \(\frac{30}{80}=0.375\). 2. Applying the sample proportion to the school gives \(0.375\cdot640=240\). 3. The random sample supports using \(240\) as a population estimate. 4. A different random sample could produce a different sample proportion, so the sample does not establish that the true population count is exactly \(240\).

a) About \(240\) students. b) “About \(240\) students” is justified. The random sample supports a population estimate, but sampling variation means the estimate is not an exact census count.

- Compute the center of the sampled wait times first. - Ask what random sampling lets you do with a sample statistic. - Decide whether one sample can force every other random sample to have the same mean.

1. The total wait time is \(78\) minutes, so the sample mean is \(78\div12=6.5\) minutes. 2. Because the customers were randomly sampled, the sample mean can be used as an informal estimate of the population mean. 3. Sampling variation means another random sample need not have the same mean, so the sample does not establish an exact population mean. 4. Therefore, conclusion II is better supported.

a) \(6.5\) minutes. b) II. The random sample supports using \(6.5\) minutes as an estimate, not as a guaranteed exact population mean.

- Put the sampled values in order. - Locate the middle observation rather than averaging all observations. - Use the sample result to make a cautious statement about all visits.

1. Order the values: \(27,28,29,30,31,32,33,34,35,36,38,39,41,42,45\). 2. With \(15\) values, the median is the \(8\)th value, \(34\). 3. A typical trail visit can be estimated at about \(34\) minutes.

Sample median: \(34\) minutes. A typical visit is about \(34\) minutes.

- For part a), compare each estimate with the known population proportion using absolute distance. - For part b), look at all three errors rather than only the smallest one. - Ask whether one unusually accurate random sample tells you how every future random sample must behave.

1. The errors are \(|0.44-0.48|=0.04\), \(|0.51-0.48|=0.03\), and \(|0.47-0.48|=0.01\). 2. The estimate \(0.47\) has the smallest absolute error. 3. The student’s conclusion is not supported. The other two random samples are \(0.04\) and \(0.03\) away from the true proportion. 4. Random samples can give different estimates by chance, so one especially accurate sample does not determine how accurate every random sample will be.

a) Errors: \(0.04,\ 0.03,\ 0.01\), respectively. The closest estimate is \(0.47\). b) No. The three random samples have different errors, so one sample being within \(0.01\) does not show that every random sample will be that accurate.

- Find the overdue proportion in the random sample. - Use that proportion with the total number of checked-out books. - State the result as an estimate.

1. The sample overdue rate is \(\frac{27}{180}=0.15\). 2. Apply the rate to the population: \(0.15\cdot3200=480\). 3. Estimate about \(480\) overdue books.

About \(480\) overdue books.

- Separate the numerical projection from the certainty of the wording. - A random sample supports an estimate of an unknown population value. - Use language that allows for sample-to-sample variation.

1. The sample proportion is \(\frac{12}{20}=0.60\). 2. Applying it to \(1000\) gives \(0.60\cdot1000=600\). 3. Sampling provides an estimate, not an exact population count. 4. The conclusion should say “about \(600\)” and mention that another random sample could differ.

A reasonable estimate is that about \(600\) of the \(1000\) households subscribe; the exact number is unknown.

- Use the complement to find the sampled count of interest. - Convert that count to a sample proportion. - Apply the proportion to the full school.

1. The number who have visited is \(125-95=30\). 2. The sample proportion is \(\frac{30}{125}=0.24=24\%\). 3. Estimate \(0.24\cdot1500=360\) students.

Estimated percentage: \(24\%\). Estimated number: about \(360\) students.

- Identify the category with the largest sample count. - Translate that sample result into cautious population language. - Do not claim certainty from one sample.

1. Space has the largest sample count, \(18\) of \(50\), or \(36\%\). 2. The sample suggests Space may be the exhibit where the greatest share of visitors spend the most time. 3. The result is not certain because another random sample could produce different category counts.

The sample suggests Space is the most common response, with \(36\%\) of sampled visitors. This is an estimate, not proof that Space is first for all visitors.

- Compare every sample estimate with the three proposed intervals. - Look for a cluster rather than focusing on one sample. - Use the center of the sample means as supporting evidence.

1. All six sample means lie from \(7.1\) to \(7.6\) hours. 2. Their mean is \(\frac{44.0}{6}\approx7.33\) hours. 3. The evidence strongly supports a population mean between \(7.0\) and \(7.7\) hours among the three choices.

Between \(7.0\) and \(7.7\) hours. The sample means cluster from \(7.1\) to \(7.6\), with an average of about \(7.33\) hours.

- Distinguish a sample statistic from the unknown population value. - Use the sample result as evidence, not as an exact identity. - State both the supported estimate and the limit of the claim.

1. The sample median \(8.4\) pounds is an estimate of the population median. 2. It supports saying that a typical middle value for all backpacks may be near \(8.4\) pounds. 3. It does not prove that the population median is exactly \(8.4\) pounds or that half of every possible sample lies on each side.

The population median is reasonably estimated to be near \(8.4\) pounds, but the sample does not prove the exact population median.

- Average the four percentage estimates before making the population prediction. - Compare the largest and smallest sample percentages. - Use the variation among the samples to decide whether the final population prediction can be exact.

1. The mean percentage is \(\frac{68+72+65+71}{4}=69\%\). 2. The range is \(72\%-65\%=7\) percentage points. 3. The nonzero range shows sample-to-sample variation: repeated random samples do not have to give the same estimate. 4. Applying the mean percentage gives \(0.69\cdot1800=1242\) students. 5. Because the percentage is estimated from samples that themselves vary, \(1242\) is a population prediction rather than an exact count.

a) \(69\%\). b) \(7\) percentage points; the repeated random samples show sampling variation. c) About \(1242\) students. The count is an estimate because different random samples produce different percentage estimates.

- Add all sampled household amounts. - Divide by the number of sampled households. - Include both the per-household and per-day units.

1. The sample total is \(2100\) gallons. 2. The sample mean is \(2100\div14=150\) gallons. 3. Estimate the population mean as about \(150\) gallons per household per day.

About \(150\) gallons per household per day.

- Locate the middle position in the ordered sample. - Use the sample median as evidence about the population median. - Keep sampling uncertainty in the wording.

1. With \(13\) ordered values, the median is the \(7\)th value. 2. The \(7\)th value is \(5\). 3. Estimate the population median rating as about \(5\). 4. Another random sample may have a different middle value, so the exact population median remains unknown.

Estimated population median: about \(5\). It is not guaranteed to be exact because it comes from one random sample.

- Use the random sample proportion as an estimate for the larger population. - Keep the conclusion approximate because a different random sample could give a different proportion. - Avoid turning the sample estimate into an exact population statement.

1. The sample was selected randomly, so its proportion can be used for an informal population inference. 2. The sample result is \(62\%\), so a defensible conclusion is that about \(62\%\) of all active cardholders would use longer Saturday hours. 3. The sample does not justify claiming that exactly \(62\%\), or exactly \(744\) people, in the full population would do so.

About \(62\%\) of the library's active cardholders would likely use longer Saturday hours.

- Focus on the difference in sample size, not on the population being sampled. - Think about which sample size should be less affected by a few unusual observations. - The question asks about usual stability across repeated random samples.

1. Both researchers use random samples from the same population. 2. Larger random samples usually have less sample-to-sample variation in their means than smaller random samples. 3. Therefore, Researcher B's sample means would usually vary less.

Researcher B. Samples of size \(80\) would usually produce more stable sample means than samples of size \(15\).

- Combine the information from all four random samples. - Use the lowest and highest sample estimates to describe variation. - Do not expect one random sample mean to equal the population mean exactly.

1. Average the four sample means: \(\frac{218+231+225+226}{4}=225\). 2. The sample means range from \(218\) to \(231\), a spread of \(13\) pages. 3. A reasonable estimate of the population mean is about \(225\) pages, with individual sample estimates varying by several pages.

Estimated population mean: about \(225\) pages. The four sample means span \(13\) pages, from \(218\) to \(231\).

- Compare how widely the estimates spread for each sample size. - A more stable method produces estimates clustered in a narrower interval. - Use the range as evidence.

1. The size-\(10\) sample proportions span \(0.80-0.20=0.60\). 2. The size-\(50\) sample proportions span \(0.62-0.42=0.20\). 3. The larger samples vary less, so size \(50\) gives more stable estimates.

Sample size \(50\). Its estimates span \(0.20\), compared with \(0.60\) for samples of size \(10\).

- Combine the five percentages to locate their center. - Use the smallest and largest estimates to describe the observed variation. - Do not treat the interval as a guarantee; it summarizes these samples.

1. The mean of the five estimates is \(\frac{46+51+49+54+50}{5}=50\%\). 2. The smallest estimate is \(46\%\), and the largest is \(54\%\). 3. A central estimate is \(50\%\), with observed sample estimates from \(46\%\) to \(54\%\).

Central estimate: \(50\%\). Observed sample-estimate interval: \(46\%\) to \(54\%\).

- Combine the sample medians to find a central value. - Compare the lowest and highest medians with that center. - Keep the central value at the precision produced by the calculation unless the problem requests rounding.

1. The mean of the six sample medians is \(\frac{13.8+14.2+14.0+13.9+14.3+14.1}{6}=14.05\) ounces. 2. The sample medians range from \(13.8\) to \(14.3\) ounces. 3. They are at most \(0.25\) ounce from \(14.05\). 4. Use \(14.05\) ounces as a central sample-based estimate of the population median.

Estimated population median: about \(14.05\) ounces. The sample medians range from \(13.8\) to \(14.3\) ounces and are within \(0.25\) ounce of \(14.05\).

- Compare the three sample percentages before combining them. - Pool favorable counts and total sample counts separately. - For part c), think about what is lost when someone chooses a sample only because its result is largest.

1. The three sample percentages are \(\frac{54}{90}=60\%\), \(\frac{57}{90}\approx63.3\%\), and \(\frac{51}{90}\approx56.7\%\). 2. The pooled samples contain \(54+57+51=162\) favorable responses out of \(270\), so the pooled proportion is \(\frac{162}{270}=0.60\). 3. The population prediction is \(0.60\cdot1350=810\) students. 4. The three samples show ordinary sample-to-sample variation. Choosing only the largest result selectively favors one random fluctuation, while pooling all \(270\) observations uses all of the random-sample evidence.

a) \(60\%\), about \(63.3\%\), and about \(56.7\%\). b) About \(810\) students. c) Pooling all three samples uses all \(270\) randomly sampled students instead of selecting the sample with the most favorable random fluctuation.

- Combine counts before relying on an average of percentages. - Check the support fraction in each sample separately. - Notice when unequal sample sizes do and do not affect the result.

1. The combined support count is \(36+72=108\), and the combined sample size is \(60+120=180\). 2. The combined proportion is \(\frac{108}{180}=0.60=60\%\). 3. Each sample separately has \(60\%\) support, so averaging the two percentages also gives \(60\%\), even though the sample sizes differ.

Combined estimate: \(60\%\). Averaging works because both samples have exactly the same sample percentage.

- Combine evidence across all polls. - Also inspect whether every poll points in the same direction. - A prediction can be supported without being certain.

1. Across the three polls, Riley receives \(54+49+56=159\) of \(300\) votes, or \(53\%\). 2. Morgan receives \(141\) votes, or \(47\%\). 3. Riley leads overall, but one poll shows Morgan ahead. 4. Predict Riley as the more likely winner, while noting that sample variation makes the outcome uncertain.

Riley is the better-supported prediction, with \(159\) of \(300\) sampled votes (\(53\%\)). The prediction should be cautious because one random poll favored Morgan.

- Average the sample estimates to locate a center. - Compare each extreme estimate with that center. - Keep the computed values unless a rounding precision is specified.

1. The mean of the sample means is \(\frac{32.2}{8}=4.025\). 2. The farthest estimates are \(3.7\) and \(4.4\). 3. Their distances from \(4.025\) are \(0.325\) and \(0.375\). 4. The central estimate is \(4.025\) trips per week, and the largest observed distance is \(0.375\) trip.

Central estimate: \(4.025\) trips per week. Largest observed distance from the center: \(0.375\) trip.

- Compare how the unusually large value affects the two measures of center. - For part b), choose the statistic that better reflects the main cluster of delivery times. - For part c), separate representativeness of a random sample from certainty about the exact population value.

1. The total is \(241\), so the mean is \(\frac{241}{9}\approx26.8\) minutes. 2. In order, the times are \(17,18,19,19,20,20,21,22,85\), so the median is \(20\) minutes. 3. The \(85\)-minute value pulls the mean well above the main cluster, so the median better represents a typical delivery in this sample. 4. Random sampling supports using the sample median as an informal population estimate, but another random sample could have a different median.

a) Mean: \(\frac{241}{9}\approx26.8\) minutes; median: \(20\) minutes. b) The median is more appropriate because the \(85\)-minute outlier pulls the mean upward. c) It is still a sample-based estimate; another random sample could produce a different median.

- Average the sample percentages for a central rate. - Apply both the central rate and the extreme observed rates to the population. - Treat each projected count as sample-based; no extra rounding of the central projection is requested.

1. The average rate is \(\frac{58+64+60+62}{4}=61\%\). 2. Applying the average rate gives \(0.61\cdot5000=3050\), so the central sample-based estimate is about \(3050\) residents. 3. The lowest sample rate gives \(0.58\cdot5000=2900\). 4. The highest sample rate gives \(0.64\cdot5000=3200\).

Central estimate: about \(3050\) residents. Counts based on the observed sample rates range from about \(2900\) to \(3200\).

- Compare sample sizes separately from the percentages obtained. - Think about which sample size is more affected by a few changed outcomes. - More stable does not mean guaranteed to be correct.

1. Sample A uses \(25\) observations. 2. Sample B uses \(200\) observations. 3. Larger random samples generally show less sample-to-sample variation. 4. The \(62\%\) estimate from Sample B would usually be expected to be more stable, even though stability does not guarantee it is closer in this one case.

The \(62\%\) estimate from Sample B, because it is based on the larger random sample of \(200\).

- A sample of \(100\) makes each count easy to interpret as a percent. - Apply each category’s share to the full population. - Round each projection to the requested place and check the combined total.

1. Because the sample size is \(100\), the counts are the percentage values. 2. Drive: \(0.46\cdot8500=3910\approx3900\). 3. Transit: \(0.28\cdot8500=2380\approx2400\). 4. Walk or bike: \(0.19\cdot8500=1615\approx1600\). 5. Other: \(0.07\cdot8500=595\approx600\). 6. The rounded estimates total \(8500\).

Drive: about \(3900\) Transit: about \(2400\) Walk or bike: about \(1600\) Other: about \(600\)

- Find the center of the ten sample estimates. - Use both extremes to describe how much the estimates change. - Relate the extremes back to the chosen center.

1. The average of the ten estimates is \(\frac{710}{10}=71\). 2. The estimates range from \(68\) to \(74\), a spread of \(6\). 3. A reasonable population-mean estimate is \(71\), with sample estimates differing by up to \(3\) from that center.

Estimated population mean: \(71\). The sample estimates range from \(68\) to \(74\), or within \(3\) of \(71\).

- Average the randomly selected Saturdays first. - Use the estimated one-Saturday mean to make the season prediction. - For part c), ask whether the unsampled Saturdays are known or merely being estimated.

1. The sample total is \(3292\), so the sample mean is \(3292\div8=411.5\) customers per Saturday. 2. Using that sample mean for a \(24\)-Saturday season gives \(411.5\cdot24=9876\) customers. 3. The sampled Saturdays were random, so the calculation gives a reasonable sample-based prediction for the season. 4. It is not an exact count: unsampled Saturdays can differ, and another random sample can produce a different estimated mean.

a) About \(411.5\) customers per Saturday. b) About \(9876\) customers for the season. c) The “exactly” claim is not justified. The total is a prediction based on a random sample, not a census of all \(24\) Saturdays.

- Compute the width of each interval. - Compare those widths in sample-size order. - Describe the pattern without claiming that every large sample is perfect.

1. The ranges are \(0.40\), \(0.14\), and \(0.05\), respectively. 2. As sample size increases from \(20\) to \(400\), the estimates become more tightly clustered. 3. Larger random samples show less sample-to-sample variation in this simulation.

Larger samples vary less: the estimate ranges shrink from \(0.40\) to \(0.14\) to \(0.05\).

- Compute the first fraction separately. - For the combined estimate, add counts and sample sizes rather than averaging fractions without weights. - Compare how many observations support each estimate.

1. The first-sample estimate is \(\frac{14}{25}=0.56=56\%\). 2. Combining all samples gives \(\frac{14+31+58}{25+50+100}=\frac{103}{175}\approx0.589=58.9\%\). 3. The combined estimate uses all \(175\) observations and therefore uses more information.

First sample: \(56\%\). All samples combined: \(\frac{103}{175}\approx58.9\%\). The combined estimate uses more information.

- Convert each sample count to a percentage before comparing the two samples. - Notice that two random samples from the same population can give different percentages. - Distinguish an estimate based on samples from an exact fact about the entire population.

1. The first sample percentage is \(\frac{186}{200}=0.93=93\%\). 2. The second sample percentage is \(\frac{194}{200}=0.97=97\%\). 3. Because the samples have equal sizes, their combined sample percentage is also \(95\%\). 4. A sample percentage is an estimate of a population percentage, not proof of its exact value. The two random samples also differ from each other, showing sample-to-sample variation. 5. The samples suggest that the population percentage is around the mid-90s, but they do not determine whether the true percentage is exactly \(95\%\) or whether it is above \(95\%\).

a) First sample: \(93\%\). Second sample: \(97\%\). b) The student treats a sample-based estimate as an exact population value. The two samples show sampling variation and together give a \(95\%\) sample estimate, but they do not determine the exact population percentage or settle whether it is greater than \(95\%\).

- Apply the central percentage and both endpoint percentages to the same population. - Percentage points change the rate before converting to a count. - Round each sample-based projection to the requested place.

1. Central projection: \(0.52\cdot4800=2496\approx2500\). 2. Lower projection: \(0.48\cdot4800=2304\approx2300\). 3. Upper projection: \(0.56\cdot4800=2688\approx2700\). 4. The rounded range reflects the observed \(\pm4\)-percentage-point sample variation.

Central estimate: about \(2500\) residents. Counts based on the observed variation: about \(2300\) to \(2700\) residents.

- Apply each percentage to the same population. - Keep the central estimate separate from the observed variation endpoints. - Round each count to the requested place because the percentages are sample-based.

1. Central projection: \(0.37\cdot6200=2294\approx2300\). 2. Lower endpoint: \(0.34\cdot6200=2108\approx2100\). 3. Upper endpoint: \(0.41\cdot6200=2542\approx2500\). 4. The observed sample estimates correspond to about \(2100\) to \(2500\) residents.

Central estimate: about \(2300\) residents. Observed endpoint counts: about \(2100\) to \(2500\) residents.

- Compare the widths of the three estimate sets. - Precision is indicated by tight clustering, not merely one convenient value. - Also check where the tight cluster is centered.

1. Set A has range \(0.76-0.21=0.55\). 2. Set B has range \(0.51-0.46=0.05\). 3. Set C has range \(0.68-0.32=0.36\). 4. Set B is tightly clustered around \(0.49\), so it gives the most precise evidence.

Set B. Its estimates range only from \(0.46\) to \(0.51\), a spread of \(0.05\), and cluster near \(0.49\).

- Look beyond the percentages to the number of observations behind each bar. - Convert each percentage to a count to see the scale of the samples. - Decide whether the caption claims more than the displayed information supports.

1. Event A's \(90\%\) represents \(0.90\times10=9\) satisfied respondents. 2. Event B's \(80\%\) represents \(0.80\times500=400\) satisfied respondents. 3. The percentages describe the observed shares, but the graph hides the very different sample sizes. 4. The bar heights alone do not justify the claim about stronger evidence. 5. The display should state the sample size and satisfied count for each event alongside the percentages.

The caption is not supported by the graph. Event A is based on only \(10\) responses, while Event B is based on \(500\). Add both sample sizes and counts to the display.

- Compute both centers using all five sample estimates. - Ask which center moves more because one estimate is much higher than the others. - Use the less affected center for the requested cautious typical estimate, while still acknowledging the unusual sample.

1. The mean is \(\frac{31+34+33+32+49}{5}=35.8\%\). 2. In order, the estimates are \(31\%,32\%,33\%,34\%,49\%\), so the median is \(33\%\). 3. The unusually high \(49\%\) estimate pulls the mean upward but does not change the median as strongly. 4. A cautious typical estimate from these samples is therefore about \(33\%\), while the unusual sample also signals meaningful sampling variation.

Mean: \(35.8\%\). Median: \(33\%\). The \(49\%\) result pulls the mean upward. A cautious typical estimate is about \(33\%\), with the variation among samples noted.

- For the pooled estimate, combine people rather than treating the two percentages as equally weighted observations. - Compare the two sample sizes before averaging percentages. - For part c), ask whether every library visitor had a comparable chance to enter Sample B.

1. The pooled count is \(18+44=62\) study-room users out of \(40+80=120\) sampled visitors. 2. The pooled estimate is \(\frac{62}{120}=\frac{31}{60}\approx51.7\%\). 3. The simple average of \(45\%\) and \(55\%\) is \(50\%\), but that gives the \(40\)-person sample the same weight as the \(80\)-person sample. 4. If Sample B came only from the study-room hallway, it would overrepresent visitors likely to use study rooms. Pooling a biased convenience sample with a random sample would not create a valid random-sample estimate for all visitors.

a) \(\frac{31}{60}\approx51.7\%\). b) A simple average incorrectly gives the unequal sample sizes equal weight. c) No. A hallway convenience sample is biased toward study-room users, and pooling it with a random sample does not remove that bias.

- Look at the direction of every change, not only the total difference. - Random variation does not usually have to move in one direction. - Use cautious language because each monthly value is still a sample estimate.

1. The estimates rise in every month from \(47\%\) to \(62\%\). 2. Random variation can move estimates up or down, but five consecutive increases form a consistent trend. 3. The data suggest that mobile-ticket use may actually be increasing, although more information is needed to separate trend from sampling variation.

The steady rise suggests a real increase may be occurring, not just random sample variation. The evidence is suggestive rather than conclusive.

- Use the sample to estimate what fraction of the population is tagged. - Relate the known tagged count to that estimated fraction. - Identify what must be true about mixing and selection for the inference to be reasonable.

1. The sample proportion tagged is \(\frac{8}{50}=0.16\). 2. Treat the same proportion as an estimate for the whole pond: \(\frac{40}{N}\approx0.16\), where \(N\) is the total population. 3. Solving gives \(N\approx40\div0.16=250\). 4. The estimate assumes the tagged turtles mixed thoroughly and were as likely to be sampled as untagged turtles.

The estimated pond population is about \(250\) turtles. The estimate assumes tagged and untagged turtles were equally likely to appear in the random sample.