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Draw inferences from samples

Learning goals

  • Uses sample statistics to estimate population characteristics informally.
  • Accounts for sample-to-sample variability.
  • Compares sample evidence with population claims.
  • States conclusions with appropriate uncertainty.

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Draw inferences from samples 1
A random sample of \(40\) riders on a bus route has a mean wait time of \(11\) minutes. What population quantity is the sample mean of \(11\) minutes intended to estimate?

Hints

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

Solution

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.

Answer

It estimates the population mean wait time for all riders on the bus route.
Draw inferences from samples 2
A school randomly checks \(10\) of its \(100\) desks. It finds \(3\) desks with loose screws. About how many desks in the school might have loose screws?

Hints

- What fraction of the checked desks had loose screws? - How many groups of \(10\) desks are in the school?

Solution

1. In the sample, \(3\) of \(10\) desks have loose screws. 2. The school has \(10\) times as many desks as were checked. 3. Estimate \(3\cdot10=30\) desks with loose screws in the school.

Answer

About \(30\) desks.
Draw inferences from samples 3
On \(4\) randomly chosen days, students wait \(8\), \(10\), \(10\), and \(12\) minutes for the school bus. Use these times to estimate the average wait for the school bus.

Hints

- What summary gives the average of the four recorded times? - How could the average of a random sample help estimate the average of all trips?

Solution

1. The sample mean is \(\frac{8+10+10+12}{4}=10\) minutes. 2. Use the sample mean as an estimate of the average wait on school days.

Answer

About \(10\,\text{minutes}\) on average.
Draw inferences from samples 4
In a random sample of \(20\) students, \(8\) say they would join an after-school art club. Does this prove that exactly \(40\%\) of all students would join? Explain.

Hints

- Was every student asked? - Can another random sample give a different percentage?

Solution

1. In the sample, \(8\div20=40\%\) would join. 2. The percentage in the entire school may differ from a percentage measured in one random sample.

Answer

No. \(40\%\) is the sample percentage, which is an estimate rather than an exact known percentage for the whole school.
Draw inferences from samples 5
A school draws two different random groups of \(10\) students. In one group, \(3\) students walk to school. In the other, \(5\) students walk. Can both groups have been selected fairly? Explain.

Hints

- Does picking students at random guarantee the same mix every time? - What would make a selection unfair, apart from getting a different count?

Solution

1. Each group is a random sample of the school. 2. Random samples do not always contain the same number of students with a given habit. 3. Different results alone do not show that either selection was unfair.

Answer

Yes. Two fair random samples can give different results just by chance.
Draw inferences from samples 6
A school newspaper claims that most students support a longer lunch break. In a random sample of \(30\) students, \(12\) support it. Does this sample support the newspaper's claim? Explain.

Hints

- What share of a group counts as 'most'? - Compare the sample supporters with half the sample.

Solution

1. Most would mean more than half the students. 2. The sample has \(12\) supporters out of \(30\), fewer than half. 3. The sample does not support the claim, though it does not establish the exact schoolwide percentage.

Answer

No. Only \(12\) of the \(30\) sampled students support the idea, which is less than half. The sample does not support a majority claim.
Draw inferences from samples 7
Ellis asks \(5\) friends whether they like the school's new breakfast menu. All \(5\) say yes. Ellis concludes, “Every student in the school likes the menu.” What is wrong with that conclusion?

Hints

- Who was asked, and who was not? - Does agreement among a few students show what all students think?

Solution

1. Ellis asked only a small group of friends, not a random sample of the school. 2. Even unanimous answers from those friends do not show what every student thinks.

Answer

The five friends do not represent every student. Their answers cannot prove that everyone in the school likes the menu.
Draw inferences from samples 8
In a sample of \(60\) library visitors, \(18\) used the self-checkout station. Find the relative frequency of self-checkout use. Assuming the sample rate is representative, estimate how many of a similar group of \(300\) visitors would use self-checkout. Explain why the second result is an estimate.

Hints

- 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.

Solution

1. The sample relative frequency is \(\frac{18}{60}=0.30=30\%\). 2. Applying the same rate to \(300\) visitors gives \(0.30\cdot300=90\). 3. The result is an estimate because a new group's observed relative frequency may differ from the sample rate.

Answer

Relative frequency: \(0.30=30\%\). Estimated users among \(300\): about \(90\).
Draw inferences from samples 9
A random sample of \(80\) students from a school of \(640\) students finds that \(30\) usually bring lunch from home. a) Use the sample to predict how many students in the school usually bring lunch from home. b) One student writes, “Exactly \(240\) students bring lunch from home.” Another writes, “About \(240\) students bring lunch from home.” Which statement is justified by the sample? Explain.

Hints

- 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.

Solution

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 \(240\).

Answer

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.
Draw inferences from samples 10
A random sample of \(12\) customers at a service desk had wait times, in minutes, \(4,\ 7,\ 5,\ 9,\ 6,\ 8,\ 5,\ 10,\ 3,\ 7,\ 6,\ 8\). a) Find the sample mean. b) Which conclusion is better supported? I. The mean wait time for all customers that day was exactly the sample mean. II. The sample mean is a reasonable estimate of the mean wait time for all customers that day, but another random sample could give a different estimate. Explain your choice.

Hints

- 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.

Solution

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.

Answer

a) \(6.5\) minutes. b) II. The random sample supports using \(6.5\) minutes as an estimate, not as a guaranteed exact population mean.
Draw inferences from samples 11
A random sample of \(15\) hiking-trail visits gives these durations, in minutes: \(28,\ 35,\ 31,\ 42,\ 33,\ 29,\ 38,\ 34,\ 30,\ 45,\ 32,\ 36,\ 41,\ 27,\ 39\). Find the sample median and use it to make an informal statement about a typical visit.

Hints

- 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.

Solution

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.

Answer

Sample median: \(34\) minutes. A typical visit is about \(34\) minutes.
Draw inferences from samples 12
The true proportion of blue beads in a large container is later found to be \(0.48\). Three earlier random samples produced estimates \(0.44,\ 0.51,\) and \(0.47\). a) Find the absolute error of each estimate and identify the closest estimate. b) Milica says, “Because one random sample was only \(0.01\) away from the true proportion, random samples from this container should always be within \(0.01\) of the true proportion.” Is that conclusion supported by these three samples? Explain.

Hints

- 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.

Solution

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. Milica'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.

Answer

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.
Draw inferences from samples 13
A library randomly checks \(180\) checked-out books and finds \(27\) overdue. There are \(3200\) checked-out books in all. Estimate the number that are overdue.

Hints

- 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.

Solution

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.

Answer

About \(480\) overdue books.
Draw inferences from samples 14
In a random sample of \(20\) households, \(12\) subscribe to a streaming service. Bram concludes, “Exactly \(600\) of the \(1000\) households subscribe.” Rewrite the conclusion so it matches what a sample can support.

Hints

- 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.

Solution

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.

Answer

A reasonable estimate is that about \(600\) of the \(1000\) households subscribe; the exact number is unknown.
Draw inferences from samples 15
A random sample of \(125\) students finds that \(95\) have not visited the school nurse this semester. Estimate both the percentage who have visited and the number who have visited among \(1500\) students.

Hints

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

Solution

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.

Answer

Estimated percentage: \(24\%\). Estimated number: about \(360\) students.
Draw inferences from samples 16
A random sample of \(50\) visitors names the exhibit where they spent the most time: Space: \(18\) Ocean: \(14\) Dinosaurs: \(11\) Weather: \(7\) What informal inference can be made about the most common response among all visitors? Include an appropriate limitation.

Hints

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

Solution

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.

Answer

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.
Draw inferences from samples 17
Six random samples estimate the average amount of sleep for students, in hours: \(7.1,\ 7.4,\ 7.3,\ 7.6,\ 7.2,\ 7.4\). Is the population mean more likely to be below \(6.5\) hours, between \(7.0\) and \(7.7\) hours, or above \(8.0\) hours? Justify using the samples.

Hints

- 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.

Solution

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.

Answer

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.
Draw inferences from samples 18
A random sample of \(21\) backpack masses, in pounds, has a median of \(8.4\). What does this sample support saying about the median backpack mass for all students? What does it not prove?

Hints

- Who was included in the backpack measurements? - Would a different random sample necessarily have the same middle value? - What can the measured backpacks tell you about backpacks that were not measured?

Solution

1. The sample median is \(8.4\) pounds. 2. It is reasonable to estimate that the population median backpack mass is near \(8.4\) pounds. 3. The sample does not prove that the population median is exactly \(8.4\) pounds; a different random sample might have a different median.

Answer

The population median is reasonably estimated to be near \(8.4\) pounds, but the sample does not prove the exact population median.
Draw inferences from samples 19
A random sample of \(14\) households has daily water use, in gallons, \(142,\ 155,\ 131,\ 168,\ 149,\ 157,\ 138,\ 161,\ 146,\ 152,\ 135,\ 159,\ 144,\ 163\). Estimate the mean daily use for all households.

Hints

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

Solution

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.

Answer

About \(150\) gallons per household per day.
Draw inferences from samples 20
A random sample of \(13\) customer ratings is \(2,\ 3,\ 4,\ 4,\ 4,\ 5,\ 5,\ 5,\ 5,\ 5,\ 5,\ 5,\ 5\). Find the sample median and use it to estimate the median rating for all customers. Explain why the estimate should not be stated as exact.

Hints

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

Solution

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.

Answer

Estimated population median: about \(5\). It is not guaranteed to be exact because it comes from one random sample.
Draw inferences from samples 21
A library randomly samples \(100\) of its \(1200\) active cardholders. In the sample, \(62\%\) say they would use longer Saturday hours. Write a defensible conclusion about all active cardholders. Do not state an exact population count.

Hints

- 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.

Solution

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.

Answer

About \(62\%\) of the library's active cardholders would likely use longer Saturday hours.
Draw inferences from samples 22
Two researchers repeatedly take random samples from the same large population. Researcher A uses samples of size \(15\). Researcher B uses samples of size \(80\). Whose sample means would usually be expected to vary less from sample to sample? Explain.

Hints

- 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.

Solution

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.

Answer

Researcher B. Samples of size \(80\) would usually produce more stable sample means than samples of size \(15\).
Draw inferences from samples 23
Two random samples of \(20\) students are taken from the same school. In the first sample, \(8\) eat school lunch. In the second, \(12\) eat school lunch. Find both sample percentages. Does their difference mean a mistake was made?

Hints

- Use each sample's own total when finding its percentage. - Consider whether two fair samples must contain the same mix of students.

Solution

1. The first sample percentage is \(\frac{8}{20}=40\%\). 2. The second sample percentage is \(\frac{12}{20}=60\%\). 3. Random samples can differ by chance, so the unequal percentages do not by themselves show an error.

Answer

The percentages are \(40\%\) and \(60\%\). No: different random samples can give different results.
Draw inferences from samples 24
Three random groups of \(10\) students each are asked whether they use the library each week. No student is in more than one group. The numbers answering yes are \(4\), \(5\), and \(6\). What fraction of all \(30\) sampled students answered yes? Why might the three groups differ?

Hints

- Add the yes counts before finding the overall fraction. - Should the denominator be one group's size or the size of all three groups? - Think about why distinct sets of students might answer differently.

Solution

1. There are \(4+5+6=15\) yes answers among \(30\) students. 2. The combined sample fraction is \(\frac{15}{30}=\frac{1}{2}\). 3. Randomly selected groups may contain different students with different habits.

Answer

\(\frac{1}{2}\) of the combined sample answered yes. The groups can differ by chance even when each is chosen fairly.
Draw inferences from samples 25
A middle school has \(600\) students. In a random sample of \(50\) students, \(10\) bring a reusable water bottle. A principal says, “About \(300\) students bring one.” Estimate how many students bring a bottle, using the sample. Does the estimate support the principal's claim?

Hints

- What fraction of the random sample brought a bottle? - Apply that fraction to the number of students in the school. - Compare your estimate with the principal's number without treating the estimate as exact.

Solution

1. The sample fraction is \(\frac{10}{50}=\frac{1}{5}\). 2. Estimate \(\frac{1}{5}\cdot600=120\) students schoolwide. 3. The estimate is much lower than \(300\), so this sample does not support the claim, though it is not a census.

Answer

About \(120\) students. This does not support the claim of about \(300\) students.
Draw inferences from samples 26
At a school movie night, \(10\) people are randomly surveyed. At the town fair, \(500\) people are randomly surveyed. The bar graph shows the share of satisfied visitors. How many people in each sample were satisfied? A reporter says, “More of the surveyed people were satisfied at movie night because its bar is taller. The bars also show exactly how everyone at each event felt.” Explain both mistakes.
Figure for problem 541309

Hints

- Read the percentage for each event from its bar. - How can a percentage and a sample size give a number of people? - Check whether the percentage comparison and the count comparison agree. - Which visitors answered the questions, and which visitors does the reporter make claims about?

Solution

1. The graph shows \(90\%\) at movie night and \(80\%\) at the fair. 2. At movie night, \(0.90\cdot10=9\) people were satisfied. At the fair, \(0.80\cdot500=400\) people were satisfied. 3. Although movie night has the higher percentage, more surveyed people were satisfied at the fair: \(400\) compared with \(9\). The two samples have different sizes. 4. The percentages describe the surveyed visitors. They may help estimate satisfaction among all visitors, but they are not known to be the exact percentages for everyone who attended.

Answer

Movie night: \(9\) satisfied people surveyed. Town fair: \(400\). The movie-night bar is taller, but more surveyed people were satisfied at the fair. The percentages are sample results, not exact rates for all attendees.
Draw inferences from samples 27
Four random samples of \(25\) library books give these mean numbers of pages: \(218,\ 231,\ 225,\ 226\). Use the samples to estimate the mean number of pages for all books in the library, and describe the sample-to-sample variation.

Hints

- 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.

Solution

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.

Answer

Estimated population mean: about \(225\) pages. The four sample means span \(13\) pages, from \(218\) to \(231\).
Draw inferences from samples 28
A computer takes \(20\) random samples of size \(10\) and \(20\) random samples of size \(50\) from the same population. The sample proportions have these ranges: Size \(10\): \(0.20\) to \(0.80\) Size \(50\): \(0.42\) to \(0.62\) Which sample size gives more stable estimates, and what evidence supports the conclusion?

Hints

- 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.

Solution

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.

Answer

Sample size \(50\). Its estimates span \(0.20\), compared with \(0.60\) for samples of size \(10\).
Draw inferences from samples 29
Five random samples estimate the percentage of households that compost: \(46\%,\ 51\%,\ 49\%,\ 54\%,\ 50\%\). Give a central estimate and a simple interval that describes how much the sample estimates varied.

Hints

- 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.

Solution

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 \(54\%\).

Answer

Central estimate: \(50\%\). Observed sample-estimate interval: \(46\%\) to \(54\%\).
Draw inferences from samples 30
Six random samples of the same size produce median package masses, in ounces: \(13.8,\ 14.2,\ 14.0,\ 13.9,\ 14.3,\ 14.1\). Estimate a typical population median and describe how far the sample medians are from that estimate.

Hints

- 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.

Solution

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.

Answer

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 \(14.05\).
Draw inferences from samples 31
A school has \(1350\) students. Three nonoverlapping random samples of \(90\) students each find that \(54\), \(57\), and \(51\) students prefer a later start time. a) Find the preference percentage in each sample. Round to the nearest tenth of a percent when needed. b) Pool all three samples and estimate how many students in the school prefer a later start. c) Kojo wants to use only the second sample because it gives the largest estimate. Explain why using all three random samples is more defensible.

Hints

- 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.

Solution

1. The three sample percentages are \(\frac{54}{90}=60\%\), \(\frac{57}{90}\approx63.3\%\), and \(\frac{51}{90}\approx56.7\%\). 2. The combined samples contain \(54+57+51=162\) favorable responses out of \(270\) distinct students, so the pooled proportion is \(\frac{162}{270}=0.60\). 3. Estimate \(0.60\cdot1350=810\) students schoolwide. 4. Choosing only the largest sample result favors one random fluctuation. Pooling all three uses more evidence.

Answer

a) \(60\%\), about \(63.3\%\), and about \(56.7\%\). b) About \(810\) students. c) Pooling all three samples uses the results from all \(270\) randomly sampled students instead of choosing the largest estimate.
Draw inferences from samples 32
Two random samples from the same county ask whether residents support a recycling program. Sample A: \(36\) of \(60\) support it. Sample B: \(72\) of \(120\) support it. Find the combined estimate of the support percentage. Explain why simply averaging the two percentages happens to work here.

Hints

- 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.

Solution

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.

Answer

Combined estimate: \(60\%\). Averaging works because both samples have exactly the same sample percentage.
Draw inferences from samples 33
Three random election polls of \(100\) students each report these votes: <table><tr><th>Poll</th><th>Riley</th><th>Morgan</th></tr><tr><td>1</td><td>\(54\)</td><td>\(46\)</td></tr><tr><td>2</td><td>\(49\)</td><td>\(51\)</td></tr><tr><td>3</td><td>\(56\)</td><td>\(44\)</td></tr></table> What prediction is best supported, and why should it be stated cautiously?

Hints

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

Solution

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.

Answer

Riley is the better-supported prediction, with \(159\) of \(300\) sampled votes (\(53\%\)). The prediction should be cautious because one random poll favored Morgan.
Draw inferences from samples 34
Eight random samples estimate the mean number of weekly transit trips per resident: \(3.8,\ 4.1,\ 4.0,\ 4.4,\ 3.9,\ 4.2,\ 4.1,\ 3.7\). Use the samples to give a central estimate and describe the largest observed distance from that center.

Hints

- 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.

Solution

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.

Answer

Central estimate: \(4.025\) trips per week. Largest observed distance from the center: \(0.375\) trip.
Draw inferences from samples 35
A random sample of delivery times, in minutes, is \(18,\ 20,\ 19,\ 21,\ 22,\ 17,\ 20,\ 19,\ 85\). a) Find the mean and median. b) A manager wants one statistic from this random sample to estimate a typical delivery time for all deliveries. Which statistic is more appropriate here, and why? c) Why should the manager still treat that statistic as an estimate rather than an exact population value?

Hints

- 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.

Solution

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.

Answer

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.
Draw inferences from samples 36
Four random samples of \(50\) residents give estimated support rates of \(58\%,\ 64\%,\ 60\%,\) and \(62\%\). The town has \(5000\) residents. Use the average sample rate to estimate support, and use the lowest and highest sample rates to show a range of plausible sample-based counts.

Hints

- 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.

Solution

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 \(0.64\cdot5000=3200\).

Answer

Central estimate: about \(3050\) residents. Counts based on the observed sample rates range from about \(2900\) to \(3200\).
Draw inferences from samples 37
A population proportion is estimated from two independent random samples. Sample A: \(17\) successes out of \(25\) Sample B: \(124\) successes out of \(200\) The two estimates are \(68\%\) and \(62\%\). Which estimate would usually be expected to vary less from sample to sample, and why?

Hints

- 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.

Solution

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.

Answer

The \(62\%\) estimate from Sample B, because it is based on the larger random sample of \(200\).
Draw inferences from samples 38
A random sample of \(100\) commuters gives these main travel methods: <table><tr><th>Method</th><th>Sample count</th></tr><tr><td>Drive</td><td>\(46\)</td></tr><tr><td>Transit</td><td>\(28\)</td></tr><tr><td>Walk or bike</td><td>\(19\)</td></tr><tr><td>Other</td><td>\(7\)</td></tr></table> Estimate the number in each category among \(8500\) commuters. Round each estimate to the nearest \(100\) commuters.

Hints

- 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.

Solution

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 \(8500\).

Answer

Drive: about \(3900\) Transit: about \(2400\) Walk or bike: about \(1600\) Other: about \(600\)
Draw inferences from samples 39
Ten simulated random samples estimate a population mean: \(72,\ 69,\ 71,\ 74,\ 70,\ 68,\ 73,\ 71,\ 70,\ 72\). Give a reasonable estimate of the population mean and describe the variation among the sample estimates.

Hints

- 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.

Solution

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.

Answer

Estimated population mean: \(71\). The sample estimates range from \(68\) to \(74\), or within \(3\) of \(71\).
Draw inferences from samples 40
A farmers market records customer counts on \(8\) randomly selected Saturdays: \(412,\ 385,\ 440,\ 398,\ 421,\ 407,\ 396,\ 433\). a) Use the sample to estimate the mean Saturday attendance. b) Use that estimate to predict the total attendance over a \(24\)-Saturday season. c) A report states, “The season attendance will be exactly the value from part b).” Evaluate that statement.

Hints

- 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.

Solution

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.

Answer

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.
Draw inferences from samples 41
Repeated random samples estimate the fraction of packages delivered on time. Samples of size \(20\) range from \(0.55\) to \(0.95\). Samples of size \(100\) range from \(0.70\) to \(0.84\). Samples of size \(400\) range from \(0.75\) to \(0.80\). Describe the relationship between sample size and variation in these estimates.

Hints

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

Solution

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.

Answer

Larger samples vary less: the estimate ranges shrink from \(0.40\) to \(0.14\) to \(0.05\).
Draw inferences from samples 42
Four independent random samples of \(75\) students give these estimates for the percentage who attend at least one school event each month: \(68\%,\ 72\%,\ 65\%,\ 71\%\). a) Find the mean of the four sample percentages. b) Find the range of the four sample percentages. What does this range show about repeated random samples? c) Use the mean percentage to predict the number among \(1800\) students. Explain why the prediction should not be treated as an exact count.

Hints

- 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.

Solution

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.

Answer

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.
Draw inferences from samples 43
Thandi uses only the first of these three random samples to estimate a population proportion: \(\frac{14}{25},\ \frac{31}{50},\ \frac{58}{100}\). Find the estimate from the first sample and the estimate from all \(175\) observations. Which uses more information?

Hints

- Consider what the numerator and denominator of each sample fraction represent. - When using all samples, build one proportion from all favorable observations and all sampled observations. - Compare how many observations support each estimate.

Solution

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.

Answer

First sample: \(56\%\). All samples combined: \(\frac{103}{175}\approx58.9\%\). The combined estimate uses more information.
Draw inferences from samples 44
Two independent random samples of \(200\) light bulbs are tested. In the first sample, \(186\) bulbs last at least \(800\) hours. In the second sample, \(194\) bulbs last at least \(800\) hours. BrightPeak Lighting claims that “more than \(95\%\) of all bulbs last at least \(800\) hours.” a) Find the percentage in each sample. b) Eszter averages the two sample percentages and says, “That gives exactly \(95\%\), so the true population percentage must be exactly \(95\%\) and the company’s claim is false.” Explain what is wrong with this conclusion and state what the two samples do support.

Hints

- 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.

Solution

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 \(95\%\).

Answer

a) First sample: \(93\%\). Second sample: \(97\%\). b) Eszter 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 \(95\%\).
Draw inferences from samples 45
A random survey estimates that \(52\%\) of \(4800\) residents support a trail project. Repeated samples of the same size have differed from the central estimate by as much as \(4\) percentage points. Give the central count estimate and the counts corresponding to \(48\%\) and \(56\%\). Round the counts to the nearest \(100\) residents.

Hints

- 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.

Solution

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.

Answer

Central estimate: about \(2500\) residents. Counts based on the observed variation: about \(2300\) to \(2700\) residents.
Draw inferences from samples 46
A random sample estimates that \(37\%\) of residents visit a community center each month. In repeated samples of the same size, estimates ranged from \(34\%\) to \(41\%\). For a population of \(6200\), find the central count estimate and the counts from the observed endpoints. Round to the nearest \(100\) residents.

Hints

- 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.

Solution

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.

Answer

Central estimate: about \(2300\) residents. Observed endpoint counts: about \(2100\) to \(2500\) residents.
Draw inferences from samples 47
Three sets of repeated random samples estimate the same population proportion. Set A estimates: \(0.21,\ 0.63,\ 0.38,\ 0.76\) Set B estimates: \(0.46,\ 0.49,\ 0.51,\ 0.48\) Set C estimates: \(0.32,\ 0.55,\ 0.44,\ 0.68\) Which set gives the most precise evidence for a population proportion near \(0.49\)? Justify with variation.

Hints

- 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.

Solution

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.

Answer

Set B. Its estimates range only from \(0.46\) to \(0.51\), a spread of \(0.05\), and cluster near \(0.49\).
Draw inferences from samples 48
Five random samples estimate the percentage of residents who use a bike trail: \(31\%,\ 34\%,\ 33\%,\ 32\%,\ 49\%\). Find the mean and median of the five estimates. Explain how the \(49\%\) result affects the two centers, and use the center that is less affected by the unusual result as a cautious typical estimate.

Hints

- 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.

Solution

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.

Answer

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.
Draw inferences from samples 49
Two random samples estimate the fraction of library visitors who use study rooms. Sample A: \(18\) of \(40\) Sample B: \(44\) of \(80\) a) Find the pooled estimate for all sampled visitors. b) Explain why simply averaging \(45\%\) and \(55\%\) is not the best combined estimate. c) Suppose Sample B had instead been taken only from visitors entering the study-room hallway, while Sample A remained random. Would the pooled percentage still be a defensible random-sample estimate for all library visitors? Explain.

Hints

- 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.

Solution

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.

Answer

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.
Draw inferences from samples 50
A transit agency takes one random sample each month. The estimated percentages of riders using mobile tickets are January \(47\%\), February \(49\%\), March \(54\%\), April \(58\%\), May \(62\%\). Is it reasonable to treat the differences as only random sample variation? Explain what pattern suggests another possibility.

Hints

- 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.

Solution

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.

Answer

The steady rise suggests a real increase may be occurring, not just random sample variation. The evidence is suggestive rather than conclusive.
Draw inferences from samples 51
Wildlife researchers tag \(40\) turtles in a pond and release them. After the turtles have mixed back into the pond, the researchers take a random sample of \(50\) turtles and find \(8\) tagged turtles. Assuming the sample is representative, estimate the total number of turtles in the pond. State the key assumption behind the estimate.

Hints

- 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.

Solution

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.

Answer

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

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