Unit 4 test

AP Statistics Unit 4 Test

15 multiple-choice questions and 1 free-response question, every one from Unit 4: Inference for Quantitative Data: Means, written in the format of the AP Statistics Exam.

Take the multiple choice on its own, timed at the exam's pace or in practice mode with each answer checked as you give it. Or sit the whole test, free response included, with each section on its own clock; the free-response scoring guide opens when that section ends.

AP Statistics Unit 4 Test

15 multiple-choice and 1 free-response question, a shorter set laid out like the AP Statistics Exam. Each section runs on its own clock and ends when time runs out.

Section	Questions	Time
Section I: Multiple Choice	15 of the exam's 42	32 minutes 9 seconds
Section II: Free Response	1 of the exam's 4	22 minutes 30 seconds

A graphing calculator with statistical capabilities is expected on both sections.

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Section I: Multiple Choice

15 questions · 32 minutes 9 seconds

The fleeces sheared from the sheep on a large ranch have weights with mean 4.6 kilograms and standard deviation 1.2 kilograms. A wool buyer takes a random sample of 16 fleeces from each shearing and finds the mean weight of the sample. The sampling distribution of that sample mean has standard deviation 0.3 kilogram. Which is the best interpretation of 0.3 kilogram?
Answer: D.
1. A
  That describes single fleeces, whose weights vary with standard deviation 1.2 kilograms. The 0.3 kilogram describes sample means, which vary less than single fleeces because each one averages 16 weights.
2. B
  The spread of the weights inside one sample is that sample's standard deviation, which should be near 1.2 kilograms, not 0.3. The 0.3 kilogram describes how the sample mean varies from one sample of 16 to the next.
3. C
  Taking a sample does not change the fleeces. Their weights still have standard deviation 1.2 kilograms. Only the sample mean, an average of 16 weights, varies less: $1.2 / \sqrt{16} = 0.3$ .
4. D
  Correct. The standard deviation of the sampling distribution is $σ / \sqrt{n} = 1.2 / \sqrt{16} = 0.3$ kilogram. It measures how far the mean weight of a random sample of 16 fleeces typically falls from the ranch's mean of 4.6 kilograms.
Finishing times in a large city marathon are skewed to the right: most runners finish in 3.5 to 5 hours, and a long tail of runners take 6 hours or more. A running magazine will select a random sample of the more than 20,000 finishers and find the mean finishing time of the sample. For which plan can the sampling distribution of the sample mean be modeled approximately by a normal distribution?
Answer: A.
1. A
  Correct. The population of finishing times is skewed, so the sample size has to be at least 30 for the sampling distribution of $\overline{x}$ to be approximately normal. A sample of 60 meets that, and 60 is far less than 10% of the finishers.
2. B
  A large sample makes the distribution of sample means close to normal, not the data. The 60 finishing times in one sample will look like the population they came from: skewed to the right, with a long tail.
3. C
  That holds for any sample size only when the population itself is approximately normal. These times are skewed, so with 20 runners the sampling distribution of $\overline{x}$ keeps some of that skew.
4. D
  The 10% condition is about the sampled values being independent, not about shape. Meeting it does not make the sampling distribution normal; with a skewed population that takes a sample of at least 30.
An inspector measures the diameter of each ball bearing in a random sample of 6 from a large production run and will construct a 95% confidence interval for the mean diameter. The critical value she uses is $t^{*} = 2.571$ rather than $z^{*} = 1.960$ . Which statement best explains why $t^{*}$ is larger?
Answer: D.
1. A
  Every t-distribution is symmetric about 0, like the standard normal. Both cutoffs, $- 2.571$ and $2.571$ , are the same distance from 0.
2. B
  It is the other way around. With few degrees of freedom the t-distribution is more spread out than the standard normal, with more area in its tails. A narrower curve would need a smaller cutoff, not a larger one.
3. C
  The sample standard deviation does not run larger than $σ$ ; from sample to sample it comes out smaller or larger. The larger cutoff allows for that uncertainty in $s$ , which is what the heavier tails of the t-distribution describe. It is not a correction for $s$ being too large.
4. D
  Correct. Using $s$ in place of $σ$ adds uncertainty, and the t-distribution with $6 - 1 = 5$ degrees of freedom shows it with heavier tails than the standard normal. About 10.7% of its area lies beyond $\pm 1.960$ , so leaving 2.5% in each tail takes a cutoff of 2.571.
A food scientist measures the protein content of a random sample of 12 cups of plain yogurt from a large production run. The sample mean is 11.6 grams and the sample standard deviation is 1.5 grams. The conditions for inference are met. Which is a 95% confidence interval for the mean protein content of all cups in the run?
Answer: B.
1. A
  This multiplies $t^{*}$ by $s = 1.5$ itself. The margin of error uses the standard error of the mean, $s / \sqrt{12} \approx 0.433$ , not the standard deviation of single cups.
2. B
  Correct. With $12 - 1 = 11$ degrees of freedom, $t^{*} \approx 2.201$ . The standard error is $1.5 / \sqrt{12} \approx 0.433$ , so the interval is $11.6 \pm 2.201 (0.433) \approx 11.6 \pm 0.95$ , or $(10.65, 12.55)$ grams.
3. C
  This uses $z^{*} = 1.96$ . The interval uses the sample standard deviation in place of $σ$ , so its critical value comes from a t-distribution with 11 degrees of freedom, $t^{*} \approx 2.201$ .
4. D
  This divides $s$ by $n = 12$ instead of by $\sqrt{12}$ , which makes the standard error, and the interval, far too small.
A sleep researcher selected a random sample of 25 of the adults in a large city who nap regularly and recorded the length of each one's afternoon nap. Her 95% confidence interval for the mean nap length of all such adults in the city has a margin of error of 6 minutes. She plans a new random sample and wants a margin of error of about 3 minutes at the same confidence level. If the sample standard deviation stays about the same, about how many adults should the new sample include?
Answer: C.
1. A
  A smaller sample makes the margin of error larger, not smaller. With 13 adults it would be about 8.8 minutes.
2. B
  Doubling the sample divides the margin of error by about $\sqrt{2}$ , not by 2. With 50 adults it would be about 4.1 minutes.
3. C
  Correct. The margin of error is roughly proportional to $1 / \sqrt{n}$ , so halving it takes about 4 times the sample size: $4 \times 25 = 100$ . With 100 adults it would be about 2.9 minutes, a little under 3 because $t^{*}$ also shrinks slightly.
4. D
  This squares the sample size. Going from 25 to 625 multiplies $\sqrt{n}$ by 5, which cuts the margin of error to about 1.1 minutes, far below the 3 minutes she wants.
A sleep clinic wants to know whether a weighted blanket reduces the time adults spend awake during the night. For each of 20 randomly selected patients, it records the minutes awake on one night under a weighted blanket and on one night under an ordinary blanket, in random order. For each patient, $d$ = (minutes awake with the weighted blanket) minus (minutes awake with the ordinary blanket). Which hypotheses fit the clinic's question?
Answer: A.
1. A
  Correct. With $d$ defined as weighted minus ordinary, fewer minutes awake under the weighted blanket makes $d$ negative. The parameter is $μ_{d}$ , the mean difference for all patients like these, so $H_{a} : μ_{d} < 0$ .
2. B
  This points the wrong way for this order of subtraction. A blanket that reduces time awake makes weighted minus ordinary less than 0, so the alternative is $μ_{d} < 0$ .
3. C
  The clinic asks whether the weighted blanket reduces time awake, which is one direction. A two-sided alternative would also count an increase as evidence, and the question does not ask about one.
4. D
  Hypotheses are about the population mean difference $μ_{d}$ , not the sample mean difference ${\overline{x}}_{d}$ . The sample mean is known once the data are in, so there is nothing to test about it.
Eight members of a cycling club each rode the same 10-kilometer time trial twice, once in standard kit and once in a one-piece skinsuit, in random order. Their times, in seconds, are shown.
$\begin{matrix} Rider & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ Standard kit & 905 & 948 & 872 & 1010 & 931 & 889 & 962 & 918 \\ Skinsuit & 893 & 941 & 857 & 1006 & 921 & 876 & 956 & 909 \end{matrix}$
The club wants to test whether the mean time is lower in a skinsuit, using $d$ = standard kit minus skinsuit for each rider. The conditions for inference are met. What t statistic does the test produce, and with how many degrees of freedom?
Answer: C.
1. A
  This treats the two rows as independent samples. The same eight riders rode both times, so the analysis uses the eight differences. Ignoring the pairing leaves the rider-to-rider spread (standard deviations near 44 and 48 seconds) in the standard error, which swamps a 9.5-second effect.
2. B
  This divides ${\overline{x}}_{d} = 9.5$ by $s_{d} \approx 3.74$ itself. The test statistic divides by the standard error, $s_{d} / \sqrt{8} \approx 1.32$ .
3. C
  Correct. The differences (standard kit minus skinsuit) are 12, 7, 15, 4, 10, 13, 6 and 9 seconds, with ${\overline{x}}_{d} = 9.5$ and $s_{d} \approx 3.74$ . Then $t = 9.5 / (3.74 / \sqrt{8}) \approx 7.18$ , with $8 - 1 = 7$ degrees of freedom.
4. D
  This divides $s_{d}$ by $n = 8$ instead of by $\sqrt{8}$ , which makes the standard error too small and the statistic too large.
A restaurant chain's recipe says a pot of rice in its new cookers is done in 18 minutes on average. A kitchen manager suspects the mean is longer. She times a random sample of 25 pots and finds a mean of 18.6 minutes and a standard deviation of 1.7 minutes. A test of $H_{0} : μ = 18$ against $H_{a} : μ > 18$ gives $t \approx 1.76$ and a p-value of 0.045. Which statement correctly describes what the 0.045 measures?
Answer: B.
1. A
  The p-value is computed by assuming $H_{0}$ is true, so it cannot be the probability that $H_{0}$ is true.
2. B
  Correct. The p-value assumes $H_{0}$ , a mean of 18 minutes. Under that assumption it is the probability of a sample mean at least as far above 18 as the 18.6 minutes observed, in a random sample of 25 pots.
3. C
  This is about one pot, not the mean of 25. Single pots vary much more than sample means (standard deviation 1.7 against a standard error of $1.7 / \sqrt{25} = 0.34$ ), so the chance for one pot is far larger than 0.045.
4. D
  The p-value is computed by assuming the null hypothesis, a mean of exactly 18 minutes, not the alternative. A mean of more than 18 minutes does not give one distribution to compute a probability from.
On a large farm, Red potatoes have weights with mean 210 grams and standard deviation 48 grams, and Yukon potatoes have weights with mean 185 grams and standard deviation 36 grams. A buyer selects independent random samples of 36 potatoes of each variety. Across all possible pairs of samples, the difference in sample mean weights, ${\overline{x}}_{R} - {\overline{x}}_{Y}$ , varies. What are its mean and its standard deviation?
Answer: A.
1. A
  Correct. The mean is $μ_{R} - μ_{Y} = 210 - 185 = 25$ grams. The standard deviation is $\sqrt{48^{2} / 36 + 36^{2} / 36} = \sqrt{64 + 36} = 10$ grams.
2. B
  This adds the standard deviations of the two sample means, $48 / 6 + 36 / 6 = 14$ . The squared terms add under one square root, $\sqrt{64 + 36}$ , which gives 10.
3. C
  This is $\sqrt{48^{2} + 36^{2}}$ , the spread of the difference between one Red potato and one Yukon potato. Each sample mean averages 36 potatoes, so each squared standard deviation is divided by 36.
4. D
  100 is $48^{2} / 36 + 36^{2} / 36$ , the square of the standard deviation. The standard deviation is its square root, 10 grams.
A test kitchen wants to estimate how much the mean cooking time to the al dente stage differs between two brands of dried spaghetti. It selects a random sample of 35 boxes of Brand A and an independent random sample of 35 boxes of Brand B from stores across a large region, cooks one serving from each box, and records the time in minutes. Which procedure should it use?
Answer: C.
1. A
  Equal sample sizes do not make data paired. Each box comes from one brand only, so no Brand A box is matched with a Brand B box, and there are no differences to compute.
2. B
  Samples of 30 or more let the sampling distribution be approximately normal, but $σ_{A}$ and $σ_{B}$ are still unknown. With the sample standard deviations in their place, the critical value comes from a t-distribution.
3. C
  Correct. Two independent random samples, a quantitative response and unknown population standard deviations call for a two-sample t-interval for $μ_{A} - μ_{B}$ .
4. D
  The response, cooking time in minutes, is quantitative, so the parameters are means, not proportions. Two brands make two populations; they do not make the response categorical.
A grower randomly assigned 30 mushroom growing kits to two growing mixes, 14 to straw and 16 to sawdust, and recorded each kit's yield in grams:

Straw: 395, 402, 407, 412, 415, 420, 425, 428, 431, 433, 438, 440, 446, 451

Sawdust: 441, 444, 448, 449, 452, 455, 457, 458, 460, 462, 465, 467, 470, 471, 476, 588

She wants a 95% two-sample t-interval for the difference in mean yield. Which statement about the conditions is correct?
Answer: D.
1. A
  Each sample is judged on its own. With 14 and 16 kits, both under 30, each group's yields must be free of strong skewness and outliers; adding the groups together to reach 30 does not count.
2. B
  Fewer than 30 is not a failure by itself. Small samples are allowed when their distributions show no strong skewness or outliers, and the straw yields meet that. The problem is the sawdust outlier.
3. C
  The kits were randomly assigned to the mixes, so this is an experiment. The 10% condition applies when sampling without replacement from a population, and it is not needed here.
4. D
  Correct. Both groups have fewer than 30 kits, so both distributions must be free of strong skewness and outliers. For sawdust, $Q_{1} = 450.5$ and $Q_{3} = 468.5$ , so the upper fence is $468.5 + 1.5 (18) = 495.5$ , and 588 is far above it.
A sheep farmer randomly assigned 40 lambs to two feed rations, 20 to each, and recorded each lamb's average daily weight gain in grams. The farmer's 95% interval estimate of $μ_{A} - μ_{B}$ , where each $μ$ is the mean daily gain on that ration for lambs like these and the order is Ration A minus Ration B, is $(- 48.9, - 9.1)$ grams per day. Which conclusion does the interval support?
Answer: A.
1. A
  Correct. The interval does not contain 0 and all its values are negative, so $μ_{A} < μ_{B}$ : we are 95% confident that the mean daily gain on Ration B is between 9.1 and 48.9 grams higher. The rations were randomly assigned, so the difference can be credited to the ration.
2. B
  The sign of the interval gives the direction. Every plausible value of $μ_{A} - μ_{B}$ is below 0, so Ration B has the higher mean daily gain.
3. C
  The interval estimates a difference in mean gains, not the gains of single lambs. Each lamb also ate only one ration, so no lamb has a gain on both to compare.
4. D
  Every value in the interval is negative, and the interval is for A minus B. A negative difference means Ration A's mean is lower, so the interval points to Ration B.
A fitness researcher randomly assigned 24 adult volunteers to eight weeks of either yoga or pilates classes, 12 to each. She measured each volunteer's reach on a sit-and-reach test before and after the eight weeks and found each volunteer's gain (after minus before), in centimeters. She wants to know whether the mean gain differs between yoga and pilates. Which procedure should she use?
Answer: B.
1. A
  A one-sample test on the gains asks whether the mean gain differs from 0, which says nothing about yoga against pilates. The before-and-after pairing has already been used to make the gains; the comparison asked about is between two independent groups.
2. B
  Correct. Each gain is one number per volunteer, and the volunteers were randomly assigned to two separate groups. Comparing the mean gain of the yoga group with the mean gain of the pilates group is a two-sample t-test for $μ_{Y} - μ_{P}$ .
3. C
  Equal group sizes do not make volunteers into pairs. A yoga volunteer and a pilates volunteer share nothing that a pairing would use, so any matching would be arbitrary.
4. D
  Reducing each gain to improved or not throws away how much each volunteer improved. The gain is quantitative, and the question is about mean gains.
A candle maker selected independent random samples of pillar candles from its large stock: 15 with cotton wicks and 18 with wood wicks. Cotton-wick burn times had ${\overline{x}}_{C} = 41.2$ hours and $s_{C} = 3.6$ hours, and wood-wick burn times had ${\overline{x}}_{W} = 38.5$ hours and $s_{W} = 4.1$ hours. Neither sample shows strong skewness or outliers. The maker tests whether the mean burn times differ, $H_{0} : μ_{C} = μ_{W}$ against $H_{a} : μ_{C} \neq μ_{W}$ , and software gives 30.9 degrees of freedom. What are the test statistic and p-value?
Answer: D.
1. A
  This adds the two standard errors, $3.6 / \sqrt{15} + 4.1 / \sqrt{18} \approx 1.90$ . The squared terms add under one square root, $\sqrt{{3.6}^{2} / 15 + {4.1}^{2} / 18} \approx 1.34$ .
2. B
  This is the one-sided p-value. The alternative is $μ_{C} \neq μ_{W}$ , so both tails count and the p-value is twice this.
3. C
  This uses the standard normal distribution. With sample standard deviations in place of $σ_{C}$ and $σ_{W}$ , the statistic follows a t-distribution, here with 30.9 degrees of freedom, which has heavier tails.
4. D
  Correct. $t = (41.2 - 38.5) / \sqrt{{3.6}^{2} / 15 + {4.1}^{2} / 18} = 2.7 / 1.341 \approx 2.01$ , and the two-sided p-value from a t-distribution with 30.9 degrees of freedom is about 0.053.
A researcher randomly assigned 60 adult volunteers to one of two ways of keeping a daily journal, 30 to each: writing by hand in a notebook before bed, or typing a few lines into a notes app at lunch. After one month, each volunteer reported the number of days out of 30 on which they wrote an entry. The researcher's test of whether the two groups' mean numbers of days differ gave p = 0.21. At the 0.05 level of significance, which conclusion is correct?
Answer: C.
1. A
  The rule runs the other way: reject $H_{0}$ only when the p-value is at most $α$ . Here 0.21 is greater than 0.05, so the decision is to fail to reject.
2. B
  Failing to reject $H_{0}$ does not show that $H_{0}$ is true. A difference may exist that this study was too small to detect; the result is only a lack of convincing evidence.
3. C
  Correct. Because $0.21 > 0.05$ , fail to reject $H_{0}$ . The data do not give convincing evidence that the mean number of days with an entry differs between the notebook and the notes app for volunteers like these.
4. D
  The p-value is computed by assuming the means are equal, so it cannot be the probability that they are equal. It is the probability, under that assumption, of a difference in sample means at least as large as the one observed.

Section II: Free Response

1 question · 22 minutes 30 seconds

Question 3: Inference (Hypothesis Test or Confidence Interval) 10 points

A rice company packs rice into bags labeled 2 kilograms (2,000 grams). A new scale built into the packing line weighs every bag as it is sealed. The maker of the scale claims that, on average, it gives the same reading as a laboratory scale. To check, the company's quality manager selected a random sample of 15 bags from the 6,000 bags packed on one day and weighed each of the 15 bags on the line scale and then on the company's laboratory scale.

The table summarizes the readings, and the dotplot shows the 15 differences in reading (line scale minus laboratory scale).

Readings for the 15 bags, in grams
Reading (grams) Mean Standard deviation

Line scale 2007.87 6.13
Laboratory scale 2005.92 5.51
Difference (line minus laboratory) 1.95 0.77

Differences in reading for the 15 bags (line scale minus laboratory scale)

Construct a 95% confidence interval for the mean difference in reading (line scale minus laboratory scale) for all bags packed that day, and use it to evaluate the claim made by the maker of the scale.

Use the given information to respond to parts A, B, C, and D. Label any subparts (e.g., i and ii) that may be present.
1. A.
  
  i. Identify the appropriate inference procedure.
  
  ii. Explain why the procedure you identified in (i) is appropriate for the way these data were collected.
  Question 3, part A: your response. It stays in this browser tab and is never sent anywhere.
  Score part A
  - 1 point Identifies the procedure as a one-sample t-interval for a population mean difference (a matched pairs t-interval).Earns it: "one-sample t-interval for the mean difference, line minus laboratory"; "matched pairs t-interval for $μ_{d}$ "Does not: "two-sample t-interval for $μ_{L} - μ_{B}$ " (both readings come from the same bags, so the two sets of readings are not independent samples); "one-sample z-interval for a mean difference" (the standard deviation of all the differences is unknown, so the critical value comes from a t-distribution); "one-sample t-test for a mean difference" (the question asks for an interval, not a test)
  - 1 point Explains that each bag was weighed on both scales, so the two readings on a bag form a pair, and the analysis uses the one sample of 15 differences. The response must connect the procedure to the same bags being weighed twice.Earns it: "Every bag was weighed on both scales, so the readings come in pairs. Taking line minus laboratory for each bag gives one sample of 15 differences."Does not: "Because there are two scales." (two scales could also be compared with two separate samples of bags; the pairing is what matters); "Because both samples have 15 bags." (equal sample sizes do not make data paired); "Because the bags were randomly selected." (that supports the randomization condition, not the choice of a paired procedure)
  Model response
  i. A one-sample t-interval for a population mean difference (a matched pairs t-interval).
  
  ii. Each of the 15 bags was weighed on both the line scale and the laboratory scale, so the two readings on a bag form a pair. Subtracting the laboratory reading from the line reading for each bag gives one sample of 15 differences, and the interval estimates the mean of those differences for all of the day's bags.
2. B.
  
  Complete the inference procedure, including calculating the appropriate statistics.
  Question 3, part B: your response. It stays in this browser tab and is never sent anywhere.
  Score part B
  - 1 point Parameter: defines $μ_{d}$ as the true mean difference in reading (line scale minus laboratory scale), in grams, for all 6,000 bags packed that day. The order of subtraction and the population must both be clear.Earns it: " $μ_{d}$ = the true mean of (line reading minus laboratory reading) for all 6,000 bags packed that day"Does not: " $μ_{d}$ is the mean of the 15 differences." (that is the sample mean difference, 1.95 grams); " $μ$ = the mean weight of the bags" (not a difference, and no order of subtraction)
  - 1 point Randomization and 10% conditions: states that the 15 bags were randomly selected, and that 15 is no more than 10% of the 6,000 bags packed that day ( $15 \leq 600$ ). Both conditions are required.Earns it: "The 15 bags were a random sample, and 15 is less than 10% of 6,000, which is 600."Does not: "The scales were randomly assigned to the bags." (nothing was assigned; every bag was weighed on both scales); "Random: met. 10%: met." (no justification for either)
  - 1 point Sample data condition: states that there are only 15 differences, fewer than 30, so the sample of differences must be free from strong skewness and outliers, and that the dotplot of the differences shows neither. A check of the 1.5 IQR fences on the differences ( $Q_{1} = 1.4$ and $Q_{3} = 2.6$ give fences $- 0.4$ and $4.4$ , and every difference from 0.7 to 3.3 lies inside them) also earns the point. The check must be made on the differences.Earns it: "There are only 15 differences, which is less than 30, but the dotplot of the differences has no strong skew and no outliers."Does not: " $n \geq 30$ " (there are only 15 bags); "Neither the line readings nor the laboratory readings have outliers." (the interval uses the differences, so the condition is checked on the differences); "By the central limit theorem the sampling distribution is approximately normal." (15 is too few to rely on that without checking the data)
  - 1 point Critical value and standard error: identifies $df = 15 - 1 = 14$ , $t^{*} \approx 2.145$ , and the standard error $s_{d} / \sqrt{15} = 0.77 / \sqrt{15} \approx 0.20$ grams.Earns it: " $t^{*} = 2.145$ with 14 degrees of freedom, and $S E = 0.77 / \sqrt{15} \approx 0.20$ "Does not: " $z^{*} = 1.96$ " (the standard deviation of the differences is estimated from the sample, so the critical value comes from a t-distribution with 14 degrees of freedom; it would give $(1.56, 2.34)$ ); " $S E = 0.77$ " (that is the standard deviation of the differences, not the standard error of their mean)
  - 1 point Interval: reports $1.95 \pm 2.145 (0.20)$ , or about $(1.52, 2.38)$ grams. An arithmetic or transcription error can be ignored if correct work is shown.Earns it: " $1.95 \pm 2.145 (\frac{0.77}{\sqrt{15}}) = 1.95 \pm 0.43$ , so $(1.52, 2.38)$ grams"Does not: " $(- 2.42, 6.31)$ " (a two-sample interval built from the two columns of readings, which ignores the pairing); " $(0.29, 3.61)$ " (uses the standard deviation of the differences in place of the standard error)
  Model response
  Let $μ_{d}$ be the true mean difference in reading (line scale minus laboratory scale), in grams, for all 6,000 bags packed that day.
  
  The randomization condition is met because the 15 bags were randomly selected. The 10% condition is met because 15 is less than 10% of 6,000, which is 600. There are only 15 differences, fewer than 30, but the dotplot of the differences shows no strong skewness and no outliers: with $Q_{1} = 1.4$ and $Q_{3} = 2.6$ the fences are $- 0.4$ and $4.4$ , and every difference lies between 0.7 and 3.3. The sample data condition is met.
  
  With $df = 15 - 1 = 14$ , $t^{*} \approx 2.145$ , and the standard error is $0.77 / \sqrt{15} \approx 0.20$ . The interval is $1.95 \pm 2.145 (0.20) \approx 1.95 \pm 0.43$ , or $(1.52, 2.38)$ grams.
3. C.
  
  i. Interpret the interval from part B in context.
  
  ii. Does the interval support the claim that, on average, the line scale gives the same reading as the laboratory scale? Justify your answer.
  Question 3, part C: your response. It stays in this browser tab and is never sent anywhere.
  Score part C
  - 1 point Interprets the interval: we are 95% confident that the interval from 1.52 to 2.38 grams contains the true mean difference in reading (line scale minus laboratory scale) for all bags packed that day. The reference to the population mean difference, with its order and context, is required.Earns it: "We are 95% confident that the true mean of line reading minus laboratory reading, for all bags packed that day, is between 1.52 and 2.38 grams."Does not: "95% of the bags read between 1.52 and 2.38 grams higher on the line scale." (the interval is about the mean difference, not single bags); "We are 95% confident that the mean of the 15 differences is between 1.52 and 2.38 grams." (the sample mean, 1.95, is known; the interval estimates the population mean difference)
  - 1 point States that the interval does not support the claim, because 0 is not in the interval: every plausible value of the mean difference is positive, so the line scale reads higher than the laboratory scale on average, by about 1.5 to 2.4 grams.Earns it: "No. The interval does not contain 0, so a mean difference of 0 is not plausible. The line scale reads about 1.5 to 2.4 grams higher on average."Does not: "Yes, because 1.5 to 2.4 grams is tiny next to 2,000 grams." (the claim is that the mean difference is 0, and 0 is not a plausible value; whether 2 grams matters to the company is a separate question); "No, the line scale is wrong on every bag." (a claim about single bags, which the interval does not support); "No." (no reference to whether 0 is in the interval)
  Model response
  i. We are 95% confident that the interval from 1.52 to 2.38 grams contains the true mean difference in reading (line scale minus laboratory scale) for all bags packed that day.
  
  ii. No. The interval does not contain 0, so a mean difference of 0 is not plausible. Every value in the interval is positive, so the data give convincing evidence that, on average, the line scale reads higher than the laboratory scale, by somewhere between about 1.5 and 2.4 grams per bag.
4. D.
  
  Interpret the 95% confidence level in context.
  Question 3, part D: your response. It stays in this browser tab and is never sent anywhere.
  Score part D
  - 1 point Explains that if many random samples of 15 bags were taken from the day's bags and a 95% interval were built from each in the same way, about 95% of those intervals would capture the true mean difference in reading (line scale minus laboratory scale) for all bags packed that day. Both the repeated sampling and the reference to the population mean difference are required.Earns it: "If we took many random samples of 15 bags from that day and made a 95% interval from each, about 95% of the intervals would contain the true mean difference, line minus laboratory, for all the day's bags."Does not: "There is a 95% probability that the true mean difference is between 1.52 and 2.38 grams." (the confidence level describes the method over many samples, not this one interval); "95% of the differences are between 1.52 and 2.38 grams." (only 5 of the 15 differences are; the interval is about the mean, not single bags); "95% of samples will have a mean difference between 1.52 and 2.38 grams." (the level is about intervals capturing the parameter, not sample means landing in this interval)
  Model response
  If the quality manager took many random samples of 15 bags from the bags packed that day and built a 95% confidence interval from each sample in this way, about 95% of those intervals would contain the true mean difference in reading (line scale minus laboratory scale) for all of that day's bags.
Your score: 0 of 10, from the boxes you ticked.

Reading (grams)	Mean	Standard deviation
Line scale	2007.87	6.13
Laboratory scale	2005.92	5.51
Difference (line minus laboratory)	1.95	0.77

AP Statistics Unit 4 Test

Section I: Multiple Choice

Section II: Free Response

Question 3: Inference (Hypothesis Test or Confidence Interval) 10 points

Your results