Free math worksheets

- Look for x-values that have a marker from each sample. - Overlap in this question means a displayed value occurs in both samples.

1. Compare the plotted x-values occupied by both samples. 2. Both samples have dots at \(4\) and \(5\). 3. Therefore, the displayed samples overlap at those two values.

Yes. The displayed samples overlap at \(4\) and \(5\).

- Measure the center difference using the shared MAD as the unit. - Describe what the ratio compares rather than translating it into a memorized overlap label. - Ask whether a mean and one spread statistic tell you the positions of all individual scores.

1. The mean difference is \(74-68=6\) points. 2. The separation is \(6\div4=1.5\) MADs. 3. This means the centers differ by one and a half times the shared typical absolute deviation. 4. Mean and MAD summaries do not show where individual scores fall, so they cannot determine the exact amount of overlap.

a) \(1.5\) MADs. b) The center difference is \(1.5\) times the shared MAD. c) No. Exact overlap is not determined by the means and MADs alone.

- Read each median from the line inside its box. - Subtract the first quartile from the third quartile for each IQR. - Compare the boxes and displayed whisker intervals separately; whiskers are not guaranteed data extrema.

1. City A has median \(40\) miles per hour and IQR \(44-36=8\) miles per hour. 2. City B has median \(52\) miles per hour and IQR \(56-48=8\) miles per hour. 3. The medians differ by \(52-40=12\) miles per hour, while the two samples have equal middle-half variability. 4. The boxes do not overlap because City A's third quartile is \(44\) and City B's first quartile is \(48\). 5. The displayed whisker intervals overlap from \(44\) to \(48\) miles per hour.

City A: median \(40\,\text{mph}\), IQR \(8\,\text{mph}\) City B: median \(52\,\text{mph}\), IQR \(8\,\text{mph}\) The boxes do not overlap, but the displayed whisker intervals overlap from \(44\) to \(48\,\text{mph}\). City B has the higher center, and the middle halves have equal variability.

- Read the center line and both box edges for each store. - Use box width to compare the spread of the middle halves. - Check overlap of the boxes separately from overlap of the displayed whisker intervals.

1. Store A has median \(24\) orders and IQR \(26-22=4\) orders. 2. Store B has median \(34\) orders and IQR \(38-30=8\) orders. 3. Store B's sample median is \(34-24=10\) orders higher. 4. The boxes do not overlap, although the displayed whisker intervals overlap from \(26\) to \(30\) orders. 5. Store B appears more variable in the middle half because its IQR is twice Store A's IQR.

Store A: median \(24\), IQR \(4\) Store B: median \(34\), IQR \(8\) Store B has the higher center and greater middle-half variability. The boxes do not overlap, but the displayed whisker intervals overlap from \(26\) to \(30\) orders.

- Measure the distance between the means using the common MAD. - Look for horizontal values that contain dots from both samples. - A standardized center difference describes separation; it does not create boundaries around either population.

1. The mean difference is \(38-30=8\), and \(8\div4=2\) MADs. 2. The plots share displayed values at \(32\), \(34\), and \(36\), so the samples overlap. 3. Machine B's dots are generally farther to the right, showing a higher center with the same variability. 4. Therefore, statement b is best supported. A two-MAD center difference does not prove that populations cannot overlap.

b) The displayed samples overlap, but Machine B's distribution is shifted higher.

- Find the distance between the two sample means. - Compare that distance separately with each stated MAD. - Do not average unequal MADs into a new common variability unit.

1. The mean difference is \(132-120=12\) gigabytes. 2. Relative to Plan A's MAD, the gap is \(12\div3=4\) MADs. 3. Relative to Plan B's MAD, the gap is \(12\div5=2.4\) MADs. 4. The variability measures differ, so report both comparisons rather than averaging them. 5. The center difference is large relative to both stated MADs.

The means differ by \(12\) gigabytes, which is \(4\) Plan A MADs and \(2.4\) Plan B MADs. The center difference is large relative to both variability measures.

- Convert the required number of shared variability units into an actual distance. - Add that distance to Sample A's mean because Sample B must be higher. - “Least possible” means use exactly the required separation.

1. Two shared MADs equal \(2\cdot6=12\). 2. The least sample mean at least that far above \(50\) is \(50+12=62\).

The least possible sample mean is \(62\).

- Read each median and subtract the first quartile from the third quartile. - The box itself represents the middle half of a sample. - Compare the displayed whisker intervals separately; they are not guaranteed to be the full data ranges.

1. Population A has median \(18\) and IQR \(22-14=8\). 2. Population B has median \(24\) and IQR \(28-20=8\). 3. The middle halves overlap from \(20\) to \(22\). 4. The displayed whisker intervals overlap from \(16\) to \(26\). 5. Population B has the higher center, while the equal IQRs indicate equal middle-half variability in these samples.

Population A: median \(18\), IQR \(8\) Population B: median \(24\), IQR \(8\) The boxes overlap from \(20\) to \(22\), and the displayed whisker intervals overlap from \(16\) to \(26\). Population B has the higher center, and the samples have equal middle-half variability.

- First put the raw center difference on the variability scale. - Separate what the ratio actually measures from what the student claims about individual observations. - Ask whether the full distributions can be reconstructed from only a mean and MAD.

1. The center difference is \(80-70=10\). 2. In MAD units, the separation is \(10\div8=1.25\). 3. The ratio says the sample means differ by \(1.25\) times the shared MAD, so the raw difference is not large compared with the stated variability. 4. However, mean and MAD summaries do not show the individual observations, so they cannot prove that the samples have little overlap or determine the exact amount of overlap.

a) \(1.25\) MADs. b) The center difference is \(1.25\) times the shared MAD. c) The summaries do not determine the exact amount of overlap, so the student's “proves little overlap” claim is not justified.

- Compare the measures of center separately from the measures of spread. - Decide what a smaller spread means in this context. - Do not use the mean alone to judge consistency.

1. The mean errors are equal, so the center difference is \(14-14=0\,\text{cm}\). 2. Team R has the smaller MAD because \(1.5<4.5\). 3. The equal centers show no separation between the typical errors, while Team R's smaller MAD shows less variability.

The teams have the same mean error. Team R is more consistent because its MAD is smaller.

- Compare the box widths within and across the two panels. - Locate the numerical intersection of the two boxes in each panel. - When variability is held constant, a larger center shift generally reduces visible overlap.

1. In Pair 1, the medians are \(30\) and \(34\), and both IQRs are \(8\). The boxes \([26,34]\) and \([30,38]\) overlap from \(30\) to \(34\). 2. In Pair 2, the medians are \(30\) and \(42\), and both IQRs are \(8\). The boxes \([26,34]\) and \([38,46]\) do not overlap. 3. Pair 1 has more middle-half overlap because its centers are closer while the box widths are the same.

Pair 1 has more overlap between its middle halves: its boxes intersect from \(30\) to \(34\). Pair 2's boxes do not overlap because their medians are farther apart while the IQRs remain equal.

- Read the line inside each box for the median. - Compare the widths of the boxes to compare IQRs. - Identify overlap of the boxes and displayed whisker intervals directly from the display.

1. Cast X has median \(45\) minutes and IQR \(51-39=12\) minutes. 2. Cast Y has median \(51\) minutes and IQR \(53-49=4\) minutes. 3. Cast X is more variable in the middle half because its IQR is three times as large. 4. The boxes overlap from \(49\) to \(51\) minutes, and the displayed whisker intervals overlap from \(45\) to \(57\) minutes. 5. Cast Y has the higher median, and visible overlap is present in both the boxes and whisker intervals.

Cast X is more variable: its IQR is \(12\) minutes compared with \(4\) minutes for Cast Y. Cast Y's median is \(6\) minutes higher. The boxes overlap from \(49\) to \(51\) minutes, and the displayed whisker intervals overlap from \(45\) to \(57\) minutes.

- Compare the median lines and box edges first. - Use the whisker endpoints to compare the displayed whisker spans. - Keep conclusions about the middle half separate from conclusions about the whiskers.

1. Both samples have median \(30\). 2. Both samples have IQR \(40-20=20\). 3. Sample A's displayed whisker span is \(100-0=100\), while Sample B's is \(50-10=40\). 4. Their boxes are identical, \([20,40]\), so the middle-half intervals overlap completely. 5. Equal medians and IQRs do not imply equal whisker spans.

Both medians are \(30\), and both IQRs are \(20\). Sample A's displayed whisker span is \(100\), while Sample B's is \(40\). Their middle-half intervals overlap completely because the boxes are identical.

- Read the median lines and box edges. - Compare the positions of the boxes as well as their widths. - Containment of one box in another indicates a narrower middle-half interval, not identical data.

1. Sample A has median \(24\) and IQR \(34-14=20\). 2. Sample B has median \(26\) and IQR \(30-20=10\). 3. Sample B's box \([20,30]\) is entirely inside Sample A's box \([14,34]\). 4. The medians differ by only \(2\), while Sample B's middle-half spread is half as large. 5. The samples have similar centers, but Sample B is much more consistent in the middle half.

Sample A: median \(24\), IQR \(20\) Sample B: median \(26\), IQR \(10\) Sample B's box is contained within Sample A's. Their centers are similar, but Sample B has much less middle-half variability.

- Use the MAD that is shared by both samples as the variability unit. - Convert two shared units into a raw center difference. - The phrase “greater than \(20\)” tells you which direction to move from Sample A's mean.

1. Two shared-MAD units correspond to a center difference of \(2\cdot4=8\). 2. Because Sample B's mean is greater than \(20\), its mean must be \(20+8=28\). 3. Therefore, choice c is correct.

c) \(28\)

- Verify that the median gaps are equal in the two panels. - Compare the numerical intersections of the boxes. - Hold the center difference constant and focus on how box width changes overlap.

1. In Pair A, each IQR is \(8\). The boxes \([36,44]\) and \([44,52]\) meet only at \(44\). 2. In Pair B, each IQR is \(24\). The boxes \([58,82]\) and \([66,90]\) overlap from \(66\) to \(82\). 3. The median gap is the same in both pairs, but Pair B has much wider middle-half intervals. 4. Pair B therefore shows more overlap between its middle halves.

Pair B has more middle-half overlap. Pair A's boxes meet only at \(44\), while Pair B's boxes overlap from \(66\) to \(82\). The same center gap produces more overlap when variability is greater.

- Divide the distance between the means by the common MAD. - Compare where each frequency polygon reaches its largest values. - Look for class intervals in which both samples have observations.

1. The mean difference is \(32-20=12\). 2. The standardized separation is \(12\div4=3\) MADs. 3. The frequency polygons are centered in different regions, with Sample B shifted to the right. 4. The polygons still occupy some common class intervals around the high end of Sample A and the low end of Sample B. 5. A three-MAD separation suggests strong overall separation, but the display confirms that overlap need not be zero.

The means are \(3\) MADs apart. The frequency polygons show strong separation of the centers with limited visual overlap in nearby class intervals.

- First compare the two MAD values themselves. - Use the same raw center gap with each actual MAD. - No automatic pooled or averaged MAD is required.

1. The mean difference is \(58-50=8\). 2. Relative to Sample A's MAD, the gap is \(8\div4=2\) MADs. 3. Relative to Sample B's MAD, the gap is \(8\div4.5\approx1.78\) MADs. 4. The MADs are close but not identical, so reporting the two comparisons directly is appropriate. Choice b is correct.

b) The MADs are similar but not equal; the mean gap is \(2\) Sample A MADs and about \(1.78\) Sample B MADs.

- Ask whether the two spread measures are actually similar enough to act like one common scale. - Compare the \(12\)-unit center gap with each MAD separately. - Distinguish a summary-statistic comparison from a claim about visible or exact data overlap.

1. The mean difference is \(32-20=12\). 2. The MADs differ substantially, so automatically averaging them does not create a standard common variability unit for this Grade 7 comparison. 3. Relative to Sample A's MAD, the center gap is \(12\div2=6\) MADs. Relative to Sample B's MAD, it is \(12\div10=1.2\) MADs. 4. Means and MADs alone also do not determine the exact amount of data overlap. A display or the raw data would be needed for an exact overlap statement.

The student should not automatically average the unequal MADs, and the summaries do not determine exact overlap. The \(12\)-unit mean gap is \(6\) Sample A MADs and \(1.2\) Sample B MADs.

- Find the distance between the two means. - Compare that distance separately with each stated MAD. - If two variability measures are not equal, report both comparisons rather than inventing a pooled value.

1. The mean difference is \(52-40=12\) dollars. 2. Relative to Neighborhood P's variability, the gap is \(12\div5=2.4\) MADs. 3. Relative to Neighborhood Q's variability, the gap is \(12\div7\approx1.71\) MADs. 4. The two MADs are similar but not equal, so they should be reported directly rather than averaged into a new variability unit. 5. The samples suggest a noticeable difference in typical monthly cost.

The mean difference is \(\$12\), which is \(2.4\) Neighborhood P MADs and about \(1.71\) Neighborhood Q MADs. The samples suggest a noticeable center difference; no averaged MAD is needed.

- Decide which quantity is the distance being measured and which quantity is the unit. - Ask how many groups of \(4\) fit into the center difference. - For part b), distinguish summary statistics from the full distributions of observed values.

1. The difference in means is \(102-90=12\) points. 2. The question asks how many MAD-sized units fit into the center difference, so the center difference is the quantity being measured and the MAD is the unit. 3. The correct ratio is \(12\div4=3\) MADs. 4. The ratio compares center separation with variability, but it does not locate the individual observations. Therefore it does not determine exact overlap.

a) The student reversed the ratio. The sample means are \(3\) MADs apart. b) No. The ratio describes center separation relative to variability, not the exact overlap of the observations.

- Find the balance point and average absolute deviation for each dot plot. - Divide the difference between the means by the common MAD. - Compare the ratio with the horizontal region occupied by both distributions.

1. Path A has mean \(35\) bicycles and MAD \(\frac{40}{8}=5\) bicycles. 2. Path B has mean \(42\) bicycles and MAD \(\frac{40}{8}=5\) bicycles. 3. The mean difference is \(42-35=7\) bicycles, so the separation is \(7\div5=1.4\) MADs. 4. The two dot plots occupy the same horizontal region from about \(34\) to \(43\), so substantial visual overlap remains. 5. The ratio and the display both suggest a moderate shift rather than clear separation.

Path A: mean \(35\), MAD \(5\) Path B: mean \(42\), MAD \(5\) The means are \(1.4\) MADs apart. The dot plots overlap visibly from about \(34\) to \(43\), so the samples show a moderate difference with substantial overlap.

- Compute the mean and average absolute deviation for each displayed sample. - Check whether any part of the two plotted sample ranges occupies the same horizontal region. - Distinguish what the samples show from what can be concluded with certainty about entire populations.

1. Shop A has mean \(20\) minutes and MAD \(\frac{12}{6}=2\) minutes. 2. Shop B has mean \(27\) minutes and MAD \(\frac{12}{6}=2\) minutes. 3. The means are \((27-20)\div2=3.5\) MADs apart. 4. The displayed samples do not overlap: Shop A's largest time is \(23\) minutes, and Shop B's smallest time is \(24\) minutes. 5. The samples suggest strong population separation, but they do not prove that the full populations never overlap.

Shop A: mean \(20\) minutes, MAD \(2\) minutes Shop B: mean \(27\) minutes, MAD \(2\) minutes The means are \(3.5\) MADs apart, and the displayed samples do not overlap. This suggests strong separation, but it does not prove zero overlap in the populations.

- Convert the number of variability units into an actual distance. - Use the direction of the comparison to decide whether to add or subtract. - Interpret the result as a comparison of center separation with variability, not as an exact overlap percentage.

1. A separation of \(2.5\) MADs equals \(2.5\cdot4=10\) units. 2. Sample B has the higher mean, so its mean is \(64+10=74\). 3. A \(2.5\)-MAD center separation is large relative to the common typical absolute deviation.

Sample B's mean is \(74\). The sample centers are \(2.5\) common MADs apart, a large separation relative to the variability.

- Find the full distance between the centers. - Divide that distance into the stated number of equal variability units. - The word “same” makes one shared MAD well-defined.

1. The difference in means is \(24-18=6\). 2. If \(6\) equals \(2\) MADs, then one MAD is \(6\div2=3\). 3. The sample centers are separated by two times the shared typical absolute deviation.

The shared MAD is \(3\). The sample centers are \(2\) MADs apart, a noticeable separation relative to the variability.

- Find the median, then find the median of each half of each ordered list. - Use the quartile intervals to compare the middle halves. - Use the minimum and maximum only when comparing the full sample ranges.

1. Group A has median \(6\), first quartile \(4\), and third quartile \(8\), so its IQR is \(8-4=4\). 2. Group B has median \(14\), first quartile \(12\), and third quartile \(16\), so its IQR is \(16-12=4\). 3. The middle halves are \([4,8]\) and \([12,16]\), so they do not overlap. 4. The full sample ranges are \([2,10]\) and \([10,18]\), so they meet at \(10\). 5. The samples have equal middle-half variability and clearly different centers, but their full ranges are not separated by a gap.

Group A: median \(6\), IQR \(4\) Group B: median \(14\), IQR \(4\) The middle halves do not overlap. The full sample ranges meet at \(10\).

- Raw center differences are not directly comparable when spreads differ. - Put each difference in units of its own variability. - The larger standardized ratio indicates clearer separation.

1. Pair A's standardized separation is \(10\div5=2\) MADs. 2. Pair B's standardized separation is \(12\div8=1.5\) MADs. 3. Although Pair B has the larger raw difference, Pair A has the larger difference relative to variability. 4. Pair A therefore has clearer likely separation.

The reasoning is incorrect. Pair A is \(2\) MADs apart, while Pair B is \(1.5\) MADs apart. Pair A has clearer separation relative to its variability.

- Find each dot plot's balance point before measuring absolute deviations. - Divide the distance between the means by the common MAD. - Compare the standardized result with the horizontal region occupied by both samples.

1. Line A has mean \(8\) and MAD \(\frac{12}{5}=2.4\). 2. Line B has mean \(13\) and MAD \(\frac{12}{5}=2.4\). 3. The mean difference is \(13-8=5\). 4. The separation is \(5\div2.4\approx2.1\) MADs. 5. The displayed sample ranges overlap from \(9\) to \(12\), so the plots show a noticeable shift with limited, not zero, overlap.

Line A: mean \(8\), MAD \(2.4\) Line B: mean \(13\), MAD \(2.4\) The means are about \(2.1\) MADs apart. The sample ranges overlap from \(9\) to \(12\), so the distributions are noticeably shifted but not completely separated.

- The numerator is the same for both pairs, so focus on the different MAD units. - Divide each center difference by its pair's shared MAD. - Interpret the result as relative separation, not as a direct count of overlapping observations.

1. Pair X has separation \((18-10)\div2=4\) MADs. 2. Pair Y has separation \((38-30)\div5=1.6\) MADs. 3. Pair X has the greater relative separation. 4. The same raw center gap represents more variability units when the shared MAD is smaller, so equal raw differences need not have the same statistical size relative to spread.

a) Pair X: \(4\) MADs; Pair Y: \(1.6\) MADs. b) Pair X. c) The shared variability units differ, so the same raw gap represents different numbers of MADs.

- Find the balance point and average absolute deviation for each dot plot. - Divide the difference between the means by the common MAD. - Look for values and horizontal regions occupied by both samples.

1. Group A has mean \(55\) minutes and MAD \(\frac{24}{6}=4\) minutes. 2. Group B has mean \(61\) minutes and MAD \(\frac{24}{6}=4\) minutes. 3. The mean difference is \(61-55=6\) minutes, so the separation is \(6\div4=1.5\) MADs. 4. The displayed samples share values at \(57\) and \(59\) minutes, and their full ranges overlap from \(55\) to \(61\) minutes. 5. The numerical and visual evidence both indicate a moderate shift with substantial overlap.

Group A: mean \(55\) minutes, MAD \(4\) minutes Group B: mean \(61\) minutes, MAD \(4\) minutes The means are \(1.5\) MADs apart. The samples share values at \(57\) and \(59\) minutes, so they overlap substantially.

- Find the center difference first. - Divide that difference separately by each sample's MAD. - Ask whether averaging unequal variability measures has been defined as a method in this problem.

1. The mean difference is \(54-48=6\). 2. Relative to Sample C's MAD, the gap is \(6\div3=2\) MADs. 3. Relative to Sample D's MAD, the gap is \(6\div5=1.2\) MADs. 4. The MADs are unequal, so averaging them would create a new variability measure not supplied by either sample. 5. Report the two direct comparisons and note that Sample D is more variable.

The sample means differ by \(6\). That gap is \(2\) Sample C MADs and \(1.2\) Sample D MADs. Because the MADs differ, compare with each one directly rather than averaging them into a new “common MAD.”

- Find the difference between the two means. - Compare that same difference with each store's MAD separately. - Judge the manager's wording from those direct center-versus-variability comparisons.

1. The mean difference is \(39-30=9\). 2. Relative to Store M's MAD, the gap is \(9\div2=4.5\) MADs. 3. Relative to Store N's MAD, the gap is \(9\div4=2.25\) MADs. 4. The center difference is large relative to both stated variability measures, so “nearly identical” is not supported.

No. The means differ by \(9\), which is \(4.5\) Store M MADs and \(2.25\) Store N MADs. The center difference is too large relative to the stated variability for “nearly identical” to be reasonable.

- Recall what MAD measures: an average distance, not an endpoint. - Ask whether the summaries identify where every observation lies. - Compare the center gap with each MAD separately instead of averaging unequal MADs.

1. A MAD is an average distance from the mean. It is not a boundary containing a fixed percentage of the data. 2. Therefore, overlap between mean-centered segments cannot be converted into a percentage of scores or a percentage of population overlap. 3. The mean difference is \(69-62=7\). 4. Relative to Class A's MAD, the gap is \(7\div5=1.4\) MADs; relative to Class B's MAD, it is \(7\div4=1.75\) MADs. 5. Those summaries compare center separation with variability, but they still do not determine an exact overlap percentage.

The method is invalid because mean \(\pm\) MAD does not contain a fixed percentage of a distribution. Segment-length overlap therefore does not equal data overlap. The \(7\)-point center gap is \(1.4\) Class A MADs and \(1.75\) Class B MADs; exact population overlap is not determined.

- Compute each IQR and compare the medians. - Distinguish an interval of possible values from a count of observations. - Decide what quartiles reveal and what individual-data information they leave unknown.

1. Each IQR is \(22-14=8\) minutes and \(29-21=8\) minutes. 2. The median difference is \(25-18=7\) minutes. 3. The numerical intervals for the middle halves intersect on \([21,22]\). 4. However, quartile summaries do not identify the individual observations or how many observations have similar values. 5. The samples have equal middle-half variability and different centers, but the amount or percentage of actual data overlap cannot be determined from these summaries alone.

The middle-half intervals intersect on \([21,22]\), but this does not mean that exactly one minute of observed data is shared or reveal how many observations overlap. Both IQRs are \(8\) minutes, and Group B's median is \(7\) minutes higher.

- The raw center difference is the same in both studies. - Measure that difference using each study's shared MAD. - For part c), identify what information a mean and MAD leave out.

1. The center difference is \(52-40=12\). 2. In Study 1, the relative separation is \(12\div3=4\) MADs. 3. In Study 2, the relative separation is \(12\div6=2\) MADs. 4. The center difference is larger relative to variability in Study 1. 5. The means and MADs do not specify the locations of individual observations, so exact overlap cannot be determined from these summaries alone.

a) Study 1: \(4\) MADs; Study 2: \(2\) MADs. b) Study 1. c) No. Means and MADs describe centers and variability but not the exact positions of observations.

- Use the left and right edges of each box to find its IQR. - Find the intersection of the two numerical box intervals. - Distinguish the middle-half interval of a sample from the individual values or counts inside that interval.

1. Service A has IQR \(18-8=10\) minutes. 2. Service B has IQR \(22-14=8\) minutes, so Service A has the larger IQR. 3. The boxes cover \([8,18]\) and \([14,22]\). 4. Their intersection is \([14,18]\), an interval with length \(18-14=4\) minutes. 5. This interval is the numerical region common to both middle-half intervals. It does not show how many actual observations are shared or imply that \(50\%\) of either sample lies in the intersection.

Service A has the larger IQR: \(10\) minutes compared with \(8\) minutes. The boxes intersect on \([14,18]\), a \(4\)-minute interval. This is the region common to the two middle-half intervals, not a count or percentage of shared observations.

- Measure the same center difference with each proposed unit. - Notice how changing the unit changes the numerical ratio. - Ask whether the problem or curriculum has defined the average of unequal MADs as a new variability statistic.

1. The mean difference is \(36-26=10\). 2. Using the smaller MAD gives \(10\div2=5\). 3. Using the larger MAD gives \(10\div5=2\). 4. The average MAD is \(\frac{2+5}{2}=3.5\), which gives \(10\div3.5\approx2.86\). 5. The numerical description changes substantially with the chosen scale. Because the MADs are unequal, there is no automatic Grade 7 rule that their average becomes the “correct” common MAD. 6. A clearer comparison is to report the \(10\)-unit center gap relative to each actual MAD: \(5\) times one MAD and \(2\) times the other.

Smaller-MAD scale: \(5\) units Larger-MAD scale: \(2\) units Average-MAD scale: \(\frac{10}{3.5}\approx2.86\) units These different ratios show why an averaged MAD is not an automatic common variability measure. Report the center gap relative to the two actual MADs instead.

- Compare the center gap separately with each sample's MAD. - Distinguish strong center separation from a claim that overlap is impossible. - Check every option against both the center and variability information.

1. The mean difference is \(86-70=16\). 2. Relative to Sample A's MAD, the gap is \(16\div4=4\) MADs. 3. Relative to Sample B's MAD, the gap is \(16\div5=3.2\) MADs. 4. The center gap is large relative to both MADs, but mean/MAD summaries alone do not prove that no observed values can overlap. 5. Therefore, statement b is best supported.

b) The \(16\)-point center gap is \(4\) Sample A MADs and \(3.2\) Sample B MADs, so it is large relative to both variability measures, but the summaries do not prove zero overlap.

- Compare each center with the target separately. - Use the spread measures to judge consistency. - Compare the distance between the two centers with each stated MAD rather than averaging them.

1. Each mean is \(3\,\text{g}\) from the target, so the machines are equally accurate by their sample means. 2. Machine A is more consistent because \(2<3\). 3. The mean difference is \(506-500=6\,\text{g}\). 4. The center gap is \(6\div2=3\) Machine A MADs and \(6\div3=2\) Machine B MADs.

The machines are equally accurate by their means. Machine A is more consistent. Their means differ by \(6\,\text{g}\), which is \(3\) Machine A MADs and \(2\) Machine B MADs.

- Ask what each MAD measures before combining anything. - Compare the same center difference with each actual spread measure. - Neither summing nor automatically averaging unequal MADs creates a Grade 7 pooled variability statistic.

1. The mean difference is \(18-10=8\). 2. Adding the two MADs does not create one standard common variability unit. 3. Relative to Sample A's MAD, the center gap is \(8\div3\approx2.67\) MADs. 4. Relative to Sample B's MAD, the center gap is \(8\div6\approx1.33\) MADs. 5. Because the MADs differ, report those two comparisons directly rather than summing or automatically averaging the MADs.

The claim is false because \(3+6\) is not one standard common variability unit. The \(8\)-unit center gap is about \(2.67\) Sample A MADs and \(1.33\) Sample B MADs.

- Decide which quantities describe center and which describe spread. - Use the actual shared MAD as the measuring unit. - Recalculate the ratio with a variability measure rather than a center measure.

1. The student used a measure of center as the variability unit. The measuring unit must come from a spread measure. 2. The mean difference is \(30-18=12\). 3. The samples have a shared MAD of \(4\). 4. The correct separation is \(12\div4=3\) shared-MAD units.

The error is using the average mean as a variability measure. The correct separation is \(3\) shared MADs.

- Find each dot plot's balance point and average absolute deviation. - Measure the center difference using the common MAD. - Check the displayed values before making an absolute claim about overlap.

1. Variety A has mean \(24\,\text{cm}\) and MAD \(\frac{12}{6}=2\,\text{cm}\). 2. Variety B has mean \(28\,\text{cm}\) and MAD \(\frac{12}{6}=2\,\text{cm}\). 3. The mean difference is \(28-24=4\,\text{cm}\), which is \(4\div2=2\) MADs. 4. The displayed samples share heights of \(25\), \(26\), and \(27\) centimeters. 5. Different means and a two-MAD separation do not prove zero overlap; the student's statement is false.

The sample means are \(24\,\text{cm}\) and \(28\,\text{cm}\), and both MADs are \(2\,\text{cm}\). The means are \(2\) MADs apart, but the samples overlap at \(25\), \(26\), and \(27\) centimeters. The claim of no population overlap is not justified.

- Read the left and right edges of each box. - Divide the shared interval length by each full box length separately. - Distinguish geometric length on the scale from the number of observations inside an interval.

1. Sample A's box is \([30,42]\), so its IQR is \(42-30=12\). 2. Sample B's box is \([36,44]\), so its IQR is \(44-36=8\). 3. The boxes overlap on \([36,42]\), which has length \(42-36=6\). 4. The overlap is \(\frac{6}{12}=\frac12\) of Sample A's box length and \(\frac{6}{8}=\frac34\) of Sample B's box length. 5. These fractions compare numerical interval lengths. A box plot does not show how observations are distributed inside each box, so the fractions are not percentages of data points.

The boxes overlap on \([36,42]\), a length of \(6\). This is \(\frac12\) of Sample A's IQR length and \(\frac34\) of Sample B's IQR length. Those fractions describe interval lengths, not proportions of observations.

- Find each sample's mean and then its average absolute distance from that mean. - Compare the same center difference separately with each MAD. - Use the dot plots themselves, not just the summaries, for any statement about visible overlap.

1. Sample A has mean \(50\) and MAD \(\frac{60}{6}=10\). 2. Sample B has mean \(52\) and MAD \(\frac{24}{6}=4\). 3. The mean difference is \(52-50=2\). 4. Relative to Sample A's MAD, the gap is \(2\div10=0.2\) MAD. Relative to Sample B's MAD, it is \(2\div4=0.5\) MAD. 5. The centers are close relative to either variability measure. Sample B is much more tightly clustered, and the plotted values for Sample B lie within the displayed span of Sample A.

Sample A: mean \(50\), MAD \(10\) Sample B: mean \(52\), MAD \(4\) The mean gap is \(0.2\) of Sample A's MAD and \(0.5\) of Sample B's MAD. The centers are close, Sample B is less variable, and its plotted values lie within Sample A's displayed span.

- Treat the box interval and the full whisker interval as different comparisons. - Compare the nearest box edges first. - Then identify the values shared by the two whisker intervals.

1. Team A's box covers \([14,\ 22]\), and Team B's box covers \([24,\ 32]\). 2. The boxes do not overlap; the gap is \(24-22=2\) minutes. 3. Team A's whisker interval is \([10,\ 30]\), and Team B's is \([20,\ 40]\). 4. The whisker intervals overlap from \(20\) to \(30\), a length of \(30-20=10\) minutes. 5. The middle halves are separated even though the wider displayed intervals overlap.

The boxes have a \(2\)-minute gap. The whisker intervals overlap by \(10\) minutes.

- The raw distance between the two means stays fixed. - In each situation, divide that distance by the MAD shared by the two samples. - Distinguish a comparison of center separation from a claim about exact data overlap.

1. At first, the shared MAD is \(5\), so the separation is \(10\div5=2\) shared-MAD units. 2. Later, the shared MAD is \(10\), so the separation is \(10\div10=1\) shared-MAD unit. 3. The later situation has less center separation relative to the variability. It may suggest more overlap, although exact overlap is not determined by the means and MADs alone.

The separation changes from \(2\) shared-MAD units to \(1\) shared-MAD unit. The later situation has less separation relative to variability; exact overlap is not determined by these summaries alone.

- Use the relationship between the two quartiles and the middle-half spread. - Write the second interval after finding its missing endpoint. - Identify the portion common to both intervals.

1. Since the IQR is \(Q_3-Q_1\), \(26-Q_1=10\). 2. Therefore, \(Q_1=16\), so Sample B's middle-half interval is \([16,\ 26]\). 3. The intervals overlap from \(16\) to \(20\). 4. The overlap length is \(20-16=4\).

Sample B has \(Q_1=16\), and the middle-half intervals overlap by \(4\).

- First find the difference between the two means. - Compare that same difference with each MAD separately. - Decide whether either comparison is close to one variability unit.

1. The mean difference is \(4.7-3.8=0.9\,\text{s}\). 2. Relative to Design A's MAD, the gap is \(0.9\div0.2=4.5\) MADs. 3. Relative to Design B's MAD, the gap is \(0.9\div0.4=2.25\) MADs. 4. The student's statement is incorrect: the means are more than two MADs apart using either actual variability measure.

The statement is incorrect. The mean gap is \(4.5\) Design A MADs and \(2.25\) Design B MADs.

- A uniform shift changes the mean but not the distances from that mean. - Compare the original center gap separately with the two MADs. - Repeat those comparisons after the shift rather than creating a pooled MAD.

1. Originally, the mean difference is \(38-32=6\). That is \(6\div4=1.5\) Sample A MADs and \(6\div3=2\) Sample B MADs. 2. Adding \(1\) to every Sample B value raises its mean to \(39\) and leaves its MAD at \(3\). 3. The new mean difference is \(39-32=7\). That is \(7\div4=1.75\) Sample A MADs and \(7\div3\approx2.33\) Sample B MADs. 4. The shift increases the center separation relative to either variability measure. Exact sample overlap is not determined by the summaries alone.

Sample B's new mean is \(39\), and its MAD remains \(3\). The center gap changes from \(1.5\) to \(1.75\) Sample A MADs and from \(2\) to about \(2.33\) Sample B MADs.

- Compare both the center gap and the MAD; neither quantity should be judged in isolation. - Form a center-difference-to-MAD ratio for each pair. - Do not convert the ratios into an exact overlap statement without the underlying distributions.

1. Pair 1 has center difference \(52-50=2\), so its relative separation is \(2\div5=0.4\) MAD. 2. Pair 2 has center difference \(70-50=20\), so its relative separation is \(20\div2=10\) MADs. 3. The student's claim is false: Pair 2 has a much larger center separation relative to its variability. 4. The ratios do not show the positions of individual observations, so they do not determine the exact amount of overlap.

Pair 1: \(0.4\) MAD; Pair 2: \(10\) MADs. The claim is false because Pair 2 has the much larger relative separation. Exact overlap cannot be determined from the means and MADs alone.

- Use the balance point of each dot plot to find its mean. - Average the absolute distances from each mean to find the MADs. - Compare the mean difference in MAD units with the values shared by the two displays.

1. Center A has mean \(14\) minutes and MAD \(\frac{12}{5}=2.4\) minutes. 2. Center B has mean \(20\) minutes and MAD \(\frac{12}{5}=2.4\) minutes. 3. The mean difference is \(20-14=6\) minutes. 4. The separation is \(6\div2.4=2.5\) MADs. 5. The plots share values at \(16\) and \(18\) minutes, but most of Center B's distribution lies to the right of Center A's. The ratio and display both suggest limited, not zero, overlap.

Center A: mean \(14\) minutes, MAD \(2.4\) minutes Center B: mean \(20\) minutes, MAD \(2.4\) minutes The means are \(2.5\) MADs apart. The plots overlap at \(16\) and \(18\) minutes, but the distributions are noticeably separated overall.

- Compute one center-difference-to-MAD ratio for each pair. - Order the ratios numerically, not the raw center differences. - For part c), ask what information about individual observations is missing from a mean and MAD.

1. Pair 1 has separation \((48-40)\div4=2\) MADs. 2. Pair 2 has separation \((77-70)\div7=1\) MAD. 3. Pair 3 has separation \((24-15)\div3=3\) MADs. 4. From smallest to largest relative separation, the order is Pair 2, Pair 1, Pair 3. 5. These ratios compare center differences with variability; they do not specify the locations of individual observations and therefore do not give exact overlap percentages.

a) Pair 1: \(2\) MADs; Pair 2: \(1\) MAD; Pair 3: \(3\) MADs. b) Pair 2, Pair 1, Pair 3. c) The ratios compare centers with variability but do not determine where the individual observations lie.

- Think about how a uniform shift affects a sample's center. - Distances from the shifted center stay the same. - Use the unchanged MAD to measure the center shift.

1. Adding \(9\) to every value raises the sample mean from \(27\) to \(36\). 2. A uniform shift does not change distances from the mean, so Sample B's MAD remains \(3\). 3. The mean difference is \(36-27=9\), and \(9\div3=3\) MADs.

Sample B has mean \(36\) and MAD \(3\). The sample means are \(3\) shared MADs apart.

- Find each center before measuring absolute distances from it. - Keep the two computed MADs separate. - Compare the mean difference with each actual MAD rather than averaging them.

1. Café A's mean is \(\frac{11+13+14+15+16+21}{6}=15\). 2. Its absolute deviations are \(4,\ 2,\ 1,\ 0,\ 1,\ 6\), so its MAD is \(\frac{14}{6}=\frac{7}{3}\). 3. Café B's mean is \(\frac{20+21+23+24+25+31}{6}=24\). 4. Its absolute deviations are \(4,\ 3,\ 1,\ 0,\ 1,\ 7\), so its MAD is \(\frac{16}{6}=\frac{8}{3}\). 5. The mean difference is \(24-15=9\). Relative to Café A's MAD, the gap is \(9\div\frac{7}{3}=\frac{27}{7}\approx3.86\) MADs; relative to Café B's MAD, it is \(9\div\frac{8}{3}=\frac{27}{8}=3.375\) MADs. 6. The center difference is large relative to both samples' typical absolute deviations.

Café A: mean \(15\), MAD \(\frac{7}{3}\) Café B: mean \(24\), MAD \(\frac{8}{3}\) The \(9\)-minute center gap is about \(3.86\) Café A MADs and \(3.375\) Café B MADs, so the centers are strongly separated relative to both variability measures.

- Find each dot plot's mean and average absolute distance from that mean. - Compare the center difference with each actual MAD separately. - Use the plotted shared values—not the summary statistics alone—to discuss visible overlap.

1. Sample P has mean \(108\) and MAD \(\frac{36}{6}=6\). 2. Sample Q has mean \(120\) and MAD \(\frac{24}{6}=4\). 3. The mean difference is \(120-108=12\). 4. The center gap is \(12\div6=2\) Sample P MADs and \(12\div4=3\) Sample Q MADs. 5. The displayed samples share values at \(114\) and \(118\), so the plot directly shows some overlap despite the noticeable center separation.

Sample P: mean \(108\), MAD \(6\) Sample Q: mean \(120\), MAD \(4\) The center gap is \(2\) Sample P MADs and \(3\) Sample Q MADs. The displayed samples overlap at \(114\) and \(118\).

- Count the ordered dots to locate each median and the medians of the lower and upper halves. - Use the quartiles to compare the middle-half intervals. - Compare that result with the complete horizontal regions occupied by both dot plots.

1. Group A has median \(\frac{7+7}{2}=7\), \(Q_1=\frac{5+6}{2}=5.5\), and \(Q_3=\frac{8+9}{2}=8.5\), so its IQR is \(8.5-5.5=3\). 2. Group B has median \(\frac{10+10}{2}=10\), \(Q_1=\frac{8+9}{2}=8.5\), and \(Q_3=\frac{11+12}{2}=11.5\), so its IQR is \(11.5-8.5=3\). 3. The middle-half intervals \([5.5,8.5]\) and \([8.5,11.5]\) meet only at \(8.5\). 4. The full samples overlap visibly from \(7\) to \(10\) minutes. 5. The samples have equal middle-half variability, different centers, and more full-distribution overlap than the quartile intervals alone show.

Group A: median \(7\), IQR \(3\) Group B: median \(10\), IQR \(3\) The middle-half intervals meet only at \(8.5\), while the full displayed samples overlap from \(7\) to \(10\) minutes.

- Use the edges of each box for the middle-half spread. - Find the interval common to both boxes. - Describe whisker endpoints and the marked outlier exactly as displayed instead of assuming they are all data extrema.

1. Crew A has IQR \(32-24=8\) minutes. 2. Crew B has IQR \(34-26=8\) minutes, so the IQRs are equal. 3. The boxes overlap from \(26\) to \(32\), a length of \(32-26=6\) minutes. 4. Crew A's displayed span from the lower whisker at \(20\) to the marked outlier at \(50\) is \(50-20=30\) minutes. 5. Crew B's displayed whisker span is \(38-22=16\) minutes.

Both IQRs are \(8\) minutes. The boxes overlap by \(6\) minutes. Crew A's displayed span from its lower whisker to the marked outlier is \(30\) minutes; Crew B's displayed whisker span is \(16\) minutes.

- Compute each mean before finding the average absolute deviation from it. - Compare the same center difference separately with each sample's MAD. - Use the plotted values themselves for the conclusion about visible overlap.

1. Sample A has mean \(10\). Its absolute deviations sum to \(2+1+0+1+2=6\), so its MAD is \(6\div5=1.2\). 2. Sample B has mean \(16\). Its absolute deviations sum to \(4+2+0+2+4=12\), so its MAD is \(12\div5=2.4\). 3. The mean difference is \(16-10=6\). 4. Relative to Sample A's MAD, the gap is \(6\div1.2=5\) MADs. Relative to Sample B's MAD, it is \(6\div2.4=2.5\) MADs. 5. The plotted samples share the value \(12\), so the strong center separation does not mean the displayed samples are completely separated.

Sample A: mean \(10\), MAD \(1.2\) Sample B: mean \(16\), MAD \(2.4\) The mean gap is \(5\) Sample A MADs and \(2.5\) Sample B MADs. The plot still shows a shared value at \(12\).

- Account carefully for negative values when finding each mean. - Compare the same \(9\,^{\circ}\text{F}\) center gap with each MAD separately. - Read the dot plots directly to decide whether the displayed samples share any values.

1. Region A has mean \(-4\,^{\circ}\text{F}\) and MAD \(\frac{18}{6}=3\,^{\circ}\text{F}\). 2. Region B has mean \(5\,^{\circ}\text{F}\) and MAD \(\frac{36}{6}=6\,^{\circ}\text{F}\). 3. The mean difference is \(5-(-4)=9\,^{\circ}\text{F}\). 4. Relative to Region A's MAD, the gap is \(9\div3=3\) MADs. Relative to Region B's MAD, it is \(9\div6=1.5\) MADs. 5. The plots share values at \(-5\,^{\circ}\text{F}\) and \(-1\,^{\circ}\text{F}\), so the displayed samples still overlap.

Region A: mean \(-4\,^{\circ}\text{F}\), MAD \(3\,^{\circ}\text{F}\) Region B: mean \(5\,^{\circ}\text{F}\), MAD \(6\,^{\circ}\text{F}\) The mean gap is \(3\) Region A MADs and \(1.5\) Region B MADs. The displayed samples share values at \(-5\,^{\circ}\text{F}\) and \(-1\,^{\circ}\text{F}\).