Justifying claims from a mean CI: practice

(a) The sample mean sits at the center: $\bar{x}$ ("x-bar") $= \frac{152.4 + 161.6}{2} = \frac{314}{2} = 157.0$ grams. The margin of error is half the width: $ME = \frac{161.6 - 152.4}{2} = \frac{9.2}{2} = 4.6$ grams. So the interval is $157.0 \pm 4.6$, built at $df = 30 - 1 = 29$.

Statement 1 is not justified. The interval estimates $\mu$, the mean weight of all apples in the harvest, not the spread of individual apples. Its half-width 4.6 g is built from the standard error $\frac{s}{\sqrt{30}}$, which is smaller than the apple-to-apple standard deviation $s$ by a factor of $\sqrt{30} \approx 5.48$. A range holding 95% of individual apples would be several times wider.

Statement 2 is not justified. Both bounds are now fixed numbers and $\mu$ is a fixed number, so this interval either captures $\mu$ or it does not. The correct version moves the 95% onto the method: if the orchard repeated this sampling many times with 30 apples each and built an interval from every sample, about 95% of those intervals would capture the true mean weight.

Statement 3 is justified. It names the confidence level, both bounds with units, the parameter, and the population. This is the sentence the exam wants.

Statement 4 is justified. Since $152.4 < 158 < 161.6$, the value 158 g lies inside the interval, so it remains a plausible value for the true mean weight. Note what it does not say: 158 is no more supported than 155 or 160, which are equally inside.

Statement 5 is not justified as written. An interval never proves anything about a single value. Say instead that 150 g lies outside the interval, so it is not a plausible value for the true mean weight at the 95% level.

Statement 6 is justified. Every value in the interval exceeds 150, and 150 sits below the lower bound of 152.4, so there is convincing evidence that the true mean weight is greater than 150 g. Statements 5 and 6 refer to the same fact about 150; only statement 6 words it as the evidence it is.

Plan. One-sample $t$ interval for a mean, since $\sigma$ is unknown and $s$ stands in for it. Random: each brand's 20 capsules are a random sample. 10%: 20 capsules is well under 10% of all capsules the brand produces. Normal or large sample: $n = 20 < 30$, so the stated symmetric dotplot with no outliers is what carries the condition. Both brands share $n = 20$ and $s = 11$, so they share a standard error and a critical value.

Common arithmetic. $SE = \frac{s}{\sqrt{n}} = \frac{11}{\sqrt{20}} = \frac{11}{4.4721} = 2.4597$ mg. With $df = 20 - 1 = 19$, the 95% column of the $t$-table gives $t^* = 2.093$, so $ME = 2.093 \times 2.4597 = 5.1481$ mg for both brands.

(a) Brand A: $493 \pm 5.1481 = (487.852,\ 498.148)$ mg. We are 95% confident this interval captures the true mean magnesium content of Brand A capsules.

(a) Justify. The labeled value 500 mg lies above the upper bound 498.148, so it is outside the interval and is not a plausible value for the true mean. There is convincing evidence that Brand A capsules average less than the labeled 500 mg, by somewhere between about 1.9 mg and 12.1 mg.

(b) Brand B: $496 \pm 5.1481 = (490.852,\ 501.148)$ mg. Now $490.852 < 500 < 501.148$, so 500 mg lies inside the interval and remains plausible. There is not convincing evidence against Brand B's label.

(c) Correct him. Failing to rule a value out is not the same as establishing it. The interval leaves every value from 490.852 mg to 501.148 mg plausible, and 494 mg and 499 mg sit in there on exactly the same footing as 500 mg. The sample mean 496 mg is in fact the single best estimate, not 500.

Notice the size of what separated the two answers. The two sample means differ by 3 mg, less than a third of one standard deviation, yet one brand's label is contradicted and the other's is not. A conclusion that flips on a 3 mg shift is a reason to report the interval itself rather than only the verdict.

(a) Each water sample is measured by both instruments, so every reading from A has one specific partner from B: the reading taken on that same sample. That is the definition of pairing, and it is the right design here, because water samples differ a lot from each other and pairing removes that variation from the comparison. The parameter is $\mu_d$, the true mean difference in nitrate reading (Instrument A minus Instrument B) for water samples measured by both instruments.

(b) Collapse to one column. $d = A - B$ gives $0.2,\ -0.2,\ 0.2,\ -0.2,\ 0.3,\ -0.2,\ 0.3,\ -0.1$ mg/L. The sum is $0.3$, so $\bar{x}_d = \frac{0.3}{8} = 0.0375$ mg/L.

(b) Standard deviation of the differences. Deviations from 0.0375 are $0.1625,\ -0.2375,\ 0.1625,\ -0.2375,\ 0.2625,\ -0.2375,\ 0.2625,\ -0.1375$, and their squares sum to $0.37875$. So $s_d = \sqrt{\frac{0.37875}{8 - 1}} = \sqrt{0.054107} = 0.232609$ mg/L.

(b) Conditions and interval. Random: the 8 water samples were randomly selected. 10%: 8 is under 10% of the water samples available. Normal or large sample: $n = 8 < 30$, so the roughly symmetric dotplot of differences with no outliers carries it. Then $SE = \frac{0.232609}{\sqrt{8}} = \frac{0.232609}{2.8284} = 0.08224$ mg/L, and with $df = 8 - 1 = 7$ the 95% column gives $t^* = 2.365$, so $ME = 2.365 \times 0.08224 = 0.194497$ mg/L.

(b) Interval: $0.0375 \pm 0.194497 = (-0.157,\ 0.232)$ mg/L. We are 95% confident this interval captures the true mean difference in reading between the two instruments.

(c) What is wrong. The interval containing 0 means a mean difference of zero has not been ruled out; it does not mean zero has been shown. The same interval leaves $-0.15$ and $+0.23$ mg/L just as plausible as 0, so agreement is one of many surviving possibilities, not a finding. Failing to detect a difference and demonstrating there is none are different claims, and only the first is supported.

(c) Correct version: because 0 lies inside the interval from $-0.157$ to $0.232$ mg/L, a true mean difference of zero remains plausible, so these data do not give convincing evidence that the two instruments read differently on average. If the lab needs to establish agreement to within some tolerance, it needs a much narrower interval, which means more paired samples, not a different conclusion from these eight.

(a) Differences $d = \text{usual} - \text{new}$: $4,\ 3,\ 4,\ 5,\ 3,\ 5,\ 3,\ 4,\ 4$ seconds. The sum is 35, so $\bar{x}_d = \frac{35}{9} = 3.8889$ seconds.

(a) Spread of the differences. Deviations from 3.8889 are $0.1111,\ -0.8889,\ 0.1111,\ 1.1111,\ -0.8889,\ 1.1111,\ -0.8889,\ 0.1111,\ 0.1111$, whose squares sum to $4.8889$. So $s_d = \sqrt{\frac{4.8889}{9 - 1}} = \sqrt{0.61111} = 0.781736$ seconds.

(a) Conditions and interval. Random: the 9 runners are a random sample of club runners, and the order of the two runs was randomized within each runner. 10%: 9 is under 10% of the club. Normal or large sample: $n = 9 < 30$, so the roughly symmetric dotplot of differences with no outliers carries it. Then $SE = \frac{0.781736}{\sqrt{9}} = \frac{0.781736}{3} = 0.260579$ s, and at $df = 9 - 1 = 8$ the 95% column gives $t^* = 2.306$, so $ME = 2.306 \times 0.260579 = 0.600894$ s.

(a) Interval: $3.8889 \pm 0.600894 = (3.288,\ 4.490)$ seconds. Because 0 is not in this interval and the whole interval is positive, there is convincing evidence that runners are faster on this course in the new shoe, by a true mean of somewhere between about 3.3 and 4.5 seconds. Because the order of the two runs was randomized within each runner, the difference is attributable to the shoe rather than to warm-up or fatigue order.

(b) Two-sample version. $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{19.069^2}{9} + \frac{18.722^2}{9}} = \sqrt{40.403 + 38.946} = \sqrt{79.349} = 8.9078$ seconds. The conservative $df = \min(9, 9) - 1 = 8$ again gives $t^* = 2.306$, so $ME = 2.306 \times 8.9078 = 20.541$ s.

(b) Interval: $(246.889 - 243.000) \pm 20.541 = 3.8889 \pm 20.541 = (-16.652,\ 24.430)$ seconds. This interval contains 0, so it would produce the conclusion that there is not convincing evidence of any difference between the two shoes.

(c) Why they disagree. The point estimate is $\frac{35}{9} = 3.8889$ seconds either way, because the mean of the differences equals the difference of the means. Only the standard error changes: 0.2606 s paired against 8.9078 s unpaired, a factor of about 34. Runners differ from one another by roughly 19 seconds on this course, and the two-sample standard error is built from that runner-to-runner spread. Subtracting within each runner cancels it, leaving only the shoe-to-shoe variation, which is well under one second.

(c) Which is valid. The paired analysis. The two-sample procedure requires the two samples to be independent, and they are not: the same nine people supply both columns, so runner 4 being slow shows up in both. The unpaired interval is not merely a wider answer to the same question, it is a procedure whose condition fails on this design. The paired conclusion in part (a) is the one to report.

(a) Each volunteer uses exactly one version, so a time from the Version A group has no particular partner in the Version B group. There is nothing to subtract within, and the two groups of 18 are independent by construction of the random assignment. The parameter is $\mu_1 - \mu_2$, the difference in true mean completion time between Version A and Version B for volunteers like these.

(b) Conditions. Random: volunteers were randomly assigned to the two versions, which is what makes the groups comparable and the two samples independent. Normal or large sample: $n_1 = n_2 = 18 < 30$, so the roughly symmetric dotplots with no outliers carry the condition for both groups. The 10% condition concerns random sampling from a population and is not the relevant check for a randomized experiment on recruited volunteers.

(b) Standard error. $SE = \sqrt{\frac{5.2^2}{18} + \frac{4.8^2}{18}} = \sqrt{\frac{27.04}{18} + \frac{23.04}{18}} = \sqrt{1.50222 + 1.28} = \sqrt{2.78222} = 1.668$ minutes.

(b) Critical value and interval. Conservative $df = \min(18, 18) - 1 = 17$, so the 95% column gives $t^* = 2.110$ and $ME = 2.110 \times 1.668 = 3.5195$ minutes. The point estimate is $24.6 - 20.1 = 4.5$ minutes, so the interval is $4.5 \pm 3.5195 = (0.981,\ 8.019)$ minutes. Technology using $df = 33.78$ reports $t^* = 2.033$ and $(1.109,\ 7.891)$, a slightly narrower interval with the same conclusion.

(b) Justify. Because 0 is not in the interval and both bounds are positive, there is convincing evidence that mean completion time is longer on Version A than on Version B, by a true mean of somewhere between about 1.0 and 8.0 minutes. Note how loose that is: the data settle the direction firmly but leave the size uncertain across an eightfold range.

(c) Cause: yes, within this study. Volunteers were randomly assigned to the versions, so the two groups differ systematically only in which version they used, and a difference this large is not plausibly explained by chance assignment alone. That licenses a causal claim about the app version for these volunteers.

(c) Generalization: no. The 36 people were volunteers, not a random sample from any wider population, so nothing here supports extending the result to the general public. Random assignment buys causation; random selection buys generalization, and only the first is present. This is the most commonly missed half of a justification on this material.

(a) Conditions. Random: two independent random samples, one from each supplier's delivery. 10%: 12 and 15 seedlings are each under 10% of the delivery they came from. Normal or large sample: $n_1 = 12$ and $n_2 = 15$ are both under 30, so the roughly symmetric dotplots with no outliers carry the condition for both groups.

(a) Standard error. $SE = \sqrt{\frac{1.80^2}{12} + \frac{1.65^2}{15}} = \sqrt{\frac{3.24}{12} + \frac{2.7225}{15}} = \sqrt{0.27 + 0.1815} = \sqrt{0.4515} = 0.671937$ cm.

(a) Critical value and interval. Conservative $df = \min(12, 15) - 1 = 11$, so the 95% column gives $t^* = 2.201$ and $ME = 2.201 \times 0.671937 = 1.478934$ cm. The point estimate is $28.40 - 27.95 = 0.45$ cm, so the interval is $0.45 \pm 1.478934 = (-1.029,\ 1.929)$ cm. Technology using $df = 22.7$ gives $t^* = 2.070$ and $(-0.941,\ 1.841)$, which contains 0 as well.

(b) What it justifies. Zero lies inside the interval, so a true mean difference of zero remains plausible, and there is not convincing evidence that the two suppliers' seedlings differ in mean height after four weeks.

(b) What it does not justify. The buyer's sentence claims the means are the same, which the interval cannot support. Every value from $-1.029$ to $1.929$ cm is plausible, so Supplier 1 running a full centimeter shorter and Supplier 2 running nearly two centimeters shorter are both still on the table alongside zero. A difference of zero is one plausible value out of many, and with samples of 12 and 15 the interval is simply too wide to separate them. Report it as no convincing evidence of a difference, never as evidence of no difference.

(c) Pairing was not available. A paired design needs a specific link joining one unit in each group, such as the same subject measured twice or two units matched on a variable before treatment. These are two separate batches of seedlings from two different suppliers, so seedling 3 from Supplier 1 has no partner among Supplier 2's fifteen. Numbering the rows 1 through 12 and 1 through 15 and subtracting would invent a pairing the design does not contain, and it would leave three of Supplier 2's seedlings unmatched.

(c) What would help instead. Pairing is not the lever here; sample size is. The interval is 2.96 cm wide against a point estimate of 0.45 cm, so the study has little chance of detecting a difference this small. Larger samples from both suppliers would shrink the standard error and narrow the interval.

(a) Conditions. Paired design, so this is a one-sample $t$ interval on the 14 differences. Random: 14 randomly selected routes, with the order of the two drives randomized within each route. 10%: 14 routes is under 10% of the company's routes. Normal or large sample: $n = 14 < 30$, so the roughly symmetric dotplot of differences with no outliers carries it.

(a) Arithmetic. $SE = \frac{s_d}{\sqrt{n}} = \frac{5.6}{\sqrt{14}} = \frac{5.6}{3.7417} = 1.496663$ minutes. With $df = 14 - 1 = 13$, the 95% column of the $t$-table gives $t^* = 2.160$, so $ME = 2.160 \times 1.496663 = 3.232792$ minutes.

(a) Interval: $4.2 \pm 3.232792 = (0.967,\ 7.433)$ minutes.

(b) Statistical justification. "We are 95% confident that the interval from 0.967 to 7.433 minutes captures $\mu_d$, the true mean reduction in driving time per route (old software minus new). Because 0 is not contained in this interval and both bounds are positive, there is convincing evidence that the new software reduces mean driving time on routes like these. Because the order of the two drives was randomized within each route, the reduction is attributable to the software rather than to which drive came first."

(c) For the manager. "On 14 of our routes, the new software saved an average of 4.2 minutes per route, and the analysis is confident the real average saving is a genuine one rather than luck of the draw. How big that saving is remains loosely pinned down: the data are consistent with anything from about 1 minute to about 7.4 minutes per route, so if the licenses only pay for themselves at 5 minutes a route, this study has not settled the question." That last clause is the part a decision-maker actually needs, and it is invisible if you report only that the result was significant.

(d) "There is a 95% probability that the true saving is between 0.967 and 7.433 minutes." Wrong, because both bounds and $\mu_d$ are fixed numbers once the data are in, so this interval either captures $\mu_d$ or it does not. The 95% describes how often intervals built this way capture the parameter across repeated samples.

(d) "The study proves the new software saves 4.2 minutes per route." Wrong twice. An interval does not prove a value, and 4.2 is the sample estimate, not the parameter; the whole point of the interval is that values from about 1 to about 7.4 minutes are all plausible for $\mu_d$.

(d) "95% of routes will save between 0.967 and 7.433 minutes." Wrong, because the interval estimates a mean, not the spread of individual routes. Its width comes from $\frac{s_d}{\sqrt{14}} = 1.4967$, while individual routes vary with $s_d = 5.6$ minutes, nearly four times as much, so plenty of individual routes will fall outside these bounds and some will get slower.

(a) The two halves of a panel come from the same sheet of steel, so they share thickness, alloy, and surface history. That link makes each coated half the natural partner of one uncoated half, and subtracting within a panel removes the panel-to-panel variation from the comparison. The parameter is $\mu_d$, the true mean reduction in corrosion depth (uncoated minus coated) for panels treated this way.

(b) Conditions. Random: 12 randomly selected panels, with the coated half chosen at random within each panel. 10%: 12 panels is under 10% of the panels available. Normal or large sample: $n = 12 < 30$, so the roughly symmetric dotplot of differences with no outliers carries it.

(b) Arithmetic. $SE = \frac{0.21}{\sqrt{12}} = \frac{0.21}{3.4641} = 0.060622$ mm. With $df = 12 - 1 = 11$, the 95% column gives $t^* = 2.201$, so $ME = 2.201 \times 0.060622 = 0.133429$ mm, and the interval is $0.34 \pm 0.133429 = (0.207,\ 0.473)$ mm.

(b) Justify. Zero lies below the lower bound of 0.207, so 0 is not a plausible value for $\mu_d$ and every plausible value is a reduction. There is convincing evidence that the coating reduces mean corrosion depth, by a true mean of somewhere between about 0.21 mm and 0.47 mm. Because the coated half was randomly assigned within each panel, the reduction is attributable to the coating.

(c) Now judge the claimed value 0.5, not 0. The claim is that $\mu_d$ is at least 0.5 mm, and the entire interval from 0.207 to 0.473 lies below 0.5. No value of 0.5 or greater is plausible at this level, so there is convincing evidence against the manufacturer's claim: the coating works, but by less than advertised.

(d) At 99%, only $t^*$ changes. The $df = 11$ row gives $t^* = 3.106$, so $ME = 3.106 \times 0.060622 = 0.188291$ mm and the interval is $0.34 \pm 0.188291 = (0.152,\ 0.528)$ mm.

(d) Answer to (b) is unchanged. Zero is still below the lower bound of 0.152, so there is still convincing evidence the coating reduces mean corrosion depth.

(d) Answer to (c) flips. The upper bound 0.528 now exceeds 0.5, so $0.152 < 0.5 < 0.528$ and a reduction of 0.5 mm is plausible at the 99% level. There is no longer convincing evidence against the manufacturer's claim.

Take the lesson from (d) seriously. Plausibility is always relative to a confidence level, so a justification that names a value must name the level too. Raising the level widens the interval, which makes the method capture $\mu_d$ more often across repeated samples but leaves more claimed values standing. Reporting "0.5 mm is not plausible" without "at 95% confidence" hides the fact that the verdict reverses one column over on the $t$-table.