Fall 2026 Rehearsal — Free-Response Worked Answers and Rubrics

One valid question is: Does using the new mulch rather than standard mulch reduce the population mean weekly irrigation volume, in liters, for tomato plots similar to these volunteer plots under the specified four-week moisture protocol? This identifies the response and a lower-mean directional comparison. Random assignment can support a causal comparison for similar plots; volunteer recruitment does not make these plots representative of all gardens.

Accept a question phrased in terms of mean(new)−mean(standard)<0, or the equivalent mean(standard)−mean(new)>0, with the response and scope clear. A question about which mulch every plant prefers or about an unmeasured outcome is not equivalent.

Use full-sun and partial-shade plots as the two blocks. Within the 60 full-sun plots, label plots 01–60 and use a random-number generator to select 30 distinct labels for new mulch; the other 30 receive standard mulch. Independently repeat the procedure within the 60 partial-shade plots. There are 30 replicates per treatment in each block, and 60 plots per treatment overall. Compare the mean water use of the two treatments within each sun block before summarizing the treatment effect.

Blocking is appropriate because sunlight affects evaporation and water use. Comparing both mulches at each sunlight level separates this known source of variation from the mulch effect and can improve precision. Do not make sunlight itself the mulch assignment.

Use the same moisture target, calibrated irrigation-volume measurement, observation schedule, cultivar, and mulch application depth wherever applicable. Give technicians a fixed watering protocol and code treatment records so the person recording volumes can be unaware of the mulch code if feasible. Other reproducible direct controls are acceptable. A claim that random assignment guarantees identical treatment groups is too strong.

Mulch type would be completely confounded with sunlight level: every new-mulch plot is sunny, while every standard-mulch plot is shaded. Sunlight can itself alter water use, so a group difference could reflect mulch, sunlight, or both. The design has no within-sunlight treatment comparison that can separate these explanations. More plots or a more precise average would not remove this structural confounding.

Accept equivalent explanations that explicitly connect sunlight both to treatment assignment and to the irrigation response. Naming “bias” without explaining this connection is insufficient.

Under the common measurement protocol, the randomized comparison supports the conclusion that new mulch causes lower mean weekly irrigation use than standard mulch for tomato plots similar to those volunteered. It does not establish an effect in every plot or representation of every garden, climate, soil type, or cultivar. The volunteers were not randomly sampled from a wider garden population.

An ethical safeguard is to obtain the owners’ informed permission, explaining random allocation to either permitted mulch, the measurements, and how any plant-health concern will be addressed. Protect identifying plot/owner information in public reports. Either a concrete informed-consent procedure or a concrete privacy/welfare safeguard earns the safeguard point; randomization by itself is not an ethical safeguard.

Label the horizontal axis “Completion time (minutes)” with equally spaced integer positions from 4 through 12. Stack 4 dots at 4, 8 at 5, 12 at 6, 8 at 7, 4 at 8, and 4 at 12. Leave positions 9, 10, and 11 empty. Each dot represents one attendee, for a total of 40 dots.

Accept horizontal rather than vertical stacking if the time axis remains evenly spaced and all 40 values are represented. Joining the six time values as evenly spaced categories would incorrectly hide the 8-to-12 gap.

The mean is $\bar x=[4(4)+5(8)+6(12)+7(8)+8(4)+12(4)]/40=264/40=6.6$ minutes. The cumulative counts are 4 through time 4, 12 through time 5, and 24 through time 6. Therefore both the 20th and 21st ordered times are 6, and the median is $(6+6)/2=6$ minutes.

Accept an equivalent expansion of all 40 times. Averaging the six distinct time labels without weighting their frequencies does not give the attendee mean.

The lower half consists of positions 1–20; its 10th and 11th values are both 5, so Q1=5 minutes. The upper half consists of positions 21–40; the 30th and 31st overall values are both 7, so Q3=7 minutes. Thus IQR=7−5=2 minutes. The lower fence is $5-1.5(2)=2$ minutes and the upper fence is $7+1.5(2)=10$ minutes. All four 12-minute observations are upper outliers. There are no lower outliers because the minimum, 4, is above 2.

Use strict below/above comparisons. Neither 4 nor 8 is an outlier; the fences are cutoffs, not observed minimum and maximum values.

The median is lower in A (6 minutes) than in B (7 minutes), so a typical attendee finishes about 1 minute sooner in A by the median. The IQR is 2 minutes in A and 9−5=4 minutes in B, so A’s middle half is less variable. The 12-minute outliers in A can pull the mean and standard deviation upward; the median and IQR are resistant to those extreme times and describe the main group more stably.

Do not infer that every A attendee is faster or that workshop membership causes the difference. The question describes these two workshop groups rather than a randomized treatment comparison.

There are 4 attendees at 8 minutes and 4 at 12, so $P(T\ge8)=8/40=0.20$. There are $4+8+12=24$ times at or below 6, giving $24/40=0.60$, or the 60th percentile under the stated at-or-below convention. Including all 12 ties at 6 is essential here; counting only strictly smaller times would give 30%, which answers a different question.

Let $\mu_d$ be the population mean old-minus-new driving-time difference, in minutes, for drivers at this depot under the comparable-route protocol. A reduction by the new method corresponds to a positive old-minus-new difference. Test $H_0:\mu_d=0$ against $H_a:\mu_d>0$.

Accept an explicitly reversed new-minus-old parameter with a less-than alternative if every later sign and tail is consistent. Hypotheses must concern the population mean difference, not the observed sample mean 4.6.

Use a paired t-test, equivalently a one-sample t-test on the 20 old-minus-new differences, with df=20−1=19. The same drivers receive both methods, so there is one sample of paired differences rather than two independent samples of driving times.

The drivers were selected by a simple random sample, and method order was randomized. For without-replacement sampling, 20 is less than 10% of 6,000 (=600). Because n=20<30, examine the distribution of differences: it is roughly symmetric and free of outliers, supporting the t procedure. Check the differences, not separate old/new marginal distributions. The context supplies independence between drivers.

The standard error is $SE=7.2/\sqrt{20}\approx1.60997$ minutes. The test statistic is $t=(4.6-0)/(7.2/\sqrt{20})\approx2.85720$. For a t-distribution with 19 degrees of freedom, $P(T\ge2.85720)\approx0.00504$. This is one upper tail, because the alternative is $\mu_d>0$.

Using the supplied rounded table, the observed t lies between 2.539 and 2.861, so $0.005<p<0.01$ is an acceptable bound. Accept a correctly reversed sign and lower tail for a reversed parameter. Doubling the tail would answer a two-sided alternative instead.

Because p≈0.00504<0.05 (and the table bound is also entirely below 0.05), reject H0. There is convincing statistical evidence that the population mean old-minus-new driving-time difference is positive for drivers at this depot under the comparable-route protocol. Equivalently, the new method lowers mean driving time under these study conditions.

The SRS supports generalization to depot drivers; it does not represent drivers at every other depot. Randomized method order and the controlled paired routes support a treatment comparison under the protocol. The result concerns an average and neither proves the parameter’s exact value nor guarantees a decrease for every driver. A conclusion following a correctly reversed definition is equivalent.

Use a two-sample t-interval for the difference between population mean delivery times, A minus B. The researcher samples distinct customer populations independently; there is no matching of an A customer to a B customer and no repeated treatment on the same customer.

Both samples are simple random samples and independent. Each sample of 100 is less than 10% of its population of more than 20,000. Both sample sizes are at least 30; A’s bounded symmetric frequency distribution and B’s stated absence of strong skewness/outliers further support the approximation. Unknown population SDs require estimated standard errors and t inference.

Calling the samples paired solely because both have 100 customers is a material design error. A one-mean procedure estimates a different parameter.

The at-most-2-day event includes the 10 one-day and 20 two-day deliveries: $P(Y\le2)=(10+20)/100=0.30$. The sample mean is $\bar y_A=[1(10)+2(20)+3(40)+4(20)+5(10)]/100=300/100=3.0$ days. Both calculations use the 100 customer observations, not the five distinct time labels as equally likely outcomes.

The 0.30 is a descriptive probability for selecting from this finite sample, not the exact unknown population probability.

The point estimate is $3.0-3.6=-0.6$ day. The estimated standard error is $\sqrt{1.101^2/100+1.4^2/100}\approx0.17811$ day. The margin of error is $1.653(0.17811)\approx0.29441$ day. Thus the 90% interval is $-0.6\pm0.29441$, or approximately $(-0.8944,-0.3056)$ day.

We are 90% confident that depot A’s population mean delivery time is between 0.3056 and 0.8944 day shorter than depot B’s, for the defined customers and orders from last month. Since every plausible A-minus-B mean difference in this interval is negative and zero is excluded, the interval provides convincing evidence that A’s population mean delivery time is lower. This observational comparison of depots does not establish that moving an order to A causes the improvement.

Accept equivalent A-minus-B interpretations with endpoints rounded reasonably, or an explicitly reversed B-minus-A interval and direction. Using the exact A SD from the frequency table changes the displayed endpoints negligibly. A conservative df=99 calculation gives a slightly wider interval and is acceptable if the alternative critical value and method are explicitly justified. Do not describe the endpoints as containing 90% of individual delivery times.

The predicted time is $\hat y=1+0.08(25)=3$ days. Since the observed time is 4 days, the residual is $4-3=+1$ day: the model underpredicts this parcel’s delivery time by 1 day. The 25-hour x value is within the observed range, so this prediction is interpolation.

The coefficient of determination is $r^2=0.75^2=0.5625$. About 56.25% of the observed variation in delivery time (the response, in days) for the 40 parcels is explained by the linear relationship with order-to-dispatch time (the explanatory variable, in hours). It does not mean that 56.25% of each parcel’s time is caused by dispatch delay, and it is not the fraction of dispatch-time variation explained by delivery time.