Fall 2026 Rehearsal — Multiple-Choice Worked Answers

The question identifies the quantitative variable (weekly cycling commute miles), the population (all registered cyclists in the town), and the estimation goal (an interval for the population mean). A single individual’s exact value and a total for the particular sample do not address this population-mean goal. The data do not support a universal causal claim about residents.

The response is support (yes/no), the parameters are the two population proportions, and the question asks whether they differ. Random sampling permits population comparisons, but the planner does not assign residence, so the study cannot identify an effect of moving districts. Daily use and strength of opinion are not measured.

The agency randomly selects whole groups (buildings) and includes every household in the selected groups, so this is cluster sampling. Stratified sampling would select households from every stratum. Not every set of the same number of households has an equal chance of selection, as it would in a household-level simple random sample. No fixed sampling interval is used.

This is voluntary response sampling: respondents select themselves. Strongly dissatisfied riders may be more likely to submit feedback than other riders, producing a systematic overestimate of dissatisfaction. That direction is a plausible risk, not a guaranteed outcome. A large sample can reduce random variability without correcting selection bias.

Blocking by starting age places similar kittens together, and random assignment of both diets within each block allows fair diet comparisons at each age. Assigning diets by age would confound diet and age. Selecting promising kittens for one diet introduces a systematic group difference, and using only one diet removes the treatment comparison.

Random assignment reduces potential confounding and supports a cause-and-effect comparison of the programs. The volunteer recruitment limits population generalization to adults similar to those studied; it does not establish representation of all adults. The conclusion concerns an average effect and does not promise benefit for every participant.

The sample mean is 4 minutes. The squared deviations sum to $4+0+0+4=8$, so $s=\sqrt{8/(4-1)}=\sqrt{8/3}\approx1.633$ minutes. Dividing by 4 gives a population-formula value of 1.414; 2.667 is the sample variance, and 4 is the range.

A stratified random sample using commute mode ensures that both distinct groups contribute randomly selected students. The eventual overall estimate should respect the population sizes of the two strata. Convenience selection, omitting bus commuters, and voluntary response do not satisfy the stated goal.

The main peak is at 1 minute, and most times lie from 0 to 3 minutes. The lone 9-minute value extends the tail to the right, with no times between 3 and 9. Equal counts at the two extremes do not make the distribution symmetric, and one dot per visit does not imply equal frequencies at each time.

Route A’s median is 12 minutes, compared with 14 for B. Both IQRs are 4 minutes: 14−10 and 16−12. A’s range is 24 minutes, compared with 10 for B. A lower median indicates a lower typical time by this measure, not that every delivery on A is faster.

The selected question names the binary enrollment response, asks about an increase in a proportion, and uses a population of employees similar to the volunteers. Random assignment supports a causal effect of receiving the invitation. Volunteer recruitment does not support a claim about every employee or company, and merely reporting a sample proportion does not address the causal investigative purpose.

Conditioning on A restricts the denominator to the 120 app users. Of them, 72 claimed a refund, so $P(R\mid A)=72/120=0.600$. Dividing by all 200 gives the joint probability, 92/200 gives the unconditional refund probability, and 72/92 reverses the conditioning.

Count the union once: $120+92-72=140$ passengers, so $P(A\cup R)=140/200=0.700$. The overlap must be subtracted because it was counted in both margins. Adding without subtraction gives the impossible 1.060; 0.360 is the intersection, and 0.340 counts the app-only and refund-only cells while excluding passengers in both.

Under independence, $P(A\cap R)=P(A)P(R)=(120/200)(92/200)=0.276$. The table gives $72/200=0.360$, so these events are not independent in this finite passenger population. The calculation is a hypothetical independence benchmark; it does not change the actual table or imply app use causes refunds.

A binomial count requires a fixed number of independent trials with two outcomes and a constant success probability. Here drawing a defective battery changes the defect proportion left in the batch. The two classifications and fixed number of draws satisfy two binomial features, but they do not repair the dependence and changing probability. The sample is too large relative to the batch to justify an independence approximation.

The mean is $\mu_X=0(0.10)+1(0.30)+2(0.40)+3(0.20)=1.7$ requests. Also $E(X^2)=0+0.30+1.60+1.80=3.70$. Thus $\sigma_X=\sqrt{3.70-1.7^2}=\sqrt{0.81}=0.90$ requests. The variance is 0.81, while 1.7 is the mean; taking $\sqrt{E(X^2)}$ alone fails to center at the mean.

The heaviest 10% lie to the right of the 90th percentile. The standard-normal cutoff is $z\approx1.28155$, so $x=500+8(1.28155)\approx510.25$ grams. The 489.75-gram cutoff marks the lightest 10%; 512.82 grams marks about the heaviest 5%, and one SD above the mean does not leave 10% in the upper tail.

There are $87+27+6+1+0=121$ simulated trials with at least 4 red results, out of 1,000. The simulated estimate is $121/1000=0.121$. The probability of exactly 4 is estimated by 0.087; 0.879 is the complement. This is an empirical estimate from the displayed simulation, not the exact binomial probability.

Within North, the walking relative frequency is 30/100=0.30. Within South, it is 30/200=0.15. Equal counts can represent different proportions when sample sizes differ. The combined walking proportion is 60/300=0.20, and it is not either district’s conditional proportion. These descriptive sample comparisons alone do not establish a population difference or causation.

Each plotted value is one mean from a newly generated sample of 36 from the same population. Repetition approximates the sampling distribution of $\bar X$; it does not plot individual visit times or change the population mean. The mean remains near 8, while the sample means vary less than individuals, with SD $6/\sqrt{36}=1$ minute.

The sample was selected randomly, and $150\le0.10(5000)=500$. However, $n\hat p=9<10$, even though the 141 observed failures exceed 10. A proportion interval uses observed success and failure counts; a total sample size above 30 alone does not establish this condition.

The binary response defines a proportion in each district, and the requested parameter is pA−pB. With two independent samples and an estimation goal, the appropriate procedure is a two-sample z-interval for this difference. A test does not produce the requested interval; t procedures concern quantitative means, and no pairing or panel count is present.

The p-value is calculated under the null model p=0.40 and counts outcomes at least as far in the alternative’s greater-than direction. It is not a probability assigned to the hypothesis itself, a household-level composting proportion, or a prediction for one future household.

The sample proportion is $118/200=0.59$. The null-model standard deviation is $\sqrt{0.50(0.50)/200}$, so $z=(0.59-0.50)/\sqrt{0.25/200}\approx2.546$. The greater-than alternative uses the upper standard-normal tail, about 0.0055. A test uses the null proportion in the denominator; substituting 0.59 yields the first distractor. The numerator is positive, and the p-value is neither the central area nor a two-sided area.

A Type II error is failing to reject a false null hypothesis. If p=0.05, the greater-than alternative is true, but leaving the line running is the fail-to-reject decision. Stopping a line when p=0.02 would be a Type I error. The other two combinations are correct decisions for their stated true proportions.

Every value in the interval for pA−pB is positive, and 0 is excluded. The interval therefore supports a higher population proportion in A, with the difference estimated at 4 to 18 percentage points. It does not compare insulation amounts for individuals or establish a causal effect of residence.

One random sample from one population is classified by two categorical variables, so the question of association calls for a chi-square test for independence. A homogeneity design uses independent samples from multiple populations or groups to compare categorical distributions. Neither a quantitative mean difference nor a single binary proportion captures the question here.

The program-B row total is 80 and the not-pass column total is 80. The null expected count is $(80)(80)/200=32$. The observed count in this cell is 36; expected counts are calculated from the margins under the null model, rather than copied from observed cells. All six expected counts in this table are greater than 5.

The before and after measurements are matched by machine, so they produce one sample of 12 quantitative differences. A one-sample t-test on those differences tests the population mean difference when its standard deviation is unknown. Treating the two columns as independent discards the pairing, and categorical procedures do not fit decibel measurements.

Both t-distributions are symmetric and bell-shaped about 0. Smaller degrees of freedom produce heavier tails and a lower peak; as degrees of freedom increase, the t-distribution approaches the standard normal. Neither distribution is uniform, and finite-df t-distributions are not exactly standard normal.

The standard error is $4/\sqrt{16}=1$ minute, so the interval is $23\pm2.131(1)$, or approximately (20.87, 25.13) minutes. The population SD is unknown, making t appropriate. The first option uses 1.96 instead of the supplied t critical value; the third divides by n instead of $\sqrt n$. The interval estimates the population mean, not a range containing 95% of individual waiting times.

The central limit theorem improves the normal approximation as sample size increases, but an extremely skewed population can require a sample much larger than 30. Random selection does not change the population’s shape or guarantee the finite-sample approximation. The claim is about the distribution of repeated sample means, not the shape of one sample.

The null expresses equal population means, μA−μB=0. A lower mean for A corresponds to μA−μB<0. Hypotheses concern population parameters, not sample means, and the null contains the equality. The direction must follow the defined subtraction order.

The observed difference is $14-12=2$ minutes. The estimated standard error is $\sqrt{4^2/40+5^2/50}=\sqrt{0.9}\approx0.94868$ minute, so $t=2/0.94868\approx2.108$. Both estimated variances are divided by their own sample sizes and then added under the square root. Reversing the subtraction would reverse the sign.

The parameter is the difference in two population means of a quantitative response, and the samples are independent. A two-sample t-interval estimates μA−μB when the population SDs are unknown. There are no matched repairs or repeated measurements to justify a paired procedure; the response is not categorical, and an interval for A alone would leave out B.

A correlation of −0.80 describes a strong negative linear association: shops with longer cooling time tend to have lower spoilage percentages. Correlation is unit-free and is not a slope or a percentage of shops. The observational association does not by itself establish that changing cooling time causes the change in spoilage.

Substitute x=6 into the model: $\hat y=40-3(6)=22\%$. This is interpolation because 6 hours is within the observed range from 2 through 8. The 18 is the amount subtracted from the intercept, not the predicted response, and prediction is not a guarantee that a particular shop’s observed spoilage will be exactly 22%.

The model predicts 22%, so the residual is observed minus predicted: $19-22=-3$ percentage points. A negative residual means the prediction was 3 percentage points too high. Reversing the subtraction yields +3, while 19 and 22 are the observed and predicted responses, not their difference.

The U-shaped residual pattern is systematic curvature rather than random scatter about zero. It suggests that a straight-line model misses part of the association. A fitted least-squares line can have residuals that sum to zero even when the model form is inadequate; that property neither proves linearity nor establishes causation.

The intercept is the predicted response when x=0: 8 minutes. Because no route distance near 0 was observed, treating this as a reliable real route time requires extrapolation. The slope, 2.5, gives the predicted change per kilometer; the fitted intercept does not establish a universal physical waiting time.

Squaring the correlation gives r²=(0.90)²=0.81. The response is price, so r² describes the fraction of its observed variation explained by the linear relationship with mass. Reversing explanatory and response variables gives the wrong contextual interpretation, and explained variation is not an individual price fraction or a causal effect.

The means are $\bar x=3$ and $\bar y=4.2$. The least-squares slope is $\sum(x-\bar x)(y-\bar y)/\sum(x-\bar x)^2=8/10=0.8$, and the intercept is $4.2-0.8(3)=1.8$. The line therefore predicts orders per minute as $1.8+0.8x$, and it passes through (3,4.2). Neither swapping coefficients nor drawing a line through selected points produces the least-squares fit.