21  Practice Midterm 1

QTM 285-1

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\DeclareMathOperator{\mathop{\mathrm{E}}}{E} \DeclareMathOperator{\mathop{\mathrm{\mathop{\mathrm{V}}}}}{V} \DeclareMathOperator{\mathop{\mathrm{\widehat{V}}}}{\widehat{V}} \DeclareMathOperator{\mathop{\mathrm{bias}}}{bias} \DeclareMathOperator*{argmin}{argmin} \newcommand{\mathcal{M}}{\mathcal{M}}

Problem 1

Suppose a friend has polled 100 of Georgia’s registered voters about their intention to vote in the 2024 election. They conducted this poll by sampling with replacement from the population of all of Georgia’s registered voters. And they report what they say is a 95% confidence interval for turnout: [.58,.62].

ExercisePart A

Has your friend calibrated their interval estimate correctly, i.e., are they really using an interval estimator that has 95% coverage? Say yes or no, then explain your answer.

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(your answer here)

ExercisePart B

If they wanted an interval estimate that was accurate to within 1%, i.e. an interval of the form \hat\theta \pm .01, roughly how many voters should they have polled? Choose the closest number from this list: 100, 1000, 10000, 100000. Circle you choice.

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(your answer here)

Problem 2

A population of m=200 people is arranged in a circle, and their responses alternate as you go around it: 1, 0, 1, 0, \ldots. So the proportion of 1s in the population is \theta=1/2. We’ll compare three ways of drawing a sample of n=100 people from it. In each case, we estimate \theta by the proportion of 1s in the sample, \hat\theta=\frac{1}{100}\sum_{i=1}^{100} Y_i, and report the interval estimate \hat\theta \pm 1.96\sqrt{\hat\theta(1-\hat\theta)/n}.

1. Draw 100 people uniformly at random with replacement.
2. Draw 100 people uniformly at random without replacement.
3. Choose one starting person uniformly at random, and take that person and the next 99 people around the circle.

Exercise

One of these intervals has coverage of roughly 95%, one has noticeably more, and one has noticeably less. Which is which? Explain.

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(your answer here)

Problem 3

A coworker has a sample of n=100 household incomes Y_1 \ldots Y_{100}, drawn with replacement from a population, and wants a 95% interval estimate for the population’s mean income. They bootstrap: they draw 10{,}000 samples of size 100 with replacement from their sample, take the mean \bar Y^\star of each, and report \bar Y \pm 1.96\,\text{se}, where \text{se} is the standard deviation of these bootstrap means. They’ve done everything right. But their boss, who has a habit of correcting people without taking the time to understand what they’re doing, glances at it and says:

For the pictures, take the sample to be simulated from a log-normal income population with median 60{,}000 and log-scale standard deviation 0.5.

You forgot to divide by \sqrt n. Rookie mistake.

To sort this out, your coworker draws two pictures on the same axis: the distribution of the incomes in their sample on the left, and the bootstrap distribution of \bar Y^\star on the right.

ExercisePart A

Is the boss right? Say what each picture is a picture of, and explain why the standard deviation of the picture on the right is, or isn’t, the right standard error for an interval estimate of the mean income.

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(your answer here)

ExercisePart B

Your coworker wants to end their explanation with a flourish: “If you want to think of the standard deviation of Y_1 \ldots Y_n—the one you’d divide by \sqrt n—in bootstrap terms, it’s the bootstrap standard deviation of an estimator you’d never actually use. It’s the standard deviation of …”

Finish the sentence. What simple estimator has bootstrap standard deviation \sigma=\sqrt{\frac{1}{n}\sum\{Y_i - \bar Y\}^2}, the same standard deviation as the sample Y_1 \ldots Y_n?

Tip. There’s more than one. Any one you can think of is fine.

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(your answer here)

Problem 4

Two polling firms each survey the same population in exactly the same way, independently of one another: each calls 1,000 of Georgia’s registered voters, drawn with replacement, and reports the proportion who say they’ll turn out. The plot below shows what a single survey of this kind produces, with two particular surveys marked and the actual turnout rate \theta drawn in green.

ExercisePart A

Typically, how far apart are the two surveys’ estimates? That is, what is the root-mean-squared length of the arrow drawn between them? Give your answer as a multiple of one survey’s standard error, and show your calculation.

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(your answer here)

ExercisePart B

The two firms report estimates 6 percentage points apart. Is that a surprising amount of disagreement?

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(your answer here)

Problem 5

You have a sample Y_1,Y_2...,Y_n drawn with replacement from a population with mean \mu and standard deviation \sigma. The usual estimate for \mu is the sample mean \bar{Y}_0 = \frac{1}{n}\sum_{i=1}^n Y_i. But one of your coworkers has a clever idea to reduce variance—instead of dividing by n, they divide by n+10. This is their estimator: \bar{Y}_{10} = \frac{1}{n+10}\sum_{i=1}^n Y_i. You try to convince them it’s a bad idea by being a bit hyperbolic—you argue that if that’s a good idea, then why not divide by n+100? They, unfortunately, love this idea. You now have to explain why this is a bad idea.

Your coworker has calculated formulas for the bias and standard deviation of the ‘divide by n+k estimator’ \bar{Y}_k=\frac{1}{n+k}\sum_{i=1}^n Y_i. \begin{aligned} \mathop{\mathrm{E}}[\bar{Y}_k] - \mu &= -\frac{k}{n+k}\mu \\ \mathop{\mathrm{sd}}[\bar{Y}_k] &= \frac{\sigma}{\sqrt{n}} \times \frac{n}{n+k}. \end{aligned}

NoteFormula Derivation. Optional Reading.

\begin{aligned} \mathop{\mathrm{E}}\sb*{\bar{Y}_k} - \mu &= \mathop{\mathrm{E}}\sb*{\frac{n}{n+k}\bar Y_0} - \mu \\ &= \frac{n}{n+k}\mathop{\mathrm{E}}[\bar Y_0] - \mu \\ &=\frac{n}{n+k}\mu - \mu \\ &= \p*{\frac{n}{n+k} - 1}\mu \\ &= \p*{\frac{n}{n+k} - \frac{n+k}{n+k}}\mu \\ &= -\frac{k}{n+k}\mu \\ & \\ \mathop{\mathrm{\mathop{\mathrm{V}}}}\sb*{\bar{Y}_k} &= \mathop{\mathrm{\mathop{\mathrm{V}}}}\sb*{\frac{n}{n+k} \bar {Y}_0} \\ &= \p*{\frac{n}{n+k}}^2 V[\bar{Y}_0] \\ &= \frac{n^2}{(n+k)^2} \frac{\sigma^2}{n} \\ &= \frac{ n\sigma^2}{(n+k)^2}=\frac{\sigma^2}{n} \times \frac{n^2}{(n+k)^2} \end{aligned}

And they’ve drawn the sampling distribution of \bar{Y}_0, \bar{Y}_{10}, and \bar{Y}_{100} (top, middle, and bottom rows respectively) for sample sizes n=100 and n=10,000 (left and right columns respectively) for \mu=0 and \sigma=1. The solid line is the estimator’s expected value \mathop{\mathrm{E}}[\bar{Y}_k], the dashed lines are \mathop{\mathrm{E}}[\bar Y_k] \pm \mathop{\mathrm{sd}}[\bar Y_k], and the dotted lines are \mathop{\mathrm{E}}[\bar Y_k] \pm 2 \mathop{\mathrm{sd}}[\bar Y_k].

On the basis of those drawings, your coworker proposes using \bar{Y}_{100} to estimate \mu.

Exercise 26.1  

Exercise

Suppose your sample size is n=100. Explain why your coworker’s original proposal to use \bar Y_{10} and your hyperbolic suggestion to use \bar Y_{100} appear, from the drawings, to be a way to increase precision without increasing bias. To explain why this is misleading, describe what happens in the case that \mu=11/10 and \sigma=1 by doing the following.

• Modifying the drawings to show the sampling distributions in this new case.
• Stating an approximation to the coverage of each of the confidence intervals \bar{Y}_{k} \pm 1.96\ \mathop{\mathrm{sd}}[\bar{Y}_k] for k=0, k=10, and k=100.

Tip. You can ‘change the location’ of the distributions in the drawing above by changing the labels on the x-axis.

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(your answer here)