CHAPTER 5

INTRODUCTION TO INFERENCE:ESTIMATION AND PREDICTION

5-1.

The sampling distribution of an estimator is the distribution model of the estimator. It can be thought of as the distribution of the values the estimator takes in all possible samples. (5 points)

The value of knowing it is that it tells how the estimator varies from sample to sample, and therefore how much credence to give the estimator. For example if $Y_1, \ldots,Y_n \sim N(\mu,1)$ is a random sample, an estimator of $\mu$ is $\hat{\mu} = \overline{Y} \sim N(\mu, 1/n),$ which is the sampling distribution. The sampling distribution quantifies precisely how we can expect the the estimator to vary. (5 points)

5-3.

We want a confidence interval for p, the population proportion who approve. Since n is large $(n\hat{p} = 5500,\;\; n(1-\hat{p}) = 4500)$, we use a 95% large sample interval. (5 points)

$$\hat{p} \pm \hat{\sigma}(\hat{p})z_{0.975} = 0.55 \pm \left... ...5)(0.45)}{10000}}\right] 1.96= (0.54, 0.56). \mbox{ (5 points)}$$

We estimate between 54 and 56% approve, which is better than 50-50. (5 points)

5-15.

a.

Estimate of proportion p₁ of whites saying guilty: $\frac{108}{100}$.
Estimate of proportion p₂ of nonwhites saying guilty': $\frac{23}{100}$.
Estimate of difference in proportions:$\frac{108}{150} - \frac{23}{100} = 0.49$. (5 points)

b.

99% confidence interval for the difference:

Since all values in the interval are positive, we can be 99% confident that the proportion of whites in the population who think Simpson is guilty is greater than the proportion of blacks. (5 points)

99% confidence means that if we take many more samples and from them compute 99% confidence intervals, then about 99% of all such intervals will really contain p₁-p₂. (5 points)

5-21.

a.

All AA batteries of that brand. (5 points)

b.

$46.7 \pm \frac{13.3}{\sqrt{152}}(1.96) = (44.6, 48.8)$. (5 points) We are 95% confident the mean lifetime of all batteries, $\mu$, is between 44.6 and 48.8. (5 points)

c.

In repeated sampling, 95% of all 95% confidence intervals will contain $\mu$. (5 points)

5-22.

a.

$46.7 \pm (1.96)(13.3) \sqrt{1 + \frac{1}{152}} = (20.5, 72.9)$. (5 points) We are 95% confident that a new battery will have a lifetime between 20.5 and 72.9 hours. (5 points)

b.

They must be the same. Otherwise, we cannot extrapolate to the future. (5 points)

c.

In repeated sampling, 95% of all level 0.95 prediction intervals will contain a new observation. (5 points)

5-23.

a.

$46.7 \pm (2.137)(13.3) = (16.1, 77.3)$. (5 points) We are 90% confident that at least 95% of all battery lifetimes are between 16.1 and 77.3 hours. (5 points)

b.

In repeated sampling, 90% of all such tolerance intervals will contain at least 95% of all battery lifetimes. (5 points)

5-24.

The estimate of the difference between population means of the treatment and control groups is

$$\hat{\mu}_t - \hat{\mu}_c = 40.7 - 39.0 = 1.7. \mbox{ (5 points)}$$

The estimated standard error is

$$\hat{\sigma}(\hat{\mu}_t - \hat{\mu}_c) = \sqrt{\frac{(10.8)^2}{50}+ \frac{(12.2)^2}{50}} = 2.3.$$

A level 0.99 confidence interval for $\mu_t - \mu_c$ is (z_0.995 = 2.5758):

$$1.7 \pm (2.3)(2.5758) = (-4.2, 7.6). \mbox{ (5 points)}$$

We are 99% confident that $\mu_t - \mu_c$ is between -4.2 and 7.6. (5 points) Since the interval contains 0, we cannot conclude there is a difference between $\mu_t$ and $\mu_c$. (5 points)

5-28.

The 95% interval will be wider since

$$L_a \gt L_b \Rightarrow t_{\nu,\frac{1+L_a}{2}} \gt t_{\nu,\frac{1+L_b}{2}} \mbox{ (5 points)}$$

5-30.

a.

Take differences of the Port A and Port B observations and analyze as paired data. (5 points)

b.

If correlation decreases sufficiently with distance, use the confidence interval for independent populations. (5 points)

5-31.

a.

If p₁ = population proportion of drug users who improve
and p₂ = population proportion of nonusers who improve,
$\hat{p}_1 = 2/3\;\;\hat{p}_2 = 3/5,$ so $\hat{p}_1$ is slightly larger than $\hat{p}_2$. (5 points)

b.

Construct a confidence interval for p₁ - p₂. (5 points)

c.

A 95% confidence interval for p₁ - p₂ is $(\frac{2}{5} - \frac{3}{5}) \pm 1.96 \sqrt{(\frac{\frac{2}{3})(\frac{1}{5})}{150} + \frac{(\frac{3}{5})(\frac{2}{5})}{100}} = (-0.06, 0.19).$ (5 points)

Result: Little evidence of a difference in proportion. (5 points)