Statistical Intervals Based on a Single Sample

Large-Sample Confidence Intervals for a Population Mean and Proportion

Large-Sample Interval for μ

• Let $X<em>1, X</em>2, …, X_n$ be a random sample from a population with mean $μ$ and standard deviation $σ$.
• The Central Limit Theorem (CLT) implies that $\bar{X}$ has approximately a normal distribution for large $n$, regardless of the population distribution.
• $\frac{\bar{X} - μ}{σ/\sqrt{n}}$ has approximately a standard normal distribution.
• A large-sample confidence interval (CI) for $μ$ with a confidence level of approximately $100(1 - α)\%$ is given by:
  $\bar{x} ± z_{α/2} \frac{s}{\sqrt{n}}$
• This CI is valid regardless of the shape of the population distribution.
• In practice, the population standard deviation $σ$ is rarely known and is replaced by the sample standard deviation $S$.
• The standardized variable becomes:
  $Z = \frac{\bar{X} - μ}{S/\sqrt{n}}$
• For large $n$, the substitution of $S$ for $σ$ adds little extra variability, and $Z$ still has approximately a standard normal distribution.

Proposition

• If $n$ is sufficiently large, the standardized variable
  $Z = \frac{\bar{X} - μ}{S/\sqrt{n}}$
  has approximately a standard normal distribution.
• The large-sample confidence interval for $μ$ is:
  $\bar{X} ± z_{α/2} \frac{S}{\sqrt{n}}$
• This interval is valid regardless of the shape of the population distribution.
• The CI can be expressed as:
  Point estimate of $μ$ ± (z critical value) × (estimated standard error of the mean).
• Generally, n > 40 is sufficient to justify the use of this interval.

Example 7.6

• Data collected on odometer readings (miles) for a sample of 50 used Porsche Boxsters from cars.com.
• Sample data: 2948, 30552, 30600, …, 84000.
• Summary quantities: $n = 50$, sample mean $\bar{x} = 45679.4$, median = 45013.5, sample standard deviation $s = 26641.675$.
• A 95% confidence level requires $z_{0.025} = 1.96$.
• The 95% confidence interval is:
  $45679.4 ± (1.96) \frac{26641.675}{\sqrt{50}} = 45679.4 ± 7384.7 = (38294.7, 53064.1)$
• Interpretation: We are 95% confident that the true population mean odometer reading lies between 38,294.7 and 53,064.1 miles.

General Large-Sample Confidence Interval

• The large-sample intervals for $\bar{X}$ and $p$ are special cases of a general large-sample CI for a parameter $\theta$.
• Suppose $\hat{\theta}$ is an estimator satisfying:
  • Approximately a normal distribution.
  • Approximately unbiased.
  • An expression for $σ_{\hat{\theta}}$, the standard deviation of $\hat{\theta}$, is available.
• Standardizing yields the random variable
  $z = \frac{\hat{\theta} - \theta}{σ_{\hat{\theta}}}$, which has approximately a standard normal distribution.
• The lower and upper confidence limits are $\hat{\theta} - z<em>{α/2}σ</em>{\hat{\theta}}$ and $\hat{\theta} + z<em>{α/2}σ</em>{\hat{\theta}}$, respectively.
• If $σ<em>{\hat{\theta}}$ involves unknown parameters, estimate them to obtain $\hat{σ</em>{\hat{\theta}}}$, then a valid CI is
  $\hat{\theta}± z<em>{α/2} \hat{σ</em>{\hat{\theta}}}$ .
• This CI is a point estimate of $θ$ ± (z critical value) × (estimated standard error of the estimator).

Confidence Interval for a Population Proportion

• Let $p$ denote the population proportion of successes.
• A random sample of $n$ individuals is selected, and $X$ is the number of successes in the sample.
• Provided $n$ is small compared to the population size, $X$ can be regarded as a binomial random variable with $E(X) = np$ and $V(X) = npq$, where $q = 1 - p$.
• If both $np ≥ 10$ and $nq ≥ 10$, $X$ has approximately a normal distribution.
• The natural estimator of $p$ is $\hat{p} = X/n$, the sample fraction of successes.
• Since $\hat{p}$ is just $X$ multiplied by the constant $1/n$, $\hat{p}$ also has approximately a normal distribution.
• The standard deviation of $\hat{p}$ is $\sqrt{\frac{pq}{n}}$, which involves the unknown parameter $p$.
• A confidence interval for $p$ with confidence level approximately $100(1 - α)\%$ is:
  $\hat{p} = \frac{\hat{p} + \frac{z^2<em>{α/2}}{2n}}{1 + \frac{z^2</em>{α/2}}{n}} ± \frac{z<em>{α/2}}{1 + \frac{z^2</em>{α/2}}{n}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n} + \frac{z^2_{α/2}}{4n^2}}$

Proposition

• Let $\tilde{p} = \frac{\hat{p} + z^2<em>{α/2}/2n}{1 + z^2</em>{α/2}/n}$. Then a confidence interval for a population proportion $p$ with confidence level approximately $100(1 - α)\%$ is:
  $\tilde{p} ± \frac{z<em>{α/2}}{1 + \frac{z^2</em>{α/2}}{n}} \sqrt{\frac{\hat{p}(1-\hat{p})}{n} + \frac{z^2_{α/2}}{4n^2}}$

• This is often referred to as the score CI for $p$.

• If the sample size $n$ is very large, then $\frac{z^2<em>{α/2}}{n}$ is generally quite negligible compared to 1, and the score interval is approximately: $\hat{p} ± z</em>{α/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

• The actual coverage probability for the traditional interval can differ considerably from the nominal probability, particularly when $p$ is not close to 0.5.

• The score interval rectifies this behavior, and its actual confidence level will be quite close to the nominal level.

• The midpoint of the score interval is always a bit closer to 0.5 than is the midpoint of the traditional interval, which is especially important when $p$ is close to 0 or 1.

• The score interval can be used with nearly all sample sizes and parameter values.

Example 7.8

• In $n = 48$ trials, 16 resulted in ignition of a substrate.
• A point estimate for $p$ is $\hat{p} = 16/48 = 0.333$.
• A 95% confidence interval for $p$ is calculated.

$\tilde{p} = \frac{0.333 + \frac{1.96^2}{2*48}}{1 + \frac{1.96^2}{48}} = 0.345$

$\frac{1.96}{1 + \frac{1.96^2}{48}} \sqrt{\frac{0.333(1 - 0.333)}{48} + \frac{1.96^2}{4*48^2}} = 0.129$

$\tilde{p} = 0.345 ± 0.129 = (0.216, 0.474)$

• The traditional interval is $\hat{p} ± z_{α/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ = $(0.200, 0.466)$

• These two intervals would be in much closer agreement if the sample size were substantially larger.

• Equating the width of the CI for $p$ to a prespecified width $w$ gives a quadratic equation for the sample size $n$ necessary to give an interval with a desired degree of precision.

$n = \frac{4z^2(\hat{p}(1-\hat{p}) + z^2/4n)}{w^2}$

• Neglecting the terms in the numerator involving $w^2$ gives

$n ≈ \frac{4z^2\hat{p}(1-\hat{p})}{w^2}$

• This latter expression is what results from equating the width of the traditional interval to $w$.
• The most conservative approach is to take advantage of the fact that $\hat{p}(1-\hat{p})$ is a maximum when $\hat{p} = 0.5$.

One-Sided Confidence Intervals (Confidence Bounds)

• One-sided confidence intervals give either a lower confidence bound or an upper confidence bound for the parameter being estimated.

• For example, a psychologist may wish to calculate a 95% upper confidence bound for true average reaction time to a particular stimulus.

• Because the cumulative area under the standard normal curve to the left of 1.645 is 0.95

• Manipulating the inequality inside the parentheses to isolate $μ$ on one side and replacing random variables by calculated values gives the inequality

  μ > \bar{x} – 1.645\frac{s}{\sqrt{n}}; the expression on the right is the desired lower confidence bound.