Estimating Population Parameters: Proportion, Mean, and Standard Deviation

Estimating a Population Proportion

Point Estimate for the Population Proportion

A point estimate is a single value statistic that serves as the best guess for an unknown population parameter. For the population proportion, the point estimate is given by:

• Formula: $\hat{p} = \frac{x}{n}$, where $x$ is the number of individuals in the sample with a specified characteristic and $n$ is the sample size.

• Example: In a poll of 1783 voters, 1123 favored the death penalty. The point estimate is $\hat{p} = \frac{1123}{1783} \approx 0.63$.

Confidence Interval for the Population Proportion

A confidence interval provides a range of plausible values for a population parameter, based on a point estimate and a margin of error. The level of confidence (e.g., 95%) indicates the expected proportion of such intervals that will contain the true parameter in repeated sampling.

• General Form: Point estimate $\pm$ margin of error

• Margin of Error Formula: $E = z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

• Confidence Interval Formula:

$\hat{p} - z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ (Lower bound) $\hat{p} + z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ (Upper bound)

• Conditions: $n\hat{p}(1-\hat{p}) \geq 10$ and $n \leq 0.05N$ (sample size is less than 5% of the population).

• Interpretation: A (1 – $\alpha$)·100% confidence interval means that this proportion of intervals from repeated samples will contain the true parameter.

• Example: For $\hat{p} = 0.63$, $n = 1783$, $z_{0.05} = 1.645$ (for 90% confidence): Lower: $0.63 - 1.645 \cdot \sqrt{\frac{0.63(1-0.63)}{1783}} \approx 0.61$ Upper: $0.63 + 1.645 \cdot \sqrt{\frac{0.63(1-0.63)}{1783}} \approx 0.65$ "We are 90% confident that the true proportion is between 0.61 and 0.65."

Determining Sample Size for Estimating a Proportion

To estimate a population proportion within a specified margin of error $E$ and confidence level, the required sample size is:

• Formula (with prior estimate $\hat{p}$): $n = \hat{p}(1-\hat{p}) \left( \frac{z_{\alpha/2}}{E} \right)^2$

• If no prior estimate: Use $\hat{p} = 0.5$ for maximum sample size: $n = 0.25 \left( \frac{z_{\alpha/2}}{E} \right)^2$

• Example: To estimate the proportion of English-only speakers within 3% (E = 0.03) at 90% confidence ($z_{0.05} = 1.645$), with prior $\hat{p} = 0.824$: $n = 0.824(1-0.824) \left( \frac{1.645}{0.03} \right)^2 \approx 436.04$ (round up to 437)

Estimating a Population Mean

Point Estimate for the Population Mean

The sample mean ($\bar{x}$) is the point estimate for the population mean ($\mu$):

• Formula: $\bar{x} = \frac{\sum x_i}{n}$

• Example: For 17 pennies with weights (in grams): $\bar{x} \approx 2.464$

Student’s t-Distribution

When estimating the mean from a normally distributed population with unknown standard deviation, the Student’s t-distribution is used:

• t-statistic: $t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$

• Properties:

  • Different for each degree of freedom ($n-1$)

  • Symmetric and centered at 0

  • More area in the tails than the normal distribution (especially for small $n$)

  • As $n$ increases, t-distribution approaches the normal distribution

Finding t-Values

The notation $t_\alpha$ refers to the t-value such that the area under the t-distribution curve to the right is $\alpha$.

Confidence Interval for the Population Mean

To construct a (1 – $\alpha$)·100% confidence interval for $\mu$:

• Conditions: Simple random sample, $n \leq 0.05N$, population normal or large $n$

• Formula:

Lower bound: $\bar{x} - t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ Upper bound: $\bar{x} + t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$

• Interpretation: For a 95% confidence interval, about 95% of such intervals from repeated samples will contain $\mu$.

• Example: For pennies, $\bar{x} = 2.464$, $s = 0.02$, $n = 17$, $t_{0.005} = 2.575$ (99% confidence): Lower: $2.464 - 2.575 \cdot \frac{0.02}{\sqrt{17}} = 2.452$ Upper: $2.464 + 2.575 \cdot \frac{0.02}{\sqrt{17}} = 2.476$ "We are 99% confident that the mean weight is between 2.452 and 2.476 grams."

Determining Sample Size for Estimating a Mean

To estimate $\mu$ within margin of error $E$ at a given confidence level:

• Formula: $n = \left( \frac{z_{\alpha/2} \cdot s}{E} \right)^2$ (round up to nearest integer)

• Example: To estimate penny weight within 0.005 grams at 99% confidence ($z_{0.005} = 2.575$, $s = 0.02$): $n = \left( \frac{2.575 \cdot 0.02}{0.005} \right)^2 = 106.09$ (round up to 107)

Estimating a Population Standard Deviation or Variance

Chi-Square Distribution

For a normally distributed population, the statistic $\chi^2 = \frac{(n-1)s^2}{\sigma^2}$ follows a chi-square distribution with $n-1$ degrees of freedom.

• Properties:

  • Not symmetric

  • Shape depends on degrees of freedom

  • All values are nonnegative

  • Becomes more symmetric as degrees of freedom increase

Finding Critical Values for the Chi-Square Distribution

Critical values $\chi^2_{\alpha}$ and $\chi^2_{1-\alpha}$ separate the middle (1 – $\alpha$)·100% of the distribution from the tails.

• Example: For 18 degrees of freedom, $\chi^2_{0.975} = 8.231$, $\chi^2_{0.025} = 31.526$

Confidence Interval for Population Variance and Standard Deviation

For a simple random sample from a normal population, a (1 – $\alpha$)·100% confidence interval for $\sigma^2$ is:

• Lower bound: $\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

• Upper bound: $\frac{(n-1)s^2}{\chi^2_{\alpha/2}}$

• For $\sigma$, take the square root of the bounds.

• Example: For Microsoft stock returns ($n=15$, $s^2=22.0659$), $\chi^2_{0.95}=6.571$, $\chi^2_{0.05}=23.685$: Lower: $\sqrt{\frac{14 \times 22.0659}{23.685}} = 13.04$ Upper: $\sqrt{\frac{14 \times 22.0659}{6.571}} = 47.01$ "We are 90% confident that the standard deviation is between 13.04 and 47.01."

Choosing the Appropriate Estimation Procedure

To determine which estimation procedure to use (proportion, mean, or standard deviation/variance), consider the parameter of interest, sample size, and whether the population is normal or approximately normal.