Confidence Intervals Notes

Confidence Intervals

Introduction

• This chapter covers constructing and interpreting confidence interval estimates for the population mean and proportion.
• It also includes determining the necessary sample size for these estimates.

Point and Interval Estimates

• Point Estimate: A single number used to estimate a population parameter.
• Confidence Interval: Provides additional information about the variability of the estimate.
• We can estimate population parameters such as $\mu$ (population mean) or $\pi$ (population proportion) using sample statistics such as $\overline{x}$ (sample mean) or p (sample proportion).

Table 1: Point Estimates

Understanding Confidence Intervals

• Confidence intervals address the uncertainty associated with point estimates.
• Interval Estimate: Gives a range of values providing more information than a point estimate.
• Such interval estimates are called confidence intervals.

Key Aspects of Confidence Intervals

• An interval gives a range of values.
• Takes into consideration variation in sample statistics from sample to sample.
• Based on observations from one sample.
• Provides information about closeness to unknown population parameters.
• Expressed in terms of a level of confidence (e.g., 95% or 99%), but can never be 100% confident.

Confidence Interval Example: Cereal Fill

• Population has $\mu = 368$ and $\sigma = 15$.
• Sample size is $n = 25$.
• From Chapter 7: $\mu \pm Z \times \sigma<em>{\overline{x}}$, where $\sigma</em>{\overline{x}} = \frac{\sigma}{\sqrt{n}}$
  • $368 \pm 1.96 \times \frac{15}{\sqrt{25}} = (362.12, 373.88)$
    • 95% of intervals formed this way will contain $\mu$.
• When $\mu$ is unknown, use $\overline{x}$ to estimate $\mu$.
  • If $\overline{x} = 362.3$, the interval is $362.3 \pm 1.96 \times \frac{15}{\sqrt{25}} = (356.42, 368.18)$
    • Since $356.42 \le \mu \le 368.18$, the interval correctly estimates $\mu$.

Practical Considerations

• In practice, only one sample of size n is taken.
• In practice, $\mu$ is unknown, so it's not known if the interval contains $\mu$.
• 95% confidence is based on using $Z = 1.96$.
• 95% of intervals formed this way may contain $\mu$.
• Based on the selected sample, one can be 95% confident the interval may contain $\mu$ (a 95% confidence interval).

General Formula for Confidence Intervals

• The general formula for all confidence intervals is:
  • Point Estimate ± (Critical Value)(Standard Error)
    • Point Estimate: The sample statistic estimating the population parameter.
    • Critical Value: A table value based on the sampling distribution and desired confidence level.
    • Standard Error: The standard deviation of the point estimate.

Confidence Level, $(1 - \alpha)$

• If the confidence level is 95%, $(1 - \alpha) = 0.95$, so $\alpha = 0.05$.
• Relative frequency interpretation:
  • 95% of all confidence intervals constructed will contain the true parameter.
• A specific interval either contains or does not contain the true parameter.
  • There is no probability involved for a specific interval.

Confidence Interval for $\mu$ ($\sigma$ Known)

Assumptions:

• Population standard deviation $\sigma$ is known.
• Population is normally distributed.
• If the population is not normal, use a large sample size (n > 30).

Confidence interval estimate:

• $\overline{x} \pm Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$
  • $\overline{x}$ is the point estimate.
  • $Z_{\alpha/2}$ is the normal distribution critical value for a probability of $\alpha/2$ in each tail.
  • $\frac{\sigma}{\sqrt{n}}$ is the standard error.

Common Levels of Confidence

Example

• A sample of 11 circuits from a normal population has a mean resistance of 2.22 ohms.
• The population standard deviation is 0.35 ohms.
• Determine a 95% confidence interval for the true mean resistance.
  • $\overline{x} \pm Z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 2.22 \pm (1.96) \frac{0.35}{\sqrt{11}} = 2.22 \pm 0.2068$

Interpretation

• We are 95% confident that the true mean resistance is between 2.0132 and 2.4268 ohms.
• Although the true mean may or may not be in this interval, 95% of intervals formed this way may contain the true mean.

Do You Ever Truly Know $\sigma$?

• Probably not!
• In real-world business situations, $\sigma$ is usually unknown.
• If $\sigma$ is known, then $\mu$ is also known (since calculating $\sigma$ requires knowing $\mu$).
• If $\mu$ is known, there's no need to estimate it.

Confidence Interval for $\mu$ ($\sigma$ Unknown)

• If the population standard deviation $\sigma$ is unknown, substitute the sample standard deviation, S.
• This introduces extra uncertainty since S varies from sample to sample.
• Use the t-distribution instead of the normal distribution.

Assumptions:

• Population standard deviation is unknown.
• Population is normally distributed.
• If the population is not normal, use a large sample (n > 30).

Use Student’s t Distribution

Confidence Interval Estimate:

• $\overline{x} \pm t_{\alpha/2} \frac{S}{\sqrt{n}}$
  • Where $t_{\alpha/2}$ is the critical value of the t-distribution with $n - 1$ degrees of freedom and an area of $\alpha/2$ in each tail.

Student’s t Distribution

• The t-distribution is a family of distributions.
• The $t_{\alpha/2}$ value depends on degrees of freedom (d.f.).
• Degrees of freedom represent the number of observations free to vary after the sample mean has been calculated.
  • $d.f. = n - 1$

Degrees of Freedom (df)

• Idea: Number of observations that are free to vary after sample mean has been calculated.
• Example: Suppose the mean of 3 numbers is 8.0. Let $X<em>1 = 7$ and $X</em>2 = 8$. Then $X<em>3$ must be 9 (i.e., $X</em>3$ is not free to vary).
• Here, $n = 3$, so degrees of freedom $= n – 1 = 3 – 1 = 2$. Two values can be any numbers, but the third is not free to vary for a given mean.

Example of t distribution confidence interval

• A random sample of $n = 25$ has $\overline{x} = 50$ and $S = 8$.
• Form a 95% confidence interval for $\mu$.
  • $d.f. = n – 1 = 24$, so $t<em>{\alpha/2} = t</em>{0.025} = 2.064$
  • The confidence interval is: $\overline{x} \pm t_{\alpha/2} \frac{S}{\sqrt{n}} = 50 \pm (2.064) \frac{8}{\sqrt{25}} = 50 \pm 3.302$
  • The confidence interval is $46.698 \le \mu \le 53.302$
• Interpreting this interval requires the approximation that the population you are sampling from is approximately a normal distribution (especially since n is only 25). This condition can be checked by creating a:
  • Normal probability plot or
  • Boxplot

Confidence Intervals for the Population Proportion, $\pi$

• An interval estimate for the population proportion ($\pi$) can be calculated by adding an allowance for uncertainty to the sample proportion (p).
• Recall that the distribution of the sample proportion is approximately normal if the sample size is large, and we must have np > 5 and n(1-p) > 5 and the standard error of the proportion is:
  • $\sigma_{\overline{p}} = \sqrt{\frac{p(1 - p)}{n}}$

Confidence Interval Endpoints

• Upper and lower confidence limits for the population proportion are calculated with the formula:
  • $p \pm Z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$
    • Where:
      • $Z_{\alpha/2}$ is the standard normal value for the level of confidence desired
      • p is the sample proportion
      • n is the sample size
        *Note: must have np > 5 and n(1-p) > 5

Example

• A random sample of 100 people shows that 25 are left-handed.
• Form a 95% confidence interval for the true proportion of left-handers.
  • $p \pm Z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}} = \frac{25}{100} \pm 1.96\sqrt{\frac{(.25)(.75)}{100}} =$
  • $= \frac{25}{100} \pm 1.96(0.0433)$
• So: We are 95% confident that $X \pm 0.0433$ contains the population proportion.
  • $0.1651 \le p \le 0.3349$

Interpretation

• We are 95% confident that the true percentage of left-handers in the population is between 16.51% and 33.49%.
• Although the interval from 0.1651 to 0.3349 may or may not contain the true proportion, 95% of intervals formed from samples of size 100 in this manner will contain the true proportion.

Determining Sample Size

Sampling Error

• The required sample size can be found to reach a desired margin of error (e) with a specified level of confidence ($1 - \alpha$).
• The margin of error is also called sampling error, it is:
  • The amount of imprecision in the estimate of the population parameter
  • The amount added and subtracted to the point estimate to form the confidence interval.
• For the Mean
  • $e = Z_{\alpha/2} \sqrt{\frac{\sigma}{n}}$
  • Now solve for n
  • $n = \frac{Z^2_{\alpha/2}\sigma^2}{e^2}$
• To determine the required sample size for the mean, you must know:
  • The desired level of confidence ($1 - \alpha$), which determines the critical value, $Z_{\alpha/2}$
  • The acceptable sampling error, e
  • The standard deviation, $\sigma$

Required Sample Size Example

• If $\sigma = 45$, what sample size is needed to estimate the mean within $± 5$ with 90% confidence?
  • $n = \frac{Z^2_{\alpha/2}\sigma^2}{e^2} = \frac{(1.645^2)(45^2)}{5^2} = 219.19$
  • so the require sample size is 220 (always round up).

If $\sigma$ is unknown

• If unknown, $\sigma$ can be estimated when using the required sample size formula:
  • Use a value for $\sigma$ that is expected to be at least as large as the true $\sigma$.
  • Select a pilot sample and estimate $\sigma$ with the sample standard deviation, S

Determining Sample Size For the Population

• $e = Z(\sqrt{\frac{\pi(1−\pi)}{n}})$
• Solve for n
  • $n = \frac{Z^2_{\alpha/2}(\pi(1−\pi)}{e^2}$
• To determine the required sample size for the proportion, you must know:
  • The desired level of confidence ($1 - \alpha$, which determines the critical value, $Z_{\alpha/2}$
  • The acceptable sampling error, e
  • The true proportion of events of interest, $\pi$
    • $\pi$ can be estimated with a pilot sample if necessary (or conservatively use 0.5 as an estimate of $\pi$)

Required Sample Size Example

• How large a sample would be necessary to estimate the true proportion defective in a large population within $± 3$, with 95% confidence?
• (Assume a pilot sample yields p = 0.12)
• Solution:
  • For 95% confidence, use $Z_{\alpha/2} = 1.96$
  • $e = 0.03$
  • $p = 0.12$, so use this to estimate $\pi$
  • $n = \frac{Z^2_{\alpha/2}(\pi(1 − \pi)}{e^2} = \frac{(1.96)^2(.12)(.88)}{(0.03)^2} = 450.74$
• So: use n = 451

Ethical Issues

• A confidence interval estimate (reflecting sampling error) should always be included when reporting a point estimate.
• The level of confidence should always be reported.
• The sample size should be reported.
• An interpretation of the confidence interval estimate should also be provided.

Final Note

• The important thing to remember is that the margin of error, confidence interval, is generally a function three things, the degree of confidence required, the sample size and the percentage being estimated.
• Thus, sampling error will decrease as:
  • The sample size (or number of interviews) gets bigger;
  • The percentage estimated approaches 0% or 100% or
  • The need to be certain about the result (e.g. the ‘‘confidence level’’) gets smaller.