CI for a single population proportion (p)

Notation

  • p is the population proportion we are trying to estimate
  • \(\widehat{p}\) is the sample proportion
  • n is the sample size

Conditions

  1. The data comes from a random sample.
  2. \(n * \widehat{p} \geq 10\) (Success Condition)
  3. \(n * (1-\widehat{p}) \geq 10\) (Failure Condition)

Note: Success and Failure conditions count the # of successes and # of failures respectively.

Formula

\(\widehat{p} \pm z^* \times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\)

\(z^*\) is the appropriate value from the normal distribution that gives us the Confidence % that we want

  • 95% Confidence → \(z^*\) = 1.96
  • 80% Confidence → \(z^*\) = 1.282
  • 90% Confidence → \(z^*\) = 1.645
  • 99% Confidence → \(z^*\) = 2.576

CI for a difference in population proportions (\(p_1 - p_2\))

Notation

  • \(p_1\) is the population proportion for group 1
  • \(p_2\) is the population proportion for group 2
  • \(\widehat{p}_1\) is the sample proportion for group 1
  • \(\widehat{p}_2\) is the sample proportion for group 2
  • \(n_1\) is the sample size for group 1
  • \(n_2\) is the sample size for group 2

Conditions

  1. Data for both groups comes from a random sample.
  2. \(n_1 * \widehat{p}_1 \geq 10\) (Success Condition Grp 1)
  3. \(n_1 * (1-\widehat{p}_1) \geq 10\) (Failure Condition Grp 1)
  4. \(n_2 * \widehat{p}_2 \geq 10\) (Success Condition Grp 2)
  5. \(n_2 * (1-\widehat{p}_2) \geq 10\) (Failure Condition Grp 2)

Note: Success and Failure conditions count the # of successes and # of failures respectively.

Formula

\((\widehat{p}_1 - \widehat{p}_2) \pm z^* \times \sqrt{\frac{\widehat{p}_1(1-\widehat{p}_1)}{n_1}+\frac{\widehat{p}_2(1-\widehat{p}_2)}{n_2}}\)

\(z^*\) is the appropriate value from the normal distribution that gives us the Confidence % that we want

  • 95% Confidence → \(z^*\) = 1.96
  • 80% Confidence → \(z^*\) = 1.282
  • 90% Confidence → \(z^*\) = 1.645
  • 99% Confidence → \(z^*\) = 2.576