WK10 - Statistical inference: population proportion: Confidence interval for a proportion

Proportions: Statistical Inference

Introduction to Proportions

• Quantitative variables can be analyzed to determine if values exceed a threshold.
• Focus shifts to the proportion of individuals or objects with a specific characteristic.
• "Successes" represent individuals with the characteristic of interest.
• The population proportion is denoted as $P$.

Point Estimate and Sampling Distribution

• Objective: to set up statistical inference for proportion, determining point estimate and sampling distribution.
• $x$: Number of successes in a sample.
• $n$: Total number of units in a sample.
• The point estimate of the population proportion: $\hat{p} = \frac{x}{n}$. $\hat{p}$ is a single best guess for the true population proportion $P$.

Examples of Calculating Sample Proportion

• Example 1: 107 out of 130 students think A.
  • $\hat{p} = \frac{107}{130} \approx 0.823$, or 82.3%.
• Example 2: 848 out of 1000 people think B.
  • $\hat{p} = \frac{848}{1000} = 0.848$, or 84.8%.
• Example 3: 58 out of 125 scientists were successful.
  • $\hat{p} = \frac{58}{125} = 0.464$, or 46.4%.

Sampling Distribution of Proportions

• $\hat{p}$ depends on the sample; to assess its reliability, we examine the sampling distribution of proportions.
• Repeated samples of the same size are taken.
• The mean of sample proportions approaches the population proportion $P$ as sample size increases.
• The standard deviation of the distribution of sample proportions: $\sqrt{\frac{P(1-P)}{n}}$.

Large Sample Size and Normality

• For large $n$, the sampling distribution of $\hat{p}$ is approximately normal.
• Mean: $P$.
• Standard Deviation: $\sqrt{\frac{P(1-P)}{n}}$.
• Due to not knowing $P$, $\hat{p}$ is used as an estimate in the standard error calculation.
• Standard Error: $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$.
• This allows the use of Z procedures.

Confidence Interval for Population Proportion

• General Formula: Point Estimate ± (Multiplier × Standard Error).
• Point Estimate: Sample proportion $\hat{p}$.
• Multiplier: Critical value $Z^*$ from the standard normal distribution for the desired confidence level.
• Confidence Interval Formula: $\hat{p} \pm Z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$.
• Conditions:
  • $n \times \hat{p} \geq 15$.
  • $n \times (1-\hat{p}) \geq 15$.
  • Both conditions correspond to having at least 15 successes and 15 failures in the sample.

Adjustments for Small Sample Sizes: Plus Four Method

• Rationale: Large sample confidence intervals can be unstable and inaccurate with small samples.
• Method: Add four imaginary observations - two successes and two failures - to stabilize inferences.
• Adjusted point estimate: $\tilde{p} = \frac{x+2}{n+4}$.
• Adjusted confidence interval formula: $\tilde{p} \pm Z^* \sqrt{\frac{\tilde{p}(1-\tilde{p})}{n+4}}$.
• Conditions for using the plus four method:
  • Sample size $n \geq 10$.
  • Confidence level $C \geq 90\%$.

Example 1: Firstborn Children Gender

• Context: Investigate if boys are more likely to be born.
• Data: 25,468 firstborn children in the USA in 2005, with 13,973 being boys.
• Parameter of interest: Proportion of firstborn boys in the USA in 2005.
• Confidence level: 95%.
• Calculations:
  • $\hat{p} = \frac{13973}{25468} \approx 0.548$.
  • $n \times \hat{p} = 25468 \times 0.548 \approx 13957$. (Number of successes).
  • $n \times (1-\hat{p}) = 25468 \times (1-0.548) \approx 11511$. (Number of failures).
• Both values (13957, 11511) are significantly greater than 15, so the large sample approach can be used.
• $Z^*$ for a 95% confidence level: 1.96.
• Confidence Interval: $0.548 \pm 1.96 \sqrt{\frac{0.548(1-0.548)}{25468}}$.
• Result: (0.542, 0.554).
• Conclusion: We are 95% confident that the proportion of firstborn boys is between 54.2% and 55.4%.
• Since the entire interval is above 0.5 (50%), there's evidence to suggest that more boys than girls are born.

Example 2: Whelks Drilling into Mussels

• Context: Investigating the proportion of whelks that spontaneously drill into mussels.
• Data: 98 whelks, with 9 drilling into mussels.
• Confidence level: 95%.
• Calculations:
  • $\hat{p} = \frac{9}{98} \approx 0.092$.
  • $n \times \hat{p} = 98 \times 0.092 = 9$. (Number of successes).
  • $n \times (1-\hat{p}) = 98 \times (1-0.092) \approx 89$. (Number of failures).
• Since n \times \hat{p} = 9 < 15, the large sample method is not suitable.
• Using the plus four method:
  • $\tilde{p} = \frac{9+2}{98+4} = \frac{11}{102} \approx 0.108$.
• $Z^*$ for a 95% confidence level: 1.96.
• Confidence Interval: $0.108 \pm 1.96 \sqrt{\frac{0.108(1-0.108)}{102}}$.
• Result: (0.045, 0.171).
• Conclusion: We are 95% confident that the proportion of whelks drilling into mussels is between 4.5% and 17.1%.

Summary of Conditions

• Large Sample Confidence Interval for Proportion:
  • $n \times \hat{p} \geq 15$ and $n \times (1-\hat{p}) \geq 15$.
• Plus Four Method Confidence Interval for Proportion:
  • Sample size $n \geq 10$.
  • Confidence level $C \geq 90\%$.