Estimating means, proportions, and variances is crucial for making inferences about populations. This topic covers methods like t-distribution for small samples, normal approximation for proportions, and chi-square distribution for variances.

These techniques help create confidence intervals, giving us a range of likely values for population parameters. Understanding when to use each method and how sample size affects our estimates is key to accurate statistical analysis.

Confidence Intervals for Means

T-Distribution for Small Samples

Use t-distribution for constructing confidence intervals when population standard deviation unknown and sample size small (typically n < 30)
Calculate degrees of freedom for t-distribution as n - 1 (n represents sample size)
Construct confidence interval for population mean using formula $\bar{x} \pm (t * (s / \sqrt{n}))$ $\overset{x}{ˉ} \pm (t * (s / n))$
- $\bar{x}$ represents sample mean
- t represents critical value from t-distribution
- s represents sample standard deviation
- n represents sample size
Determine critical value t based on desired confidence level and degrees of freedom
T-distribution approaches normal distribution as sample size increases
- Becomes more appropriate for larger samples
Margin of error in t-distribution confidence interval influenced by:
- Sample size
- Sample standard deviation
- Chosen confidence level
Example: Constructing 95% confidence interval for mean height of 25 students
- Sample mean = 170 cm, sample standard deviation = 8 cm
- Degrees of freedom = 24, t-critical value (95% confidence) = 2.064
- Confidence interval: 170 ± (2.064 * (8 / √25)) = (166.7 cm, 173.3 cm)

Estimating Proportions and Confidence Intervals

Normal Approximation for Proportions

Use normal approximation to binomial distribution for constructing confidence intervals for population proportions
- Applicable when np ≥ 5 and n(1-p) ≥ 5
- n represents sample size
- p represents sample proportion
Calculate confidence interval for population proportion using formula $\hat{p} \pm (z * \sqrt{(\hat{p}(1-\hat{p}))/n})$ $\overset{p}{^} \pm (z * (\overset{p}{^} (1 - \overset{p}{^})) / n)$
- $\hat{p}$ represents sample proportion
- z represents critical value from standard normal distribution
- n represents sample size
Compute standard error of proportion as $\sqrt{(\hat{p}(1-\hat{p}))/n}$
Margin of error in proportion confidence interval influenced by:
- Sample size
- Sample proportion
- Chosen confidence level
Consider alternative methods for small samples or extreme proportions:
- Wilson score interval
- Clopper-Pearson interval
Normal approximation method assumes sampling distribution of sample proportion approximately normal
- Generally true for large samples due to Central Limit Theorem
Example: Estimating proportion of left-handed people in population
- Sample size = 1000, left-handed individuals = 110
- Sample proportion = 0.11, z-critical value (95% confidence) = 1.96
- Confidence interval: 0.11 ± (1.96 * √((0.11 * 0.89) / 1000)) = (0.091, 0.129)

Point Estimates and Confidence Intervals for Variance

Estimating Population Variance and Standard Deviation

Use sample variance (s²) as point estimate for population variance (σ²)
Use sample standard deviation (s) to estimate population standard deviation (σ)
Construct confidence intervals for population variances using chi-square distribution
- Sampling distribution of sample variance follows chi-square distribution
Calculate confidence interval for population variance using formula $((n-1)s²)/χ²_{upper} < σ² < ((n-1)s²)/χ²_{lower}$ $((n - 1) s^{2}) / χ_{u pp er}^{2} < σ^{2} < ((n - 1) s^{2}) / χ_{l o w er}^{2}$
- n represents sample size
- s² represents sample variance
- χ² represents critical values from chi-square distribution
Set degrees of freedom for chi-square distribution in variance estimation to n - 1
Obtain confidence intervals for population standard deviations by taking square root of lower and upper bounds of variance confidence interval
Width of confidence interval for variances and standard deviations influenced by:
- Sample size
- Chosen confidence level
Assume population normally distributed for validity of confidence intervals
Example: Estimating population variance of exam scores
- Sample size = 40, sample variance = 25
- 95% confidence interval for variance: (17.76, 38.39)
- 95% confidence interval for standard deviation: (4.21, 6.20)

Central Limit Theorem and Normal Approximations

Applying Central Limit Theorem in Interval Estimation

Central Limit Theorem (CLT) states sampling distribution of sample mean approaches normal distribution as sample size increases
- Applies regardless of underlying population distribution
Apply CLT for practical purposes when sample size ≥ 30
- Assumes population not extremely skewed
CLT justifies use of normal approximations in interval estimation for means
- Applicable even when population distribution unknown or non-normal
Apply CLT to sampling distribution of sample proportion for proportions
- Allows normal approximations when np ≥ 5 and n(1-p) ≥ 5
Standard error of mean (SEM) decreases as sample size increases
- Key implication of CLT in interval estimation
CLT enables use of z-scores and t-scores in constructing confidence intervals
- Based on normal and approximately normal distributions
Understanding CLT crucial for determining appropriate interval estimation methods
- Parametric methods for large samples or normal populations
- Non-parametric or bootstrap methods for small samples or non-normal populations
Example: Applying CLT to estimate mean household income
- Sample size = 100, sample mean = $50,000, sample standard deviation =$ 10,000
- 95% confidence interval: $50,000 ± (1.96 * ($ 10,000 / √100)) = ( $48,040,$ 51,960)

2,589 studying →