Stats Recitation: Sampling Distributions, Confidence Intervals, and t-Intervals

Sampling Distributions and the Central Limit Theorem

• Parameters vs. Statistics:

  • A parameter is a numerical summary of a population (such as the population mean $m$ or $\mu$, and population standard deviation $\sigma$).

  • A statistic is a numerical summary computed from a sample (such as the sample mean $\bar{x}$ or $M$, and sample standard deviation $s$).

• Central Limit Theorem (CLT):

  • When drawing a random sample of size $n$ from any population, the sampling distribution of the sample mean $\bar{x}$ approaches a normal distribution as the sample size increases.

  • Threshold requirement: $n \ge 30$.

• Transitioning from Individual Observations to Sample Distributions:

  • When sample size $n$ is introduced, the population parameters $m$ and $\sigma$ transition to sampling distribution parameters $m_{\bar{x}}$ and $\sigma_{\bar{x}}$.

  • The mean of the sampling distribution remains equal to the population mean: $m_{\bar{x}} = m$.

  • The standard error (standard deviation of the sampling distribution) is divided by the square root of $n$: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$.

• Comparative Worked Example (Individual vs. Sample Mean):

  • Case 1: Individual Observation ($n = 1$):

  • Given parameters: Mean $m = 150$, standard deviation $\sigma = 20$.

  • Objective: Calculate the probability that an individual observation exceeds $165$, denoted as $P(X > 165)$.

  • Compute the standard normal test statistic ($Z$-score):       $Z = \frac{X - m}{\sigma} = \frac{165 - 150}{20} = \frac{15}{20} = 0.75$

  • Determine probability using cumulative distribution tables:       $P(Z > 0.75) = 1 - 0.7734 = 0.2266$

  • Case 2: Sample Mean Distribution ($n = 9$):

  • Given: Sample size $n = 9$, baseline mean $m = 150$, baseline standard deviation $\sigma = 20$.

  • Define sampling parameters:       $m_{\bar{x}} = 150$       $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{9}} = \frac{20}{3} = 6.67$

  • Objective: Calculate $P(\bar{X} > 165)$.

  • Compute the standard normal test statistic using standard error ($\sigma_{\bar{x}} = 6.667$):       $P\left(Z > \frac{165 - 150}{6.667}\right) = P(Z > 2.25)$

  • Determine probability using cumulative distribution tables:       $P(Z > 2.25) = 1 - 0.9878 = 0.0122$

Confidence Intervals for a Population Mean with Known Variance ($z$-Intervals)

• Formula and Definition:

  • When the population standard deviation $\sigma$ is known (or when sample size $n$ is large and $\sigma$ is modeled via standard normal critical values $z$):     $\bar{x} \pm Z_{\alpha/2} \left[\frac{\sigma}{\sqrt{n}}\right]$

  • Here, $\alpha = 1 - \text{Confidence Level}$, and $\frac{\alpha}{2}$ represents the area in each tail of the standard normal distribution.

• Confidence Level Trade-Off ($95\%$ vs. $90\%$):

  • Higher confidence requires capturing a larger portion of the central area under the curve, leaving smaller tails (e.g., $2.5\%$ per tail for $95\%$ confidence vs. $5\%$ per tail for $90\%$ confidence).

  • Principle: More confident $\implies$ longer (wider) interval.

• Worked Example: Estimating Apple Yield:

  • Given data: Sample size $n = 64$, sample mean $\bar{x} = 140$, standard deviation $\sigma = 20$, confidence level = $95\%$.

  • Determine significance level and critical value:     $1 - 0.95 = 0.05 \implies \alpha = 0.05$     $\frac{\alpha}{2} = \frac{0.05}{2} = 0.025$     $Z_{0.025} = 1.96$

  • Calculate margin of error:     $E = Z_{\alpha/2} \left[\frac{\sigma}{\sqrt{n}}\right] = 1.96 \left[\frac{20}{\sqrt{64}}\right] = 1.96 \left[\frac{20}{8}\right] = 1.96 [2.5] = 4.90$

  • Construct the confidence interval:     $140 \pm 4.90$     $(135.1, 144.9)$

  • Formal interpretation: We are $95\%$ confident the number of apples is between $135.1$ and $144.9$.

Sample Size Determination for Estimating a Population Mean

• Sample Size Formula:

  • Derived by setting the margin of error (or bound $B$) equal to $Z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right)$ and solving for $n$:     $n = \left(\frac{\sigma Z_{\alpha/2}}{B}\right)^2$

• Worked Example: Sample Size for Specified Precision:

  • Given: $\sigma = 20$, desired $90\%$ confidence interval within a margin of error of $B = \pm 3$.

  • Determine critical value:     $\alpha = 1 - 0.90 = 0.10$     $\frac{\alpha}{2} = \frac{0.10}{2} = 0.05$     $Z_{0.05} = 1.6449$

  • Calculate required sample size $n$:     $n = \left(\frac{20 \times 1.6449}{3}\right)^2 = \left(\frac{32.898}{3}\right)^2 = (10.966)^2 = 120.25 \approx 121$

  • Rule: Round up to the next integer to guarantee the margin of error does not exceed the bound $B$, yielding $n = 121$.

Small Sample Inference with Unknown Variance ($t$-Intervals)

• Conditions for Using the Student's $t$-Distribution:

  • Applied when the sample size $n$ is small ($n < 30$) and the population standard deviation $\sigma$ is unknown, requiring the use of the sample standard deviation $s$.

  • Degrees of freedom: $df = n - 1$.

  • Formula for the confidence interval:     $\bar{x} \pm t_{\alpha/2} \left[\frac{s}{\sqrt{n}}\right]$

• Worked Example: Small Sample Mean Estimation:

  • Given data: Sample size $n = 16$, sample mean $\bar{x} = 140$, sample standard deviation $s = 22$, confidence level = $95\%$.

  • Determine degrees of freedom and tail area:     $df = n - 1 = 16 - 1 = 15$     $1 - 0.95 = 0.05 \implies \frac{\alpha}{2} = 0.025$

  • Look up critical value for $df = 15$ with tail area $0.025$:     $t_{0.025, 15} = 2.1314$

  • Compute standard error and margin of error:     $\frac{s}{\sqrt{n}} = \frac{22}{\sqrt{16}} = \frac{22}{4} = 5.5$     $E = 2.1314 \times 5.5 = 11.7227 \approx 11.7$

  • Construct interval:     $140 \pm 11.7$

  • Upper limit: $140 + 11.7 = 151.7$

  • Lower limit: Transcribed in course records as $138.3$ (mathematical difference $140 - 11.7 = 128.3$).

Large-Sample Inferences and Sample Size for Population Proportions

• Confidence Intervals for Proportions:

  • Population parameter: $p$ (the true proportion of the population possessing a given attribute).

  • Sample proportion estimator: $\hat{p} = \frac{x}{n}$, where $x$ is the number of successes in $n$ trials.

  • Standard error of the proportion: $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.

  • Large-sample confidence interval formula:     $\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

• Practice Problem 8.110: European Travel Destinations:

  • Scenario: A survey of European trip planning.

  • Population ($POP$): All families planning a trip to Europe.

  • Parameter $p$: Percentage of the European trip planning population that want to visit France.

  • Sample data: Sample size $n = 125$ families, with $x = 15$ wanting to go to France.

  • Point estimate:       $\hat{p} = \frac{x}{n} = \frac{15}{125} = 0.12$

  • Compute $98\%$ Confidence Interval for the True Proportion:

  • Significance level:       $\alpha = 1 - 0.98 = 0.02$       $\frac{\alpha}{2} = 0.01$

  • Critical value for tail area $0.01$:       $Z_{0.01} = 2.3263$

  • Margin of error calculation:       $E = Z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = 2.3263 \sqrt{\frac{0.12(1 - 0.12)}{125}} = 2.3263 \sqrt{\frac{0.12(0.88)}{125}} = 2.3263 \times 0.02905 = 0.0676 \approx 0.07$

  • Interval calculation:       $\hat{p} \pm E = 0.12 \pm 0.07$       $[0.05, 0.19]$

  • Conclusion: We are $98\%$ confident that $0.05 \le p \le 0.19$.

• Sample Size Determination for Proportions Without Prior Estimates:

  • When no prior estimate of $\hat{p}$ is available, set $\hat{p} = 0.5$ to maximize the product $\hat{p}(1 - \hat{p}) = 0.5(0.5) = 0.25$, ensuring a conservative (maximum) sample size.

  • Sample size formula:     $n = \hat{p}(1 - \hat{p}) \left(\frac{Z}{B}\right)^2$

  • Application: Require bound of error $B = 0.05$ at $98\%$ confidence ($Z = 2.3263$):     $n = 0.5(1 - 0.5) \left(\frac{2.3263}{0.05}\right)^2 = 0.25 \left(46.526\right)^2 = 0.25 \times 2164.67 = 541.17 \approx 542$

Systematic Factors Controlling Margin of Error

• Conditions That Reduce Margin of Error:

  • The margin of error is defined as $E = Z_{\alpha/2} \left(\frac{\sigma}{\sqrt{n}}\right)$.

  • A smaller margin of error (higher precision) is achieved by:

  • Lowering the population standard deviation $\sigma$ (variability lower than expected).

  • Lowering the confidence level (which decreases critical value $Z_{\alpha/2}$ or $t_{\alpha/2}$).

  • Increasing the sample size $n$ (which increases the denominator $\sqrt{n}$).

Comprehensive Homework and Recitation Problem Sets

• Problem 1: Standard Normal Critical Value for $99\%$ Confidence:

  • Target: Find the $Z$-score corresponding to a central area of $0.99$.

  • Tail area:     $\frac{1 - 0.99}{2} = 0.005$

  • Critical value:     $Z(0.005) = 2.5758$

• Problem 2: Mining Closure Operation Days:

  • Study context: Over $30$ years, closed $\ge 6,000$ mines out of a total population of $17,000$.

  • Sample parameters: Random sample of $n = 40$ days ($40 \ge 30$, satisfying CLT), sample mean $\bar{x} = 15.07$, population standard deviation $\sigma = 1.8$, confidence level = $95\%$.

  • Significance level and critical value:     $1 - 0.95 = 0.05 \implies \frac{\alpha}{2} = 0.025$     $Z(0.025) = 1.9600$

  • Margin of error computation:     $E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.96 \cdot \frac{1.8}{\sqrt{40}} = 1.96 \cdot \frac{1.8}{6.324555} = 1.96 \times 0.2846 = 0.558$

  • Boundary calculations:     $15.07 \pm 0.558$

  • Upper boundary: $15.07 + 0.558 = 15.63$

  • Lower boundary: $15.07 - 0.558 = 14.51$

• Problem 3: Multiple Sample Mean Evaluations and Standard Errors:

  • Sample size $rs = 15$:

  • Mean $M = 38.1$, $\sigma = 1.0$, confidence level = $95\%$.

  • Standard error calculation:       $\frac{\sigma}{\sqrt{rs}} = \frac{1.0}{\sqrt{15}} = 0.2582$

  • Margin of error:       $E = 1.96 \times 0.2582 = \pm 0.51$

  • Sample size $10$:

  • Given sample mean: $M(\text{Size } 10) = 39.75$.

• Problem 4: Oocyte Counts in Age Cohort:

  • Given data: Sample size $n = 98$, subject age cohort $36\text{--}40\,\text{years old}$, sample mean $M = 9.7$, confidence level = $95\%$.

  • Resulting standard normal interval: $(8.3, 11.1)$.

  • Interpretation: We are $95\%$ confident that the true mean of all oocytes is between $8.3$ and $11.1$.

• Problem 5: Normal Distribution Mean Estimation ($n = 18$):

  • Given parameters: Sample mean $\bar{x} = 57,500$, standard deviation $\sigma = 9,500$, sample size $n = 18$, confidence level = $90\%$.

  • Critical value:     $\alpha = 1 - 0.90 = 0.10 \implies \frac{\alpha}{2} = 0.05$     $Z_{0.05} = 1.6449$

  • Margin of error computation:     $E = 1.6449 \times \frac{9,500}{\sqrt{18}} = 1.6449 \times \frac{9,500}{4.24264} = 1.6449 \times 2239.17 = 3,683$

  • Construct interval:     $57,500 \pm 3,683$

  • Upper limit: $57,500 + 3,683 = 61,183$

  • Lower limit: $57,500 - 3,683 = 53,817$

• Problem 6: Sample Size for Specified Margins of Error ($90\%$ Confidence):

  • Formula:     $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$

  • Parameters: $\sigma = 9,500$, $Z = 1.645$.

  • Case A: Bound of Error $E = 3,000$:     $n = \left(\frac{1.645 \times 9,500}{3,000}\right)^2 = \left(\frac{15,627.5}{3,000}\right)^2 = (5.209)^2 = 27.13 \implies n = 28$

  • Case B: Bound of Error $E = 1,000$:     $n = \left(\frac{1.645 \times 9,500}{1,000}\right)^2 = (15.6275)^2 = 244.2 \implies n = 245$

• Problem 7: Infant Birth Weight Sample Size Determination:

  • Parameters: Mean $m = 3,600$, standard deviation $\sigma = 600$, confidence level = $99\%$, desired bound within $E = 150\,g$.

  • Critical value for $99\%$ confidence: $Z = 2.576$.

  • Sample size computation:     $n = \left(\frac{Z \cdot \sigma}{E}\right)^2 = \left(\frac{2.576 \times 600}{150}\right)^2 = \left(\frac{1,545.6}{150}\right)^2 = (10.304)^2 = 106.17 \approx 106.7 \implies n = 107\,\text{babies}$

• Problem 8: Sampling Methodology:

  • Design requirement: Simple random sample ($SRS$) required for unbiased parameter estimation.

• Problem 12: High-Precision Mean Estimation ($n = 37$):

  • Parameters: Sample size $n = 37$, sample mean $\bar{x} = 16.93514$, population standard deviation $\sigma = 2.5$, sample standard deviation $s = 2.7$, confidence level = $99\%$.

  • Standard error computation:     $SE = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{37}} = \frac{2.5}{6.08276} = 0.41100$

  • Margin of error ($Z_{0.005} = 2.5758$):     $E = 2.5758 \times 0.41100 = 1.05866$

  • Confidence interval limits:     $\bar{x} \pm E = 16.93514 \pm 1.05866$

  • Lower boundary: $16.935 - 1.05866 = 15.8763$

  • Upper boundary: $16.935 + 1.05866 = 17.9987$

• Problem 13: Aggregated Sample Estimation ($n = 45$):

  • Parameters: Sample size $n = 45$, sum of observations $\sum X = 935.3$, standard deviation $\sigma = 2.7$, confidence level = $99\%$.

  • Sample mean:     $\bar{x} = \frac{935.3}{45} = 20.784$

  • Standard error:     $SE = \frac{\sigma}{\sqrt{n}} = \frac{2.7}{\sqrt{45}} = \frac{2.7}{6.7082} = 0.4024922$

  • Margin of error ($Z_{0.005} = 2.5758$):     $E = 2.5758 \times 0.4024922 = 1.0367513$

  • Confidence interval upper bound:     $\bar{x} + E = 20.784 + 1.0367513 = 21.8211957$

• Problem 14: Comparative Analysis of Powder Depth:

  • Confidence interval for Breckenridge: $(15.8763, 17.9987)$.

  • Comparative site confidence interval: $(19, 21)$.

  • Observational finding: The intervals do not overlap; there is a large gap between the mean powder depth intervals.

• Problem 15: Critical Value Lookups:

  • Distribution parameter: $t_{0.0001, 3} = 22.2037$.

• Problem 16: Small Sample $t$-Interval with Unknown Variance ($n = 27$):

  • Given data: Sample size $n = 27$, sample mean $\bar{x} = 211.2$, variance measure $s^2 = 6,344.37$, confidence level = $95\%$.

  • Degrees of freedom and significance level:     $df = n - 1 = 27 - 1 = 26$     $1 - 0.95 = 0.05 \implies \frac{\alpha}{2} = 0.025$

  • Look up critical value: Evaluate critical value at $df = 26$ with tail area $0.025$.

  • Confidence interval formula:     $\bar{x} \pm t_{\alpha/2, 26} \left[\frac{s}{\sqrt{n}}\right]$

• Problem 17: Small Sample Assessment ($n = 21$):

  • Given data: Sample size $n = 21$, sample mean $\bar{x} = 37.8$, sample standard deviation $s = 5.9$, confidence level = $90\%$.

  • Degrees of freedom:     $df = n - 1 = 21 - 1 = 20$

  • Significance level:     $\alpha = 1 - 0.90 = 0.10 \implies \frac{\alpha}{2} = 0.05$

  • Critical value for $df = 20$ with tail area $0.05$:     $t_{0.05, 20} = 1.7247$

  • Margin of error calculation:     $E = t_{\alpha/2} \left(\frac{s}{\sqrt{n}}\right) = 1.7247 \left(\frac{5.9}{\sqrt{21}}\right) = 1.7247 \left(\frac{5.9}{4.5825757}\right) = 1.7247 \times 1.28748 = 2.2205$

  • Confidence interval:     $37.8 \pm 2.2205$

  • Upper boundary: $37.8 + 2.2205 = 40.0205$

  • Lower boundary: $37.8 - 2.2205 = 35.5795$