Statistics 1 for Economics - 6B: T-Distribution and Inference

T-Distribution for Population Mean $\mu$ (Unknown Variance)

• Context of Application: The T-distribution is used for making inferences about the population mean $\mu$ when the population variance $\sigma^2$ is unknown.
• Standardization vs. Studentization:
  • When $\sigma^2$ is known, we use the Z-statistic: $Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim N(0, 1)$.
  • When $\sigma^2$ is unknown, we replace $\sigma$ with the sample standard deviation $s$. This yields the T-statistic: $T = \frac{\bar{X} - \mu}{s / \sqrt{n}} \sim t[df = n-1]$.
• Degrees of Freedom ($df$): The T-distribution is defined by its degrees of freedom, which for a single sample is $n - 1$.
  • As $df$ increases, the T-distribution approaches the Normal distribution ($N(0, 1)$).
  • Example: For a sample size of $n = 9$, the $df = 8$.

Confidence Intervals for a Single Mean ($\sigma^2$ unknown)

• General Formula: $\mu = \bar{x} \pm t_{\alpha/2, n-1} \left( \frac{s}{\sqrt{n}} \right)$.
• Components:
  • $\bar{x}$: Sample mean.
  • $t_{\alpha/2, n-1}$: Critical value from the T-distribution table based on the desired confidence level ($1 - \alpha$) and degrees of freedom.
  • $s$: Sample standard deviation ($s = \sqrt{s^2}$).
  • $n$: Sample size.
• Example: Medicine Tablets:
  • Data: A sample of 10 tablets yielded measurements: $11.8$, $11.7$, $12.5$, $11.6$, $12.1$, $11.7$, $11.8$, $11.4$, $11.7$, $11.7$.
  • Summary Stats: $n = 10, \bar{x} = 11.8, s^2 = 0.09111, s = 0.302$.
  • Calculation for 95% Confidence:
    • $1 - \alpha = 0.95 \implies \alpha/2 = 0.025$.
    • $df = 10 - 1 = 9$.
    • Critical value $t_{0.025, 9} = 2.262$.
    • Interval: $11.8 \pm 2.262 \times \frac{0.302}{\sqrt{10}} = 11.8 \pm 0.22$.
    • Result: $11.58 < \mu < 12.02$.

Hypothesis Testing for a Single Mean ($\sigma^2$ unknown)

• Process Steps:
  1. Assumptions and Conditions: Random sample; population variance $\sigma^2$ is unknown; population is normally distributed ($X \sim N$) or sample size is large enough to apply CLT ($n \ge 30$).
  2. Hypotheses:
     • $H_0: \mu = \mu_0$ (e.g., $H_0: \mu = 12$).
     • $H_1: \mu \neq \mu_0$.
  3. Test Statistic: $T = \frac{\bar{X} - \mu_0}{S / \sqrt{n}} \sim t[df = n - 1]$.
  4. Rejection Region: For $\alpha = 5$, reject if $|T| \ge t_{0.025, 9} = 2.262$.
  5. Sample Realization:
     • $t_{obs} = \frac{11.8 - 12}{0.302 / \sqrt{10}} = -2.09$.
  6. Decision: Since $-2.262 \le -2.09 \le 2.262$, we do not reject $H_0$.
  7. Conclusion: At the 5% significance level, there is insufficient evidence to conclude the mean active substance is different from 12 mg.

Determining the p-value with the T-Table

• Definition: The probability of observing evidence as strong or stronger than the observed result, assuming $H_0$ is true.
• Calculation (Two-sided): $p\text{-value} = P(T \le -|t_{obs}|) + P(T \ge |t_{obs}|) = 2 \times P(T \ge |t_{obs}|)$.
• Example Calculation:
  • $t_{obs} = -2.09, df = 9$.
  • $\frac{1}{2} p\text{-value} = P(T \ge 2.09)$.
  • From the T-table for $df=9$:
    • $t_{0.05, 9} = 1.833$
    • $t_{0.025, 9} = 2.262$
    • Since $1.833 < 2.09 < 2.262$, then $0.05 > \frac{1}{2} p\text{-value} > 0.025$.
    • Total $p\text{-value}$ Range: $0.10 \ge p\text{-value} \ge 0.05$.

Using the Statistical T-Table (Rules of Thumb)

• General Lookup: Find the column for significance and row for $df$.
• Missing $df$ Values:
  • Use the closest value in the table (e.g., for $df = 153$, use $df \approx 160$).
  • If exactly in the middle, use the smaller $df$ (e.g., for $df = 150$, use $df \approx 140$).
• Large Sample Approximations:
  • If $200 \le df < 400$, use $df \approx 200$.
  • If $df \ge 400$, use $df \approx \infty$, where $t_{\alpha} = z_{\alpha}$.

Determining Sample Size

• Objective: Achieve a specific margin of error ($B$) for a confidence interval.
• Formula: $n \ge \left( \frac{t_{\alpha/2, n-1} \times s}{B} \right)^2$.
• Iterative Process: Since $n$ changes, the degrees of freedom and resulting $t_{\alpha/2, n-1}$ also change. Additionally, the sample standard deviation $s$ may vary between samples, requiring recalculation of the margin.
• Accuracy Methods:
  • Smaller confidence level ($1-\alpha$) yields smaller $t$ and a smaller margin.
  • Larger $n$ yields a smaller margin.

Inference About the Difference Between Two Means ($\mu_1 - \mu_2$)

• Conditions and Assumptions:
  • Quantitative data collected from independent random samples.
  • Populations are normally distributed, OR sample sizes are large ($n_1 \ge 30$ and $n_2 \ge 30$).
  • Population variances $\sigma_1^2$ and $\sigma_2^2$ are unknown.
• General Case (Equal Variances Not Assumed):
  • Test Statistic: $T = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \sim t[df]$.
  • Degrees of Freedom (Welch-Satterthwaite): $df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(s_1^2 / n_1)^2}{n_1 - 1} + \frac{(s_2^2 / n_2)^2}{n_2 - 1}}$.
  • Confidence Interval: $(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$.

Example: Paint Drying Times

• Problem: Compare drying times of two manufacturers.
  • Manufacturer 1: $n_1 = 45, \bar{x}_1 = 185, s_1 = 20$.
  • Manufacturer 2: $n_2 = 40, \bar{x}_2 = 201, s_2 = 57$.
  • Significance level $\alpha = 10$.
• Hypothesis Test:
  • $H_0: \mu_1 - \mu_2 = 0$ vs. $H_1: \mu_1 - \mu_2 \neq 0$.
  • $df \approx 45$ (calculated via formula, result $47.499$ rounded down to nearest table entry).
  • $t_{crit} = t_{0.05, 45} = 1.679$.
  • $t_{obs} = \frac{185 - 201 - 0}{\sqrt{\frac{20^2}{45} + \frac{57^2}{40}}} = -1.69$.
  • Decision: Since $|t_{obs}| > 1.679$, reject $H_0$. There is sufficient evidence that mean drying times differ.
• 99% Confidence Interval:
  • $1 - \alpha = 0.99 \implies \alpha/2 = 0.005$.
  • $t_{0.005, 45} = 2.690$.
  • Interval: $(185 - 201) \pm 2.690 \times \sqrt{\frac{20^2}{45} + \frac{57^2}{40}} = -16 \pm 25$.

Special Case: Equal Population Variances ($\sigma_1^2 = σ_2^2$)

• Note: This is often referred to as the "pooled-variance" T-test. Keller Chapter 13.1 focuses primarily on the general case where equal variances are not assumed.
• Pooled Variance Estimate: $s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$.
• Test Statistic: $T = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{s_p^2 (\frac{1}{n_1} + \frac{1}{n_2})}}$.
• Degrees of Freedom: $df = n_1 + n_2 - 2$.
• Confidence Interval: $(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \sqrt{s_p^2 (\frac{1}{n_1} + \frac{1}{n_2})}$.

Case Study: Remplene Synthetic Fiber

• Research Question: Is the stretch of "Remplene" clothing reduced by washing?
• Experiment: 13 pieces total. 6 are washed 10 times, 7 are not washed.
  • Washed (n=6): $12.3, 13.7, 10.4, 11.4, 14.9, 12.6$.
  • Unwashed (n=7): $15.7, 10.3, 12.6, 14.5, 12.6, 13.8, 11.9$.
• SPSS Output Interpretation:
  • Group Statistics:
    • Washed Mean ($\bar{x}_1$): $12.550$, Std Dev: $1.6034$.
    • Unwashed Mean ($\bar{x}_2$): $13.057$, Std Dev: $1.7784$.
  • Levene's Test for Equality of Variances: $F = 0.135, Sig. = 0.720$. Since $Sig. > 0.05$, variances can be assumed equal.
  • T-test for Equality of Means:
    • Equal Variances Assumed: $t = -0.536, df = 11, Sig. (2-tailed) = 0.603$.
    • Equal Variances Not Assumed: $t = -0.541, df = 10.954, Sig. (2-tailed) = 0.600$.
  • Conclusion: Since the $p\text{-value} (0.603) > 0.05$, we fail to reject the null hypothesis. There is no significant evidence that washing reduces stretch.