Statistical Analysis

Lecture 7: T-Tests and Correlations

Bogdan G. Popescu
bogdan.popescu@johncabot.edu

John Cabot University

Lecture Overview

  • From z-tests to t-tests
  • Confidence intervals for population estimates
  • The Pearson correlation coefficient
  • Correlation vs. causation

T-Tests

Recall: Z-Test Assumptions

  • Population mean \(\mu\) is known
  • Population standard deviation \(\sigma\) is known
  • The second assumption is rarely met in practice

Solution: use the t-test, which replaces \(\sigma\) with the sample standard deviation \(s_x\).

The t-Distribution

  • Symmetric, bell-shaped, mean = 0
  • Does not require known \(\sigma\)
  • Defined by degrees of freedom (df = \(n - 1\))
  • Heavier tails than normal for small \(n\)
  • Converges to normal as \(n\) increases

t-Distribution vs. Normal

Figure 1: The t-distribution has heavier tails for small samples

Z-Observed vs. T-Observed

Z-observed (requires known \(\sigma\)):

\[Z_{\text{obs}} = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}\]

T-observed (uses sample \(s_x\)):

\[t_{\text{obs}} = \frac{\bar{X} - \mu}{s_x / \sqrt{n}}\]

The only difference: \(\sigma \to s_x\)

Degrees of Freedom

  • df = independent pieces of information
  • For a one-sample t-test: \(\text{df} = n - 1\)
  • Lower df = heavier tails = harder to reject \(H_0\)
  • As df \(\to \infty\), t-distribution \(\to\) normal

NHST Steps (Same as Before)

  1. Form groups in the data
  2. Define \(H_0\) (no difference)
  3. Set \(\alpha\) (default = 0.05)
  4. Choose one- or two-tailed test
  5. Find the critical value
  6. Calculate the observed value

Example: EU vs. Latin America

Does life expectancy differ between EU and Latin America?

Statistic Latin America EU
Mean life exp. 61.914 67.114
SD 5.14 5.55
n 23 27
  • \(H_0\): \(\mu_{\text{Latam}} = \mu_{\text{EU}}\)
  • \(H_1\): \(\mu_{\text{Latam}} \neq \mu_{\text{EU}}\) (two-tailed)

T-Test Result: Life Expectancy

Result Value
t-statistic -3.4364
df 47.64
p-value 0.001232
95% CI [-8.24, -2.16]

p < 0.05: reject \(H_0\). EU life expectancy is significantly higher than Latin America’s.

Visualizing the T-Test

Figure 2: Observed t falls in the rejection region

Checking Assumptions: QQ Plots

Before trusting the test, check normality of each group:

Figure 3: Latin America (n = 23)
Figure 4: EU (n = 27)

Checking Assumptions: Shapiro-Wilk

Group W statistic p-value Normal?
Latin America 0.963 0.5255 Yes
EU 0.8486 0.0011 No
  • If p > 0.05: we cannot reject normality
  • T-test is robust to mild non-normality when n > 20

Latin America vs. South America

Latin America includes Central America and Caribbean — not just South America:

Region Countries n
Latin America 23 countries (incl. Mexico, Cuba, etc.) 23
South America 12 countries (Argentina to Venezuela) 12
  • Which region you choose affects your sample size
  • Let’s see what happens with the smaller group

The Small-n Problem

South America (n=12) EU (n=27)
Mean 62.965 67.114
SD 4.17 5.55
Result S. America vs. EU Latam vs. EU
95% CI [-7.45, -0.85] [-8.24, -2.16]
CI width 6.6 6.09
p-value 0.0156 0.001232

Lesson: Sample Size Drives Power

Figure 5

  • Same underlying effect, but less precision with n = 12

Second Comparison: Urbanization

Statistic Latin America EU
Mean urb. 59.018 67.138
SD 16.3 13.12
Result Value
t-statistic -1.9176
p-value 0.06195

p = 0.062 > 0.05: fail to reject \(H_0\). No significant difference in urbanization.

Could We Be Wrong?

\(H_0\) True \(H_0\) False
Reject \(H_0\) Type I Error (\(\alpha\)) Correct
Retain \(H_0\) Correct Type II Error (\(\beta\))
  • Rejected \(H_0\) for life expectancy — Type I error possible?
  • Retained \(H_0\) for urbanization — Type II error possible?
  • Lowering \(\alpha\) reduces Type I but increases Type II

Z-Test vs. T-Test

Criterion Z-Test T-Test
Pop. variance Known Unknown
Sample size Large (n > 30) Any (esp. small)
Distribution Normal Student’s t
Practice Rarely used Frequently used

Social data typically come from samples, not populations, so the t-test is the standard tool.

Confidence Intervals

What Is a Confidence Interval?

  • Sample estimates vary from true population parameters
  • A CI gives a plausible range for the parameter
  • 95% CI: repeated sampling → 95% of CIs contain \(\mu\)

Confidence Interval Formula

\[CI = \bar{X} \pm t \times \frac{s_x}{\sqrt{n}}\]

Where:

  • \(\bar{X}\): sample mean
  • \(s_x\): sample standard deviation
  • \(n\): sample size
  • \(t\): critical value (for 95% CI \(\approx\) 1.96)

Worked Example: Latin America

\[CI = 59.018 \pm 2.074 \times \frac{16.3}{\sqrt{23}}\]

\[CI = [51.97,\; 66.07]\]

Our best estimate for mean LATAM urbanization is 59.018, but it plausibly ranges from 51.97 to 66.07.

Visualizing Confidence Intervals

Figure 6: 95% confidence intervals for regional urbanization means

Interpreting Overlapping CIs

  • The CIs overlap between the two regions
  • This is consistent with failing to reject \(H_0\)
  • Overlapping CIs suggest no significant difference
  • Non-overlapping CIs would suggest a significant gap

Correlation

What Is Correlation?

  • Measures strength and direction of a linear relationship
  • Ranges from \(-1\) (perfect negative) to \(+1\) (perfect positive)
  • 0 means no linear relationship
  • Example: urbanization and life expectancy

Scatterplot: World

Figure 7: Urbanization vs. life expectancy across all countries

Scatterplots: EU vs. Latin America

Figure 8: EU: negative correlation
Figure 9: Latin America: positive correlation

Pearson Correlation: Formula

\[r = \frac{\sum(x_i - \bar{x})(y_i - \bar{y}) / (n-1)}{s_x \cdot s_y}\]

Where:

  • \(x_i, y_i\): observed values
  • \(\bar{x}, \bar{y}\): sample means
  • \(s_x, s_y\): sample standard deviations
  • \(n\): sample size

Computing \(r\): 13 Steps

  1. Mean of X (\(\bar{x}\))
  2. Mean of Y (\(\bar{y}\))
  3. Deviation scores for X
  4. Deviation scores for Y
  5. Square deviations for X
  6. Square deviations for Y
  1. Sum squared deviations for X
  2. Sum squared deviations for Y
  3. Standard deviation of X (\(s_x\))
  4. Standard deviation of Y (\(s_y\))
  5. Cross-products
  6. Sum cross-products
  7. Pearson \(r\)

Step 1: Mean of X

X = Urbanization

\[\bar{x} = \frac{\sum x_i}{n} = 51.33\]

Country Urb. (x)
Afghanistan 18.6118
Albania 40.4442
Algeria 52.6092
American Samoa 79.7625
… …
Zimbabwe 26.3045
\(\bar{x}\) 51.33

Step 2: Mean of Y

Y = Life Expectancy

\[\bar{y} = \frac{\sum y_i}{n} = 61.259\]

Country Urb. (x) Life Exp. (y)
Afghanistan 18.6118 45.3833
Albania 40.4442 68.2861
Algeria 52.6092 57.5301
American Samoa 79.7625 68.6375
… … …
Zimbabwe 26.3045 54.2472
\(\bar{x}\) 51.33
\(\bar{y}\) 61.259

Step 3: Deviation Scores for X

\[x_i - \bar{x}\]

Country Urb. (x) \(x_i - \bar{x}\)
Afghanistan 18.6118 18.612 - 51.33
Albania 40.4442 40.444 - 51.33
Algeria 52.6092 52.609 - 51.33
American Samoa 79.7625 79.762 - 51.33
… … …
Zimbabwe 26.3045 26.304 - 51.33
\(\bar{x}\) 51.33

Step 3: Deviation Scores for X

\[x_i - \bar{x}\]

Country Urb. (x) \(x_i - \bar{x}\)
Afghanistan 18.6118 -32.7179
Albania 40.4442 -10.8855
Algeria 52.6092 1.2795
American Samoa 79.7625 28.4328
… … …
Zimbabwe 26.3045 -25.0252
\(\bar{x}\) 51.33

Step 4: Deviation Scores for Y

\[y_i - \bar{y}\]

Country Life Exp. (y) \(y_i - \bar{y}\)
Afghanistan 45.3833 45.383 - 61.259
Albania 68.2861 68.286 - 61.259
Algeria 57.5301 57.53 - 61.259
American Samoa 68.6375 68.638 - 61.259
… … …
Zimbabwe 54.2472 54.247 - 61.259
\(\bar{y}\) 61.259

Step 4: Deviation Scores for Y

\[y_i - \bar{y}\]

Country Life Exp. (y) \(y_i - \bar{y}\)
Afghanistan 45.3833 -15.8757
Albania 68.2861 7.0271
Algeria 57.5301 -3.7289
American Samoa 68.6375 7.3785
… … …
Zimbabwe 54.2472 -7.0118
\(\bar{y}\) 61.259

Step 5: Squared Deviations for X

\[(x_i - \bar{x})^2\]

Country \(x_i - \bar{x}\) \((x_i - \bar{x})^2\)
Afghanistan -32.7179 (-32.7179)²
Albania -10.8855 (-10.8855)²
Algeria 1.2795 (1.2795)²
American Samoa 28.4328 (28.4328)²
… … …
Zimbabwe -25.0252 (-25.0252)²

Step 5: Squared Deviations for X

\[(x_i - \bar{x})^2\]

Country \(x_i - \bar{x}\) \((x_i - \bar{x})^2\)
Afghanistan -32.7179 1070.4623
Albania -10.8855 118.4943
Algeria 1.2795 1.6372
American Samoa 28.4328 808.4251
… … …
Zimbabwe -25.0252 626.2598

Step 6: Squared Deviations for Y

\[(y_i - \bar{y})^2\]

Country \(y_i - \bar{y}\) \((y_i - \bar{y})^2\)
Afghanistan -15.8757 (-15.8757)²
Albania 7.0271 (7.0271)²
Algeria -3.7289 (-3.7289)²
American Samoa 7.3785 (7.3785)²
… … …
Zimbabwe -7.0118 (-7.0118)²

Step 6: Squared Deviations for Y

\[(y_i - \bar{y})^2\]

Country \(y_i - \bar{y}\) \((y_i - \bar{y})^2\)
Afghanistan -15.8757 252.038
Albania 7.0271 49.3797
Algeria -3.7289 13.9047
American Samoa 7.3785 54.4417
… … …
Zimbabwe -7.0118 49.1656

Step 7: Sum Squared Deviations for X

\[\sum_{i=1}^{n}(x_i - \bar{x})^2 = 1070.4623 + 118.4943 + \cdots + 626.2598\]

\[\sum_{i=1}^{215}(x_i - \bar{x})^2 = 1.2512211\times 10^{5}\]

Step 8: Sum Squared Deviations for Y

\[\sum_{i=1}^{n}(y_i - \bar{y})^2 = 252.038 + 49.3797 + \cdots + 49.1656\]

\[\sum_{i=1}^{215}(y_i - \bar{y})^2 = 1.5988861\times 10^{4}\]

Step 9: Standard Deviation of X

\[s_x = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{1.2512211\times 10^{5}}{215 - 1}} = \sqrt{\frac{1.2512211\times 10^{5}}{214}}\]

\[s_x = 24.18021\]

Step 10: Standard Deviation of Y

\[s_y = \sqrt{\frac{\sum(y_i - \bar{y})^2}{n-1}} = \sqrt{\frac{1.5988861\times 10^{4}}{215 - 1}} = \sqrt{\frac{1.5988861\times 10^{4}}{214}}\]

\[s_y = 8.64374\]

Step 11: Cross-Products

\[(x_i - \bar{x})(y_i - \bar{y})\]

Country \(x_i - \bar{x}\) \(y_i - \bar{y}\) \((x_i-\bar{x})(y_i-\bar{y})\)
Afghanistan -32.7179 -15.8757 (-32.7179)(-15.8757)
Albania -10.8855 7.0271 (-10.8855)(7.0271)
Algeria 1.2795 -3.7289 (1.2795)(-3.7289)
American Samoa 28.4328 7.3785 (28.4328)(7.3785)
… … … …
Zimbabwe -25.0252 -7.0118 (-25.0252)(-7.0118)

Step 11: Cross-Products

\[(x_i - \bar{x})(y_i - \bar{y})\]

Country \(x_i - \bar{x}\) \(y_i - \bar{y}\) \((x_i-\bar{x})(y_i-\bar{y})\)
Afghanistan -32.7179 -15.8757 519.4201
Albania -10.8855 7.0271 -76.4933
Algeria 1.2795 -3.7289 -4.7713
American Samoa 28.4328 7.3785 209.7904
… … … …
Zimbabwe -25.0252 -7.0118 175.472

Step 12: Sum of Cross-Products

\[\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y}) = 519.4201 + -76.4933 + \cdots + 175.472\]

\[\sum_{i=1}^{215}(x_i - \bar{x})(y_i - \bar{y}) = 2.819337\times 10^{4}\]

Step 13: Pearson \(r\)

\[r = \frac{\frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{n-1}}{s_x \cdot s_y} = \frac{\frac{2.819337\times 10^{4}}{214}}{24.18021 \times 8.64374}\]

\[r = \frac{131.7447}{209.0076} = 0.6303347\]

Computing \(r\) in R

Statistic Value
Pearson \(r\) 0.6303
t-statistic 11.85
p-value <2e-16

\(r =\) 0.63: strong positive correlation. Statistically significant at \(\alpha = 0.05\).

Interpreting Correlation Strength

\(|r|\) Range Interpretation
0.00 – 0.19 Very weak
0.20 – 0.39 Weak
0.40 – 0.59 Moderate
0.60 – 0.79 Strong
0.80 – 1.00 Very strong

Our \(r =\) 0.63 falls in the strong range.

Caveat: These are Cohen-style rules of thumb, not universal laws. What counts as “strong” depends on the field — \(r = 0.3\) may be large in psychology but small in physics.

Correlation vs. Causation

Why Correlation \(\neq\) Causation

  • Reverse causation: does X cause Y, or Y cause X?
  • Confounders: a third variable Z may drive both
  • Measurement error: both X and Y may be noisy
  • All three create endogeneity

A Classic Example

Ice cream sales and drowning deaths are positively correlated.

Does ice cream cause drowning?

No — summer temperature drives both:

  • Hot weather → more ice cream purchased
  • Hot weather → more people swimming → more drownings

The correlation is real; the causal story is wrong.

The Time-Trend Confound

Both variables trend upward over time — is the correlation just a shared trend?

Confounders: A Visual

%%{init:{'flowchart':{'useMaxWidth':true,'nodeSpacing':60,'rankSpacing':60},'themeVariables':{'fontSize':'22px'}}}%%
flowchart TD
  Z["Confounder<br/>(e.g., GDP)"] --> X["Urbanization"]
  Z --> Y["Life Expectancy"]
  X -. "correlation<br/>(not causation)" .-> Y

  • GDP drives both: richer → more urban + longer lives

Why \(r \approx\) 0.63 Is Partly an Artifact

  • Both variables rise with development over time
  • Countries that developed early score high on both
  • The \(r =\) 0.63 reflects shared development, not direct cause
  • The time-trend plots confirm this parallel movement

Reverse Causation

%%{init:{'flowchart':{'useMaxWidth':true,'nodeSpacing':60,'rankSpacing':60},'themeVariables':{'fontSize':'22px'}}}%%
flowchart LR
  X["Urbanization"] -- "?" --> Y["Life Expectancy"]
  Y -- "?" --> X

  • Does urbanization improve life expectancy?
  • Or does higher life expectancy enable urbanization?
  • With observational data alone, we cannot tell

What Would We Need?

  • Correlation tells us what moves together
  • It does not tell us what causes what
  • To establish causation, we need:
    • Randomized experiments (gold standard)
    • Quasi-experimental designs (natural experiments, instrumental variables)
  • Always ask: “What else could explain this pattern?”

Your Turn

Exercise: Africa vs. EU

Using the same dataset, answer these questions:

  1. Define an Africa subgroup (pick 10–15 countries)
  2. Run a t-test comparing life expectancy: Africa vs. EU
  3. Check normality with a QQ plot
  4. Compute and interpret the 95% CI for the difference
  5. Is the result significant? What is the effect size?
  • Bonus: compute Pearson’s \(r\) for Africa alone

Conclusion

Key Takeaways

  • T-tests replace z-tests when \(\sigma\) is unknown
  • Always check assumptions: QQ plots and Shapiro-Wilk
  • Sample size matters: small n → wide CIs, low power
  • Confidence intervals quantify estimation uncertainty
  • Pearson’s \(r\) measures linear association strength
  • Correlation \(\neq\) causation: shared trends inflate \(r\)