2025-10-06 13:32 Tags:Bordeaux

Two-sample t-test Example

What is a t-test?

A t-test is a statistical test that compares means (averages).
It asks:

“Are the observed differences in means big enough that they’re unlikely to be due to random chance?”

Question: Do mothers of low-birth-weight babies have a different average weight compared to mothers of normal-birth-weight babies?

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Step 1: Hypotheses

• Null hypothesis:

$$H_0:{\mu }_{\text{low}}={\mu }_{\text{normal}}$$

• Alternative hypothesis:

$$H_a:{\mu }_{\text{low}}\ne {\mu }_{\text{normal}}$$

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Step 2: Sample data (illustration)

• Group 1 (low BW mothers):
  $n_1=10,{\bar{x}}_1=115,s_1=10$

• Group 2 (normal BW mothers):
  $n_2=12,{\bar{x}}_2=130,s_2=12$

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Step 3: Formula for two-sample t-test (equal variances)

1. Pooled variance:

$$s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$$

2. Standard error (SE):

$$SE=s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$

3. t-statistic:

$$t=\frac{{\bar{x}}_1-{\bar{x}}_2}{SE}$$

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Step 4: Plug in numbers

1. Pooled variance:

$$s_p^2=\frac{(10-1)(10^2)+(12-1)(12^2)}{10+12-2}=\frac{900+1584}{20}=124.2$$

So pooled SD:

$$s_p=\sqrt{124.2}\approx 11.15$$

2. Standard error:

$$SE=11.15\cdot \sqrt{\frac{1}{10}+\frac{1}{12}}=11.15\cdot \sqrt{0.1833}=11.15\cdot 0.428\approx 4.77$$

3. t-statistic:

$$t=\frac{115-130}{4.77}=\frac{-15}{4.77}\approx -3.15$$

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Step 5: Decision

• Degrees of freedom:

$$df=n_1+n_2-2=20$$

• Critical value (two-sided, $\alpha =0.05$): $\pm 2.086$

• Our test statistic: $t=-3.15$

Since $|t|>2.086$, we reject $H_0$.

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✅ Conclusion

There is a significant difference in maternal weights between mothers of low vs. normal birth weight babies (p < 0.01).

Chi-square Test of Independence Example

1. What is a chi-square test?

A chi-square test is used for categorical variables (yes/no, male/female, race groups, etc.).

It answers questions like:

“Are these two categorical variables related, or are they independent?”
“Do proportions differ across groups?”

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Step 1: Hypotheses

• Null hypothesis:

$$H_0:\text{Proportion of low birth weight infants is the same across races}$$

• Alternative hypothesis:

$$H_a:\text{Proportion of low birth weight infants differs by race}$$

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Step 2: Sample data (illustration)

Race	Low BW (1)	Normal BW (0)	Total
White	20	180	200
Black	15	85	100
Other	10	90	100
Total	45	355	400

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Step 3: Expected counts

Formula:

$$E_{ij}=\frac{\text{Row total}\times \text{Column total}}{\text{Grand total}}$$

Example for White–Low BW:

$$E=\frac{200\times 45}{400}=22.5$$

Do this for each cell.

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Step 4: Compute chi-square statistic

Formula:

$${\chi }^2=\sum \frac{(O-E{)}^2}{E}$$

Let’s calculate a few cells:

• White–Low BW:

$$\frac{(20-22.5{)}^2}{22.5}=\frac{6.25}{22.5}\approx 0.278$$

• Black–Low BW:

$$\frac{(15-11.25{)}^2}{11.25}=\frac{14.06}{11.25}\approx 1.25$$

• Other–Low BW:

$$\frac{(10-11.25{)}^2}{11.25}=\frac{1.56}{11.25}\approx 0.139$$

… and similarly for the “Normal BW” cells.

Add them all up:

$${\chi }^2\approx 2.39$$

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Step 5: Decision

• Degrees of freedom:

$$df=(\text{rows}-1)(\text{cols}-1)=(3-1)(2-1)=2$$

• Critical value (α = 0.05, df = 2): 5.99

• Our test statistic: ${\chi }^2=2.39$

Since $2.39<5.99$, we fail to reject $H_0$.

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✅ Conclusion

There is no significant evidence that the proportion of low birth weight infants differs across racial groups (p > 0.05).

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Conclusion

🧩 Step 1: What are we comparing?

The table is all about bivariate comparisons — comparing an outcome (dependent variable) across one or more groups (independent variable).

Two key questions:

1. Is the outcome variable continuous (a number like weight, height, blood pressure) or categorical (like yes/no, race, low vs normal)?

2. How many groups are we comparing? (1, 2, or more?)

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🧩 Step 2: Continuous outcomes

If your dependent variable is continuous:

a) One group vs. a fixed value

• Example: Is the average maternal weight = 120 pounds?

• Test: One-sample t-test (parametric) or Wilcoxon signed-rank test (nonparametric).

• Why? Because you’re comparing the sample mean/median to a known constant.

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b) Two independent groups

• Example: Is mean maternal weight different for low vs. normal birth weight infants?

• Test: Two-sample t-test (parametric) or Mann–Whitney U (nonparametric).

• Why? Because you want to compare averages of two groups.

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c) More than two groups

• Example: Is mean infant birth weight different across 3 race categories?

• Test: ANOVA (parametric) or Kruskal–Wallis test (nonparametric).

• Why? ANOVA extends the t-test idea to more than 2 groups.

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d) Two matched/paired groups

• Example: Compare a mother’s blood pressure before pregnancy vs. during pregnancy (same person, measured twice).

• Test: Paired t-test (parametric) or Wilcoxon signed-rank test (nonparametric).

• Why? Because the data points are linked (paired).

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🧩 Step 3: Categorical outcomes

If your dependent variable is categorical (yes/no, or categories like race):

a) Two independent groups

• Example: Compare proportion of low birth weight in smokers vs. non-smokers.

• Test: z-test for proportions (parametric) or Chi-square test (nonparametric). If sample is small, Fisher’s exact test.

• Why? Because you’re comparing proportions between 2 groups.

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b) More than two groups

• Example: Does the proportion of low birth weight differ across 3 races?

• Test: Chi-square test of independence.

• Why? Because chi-square is the standard tool for comparing categorical distributions across groups.

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c) Paired/matched binary data

• Example: Did the same baby get classified as “low birth weight” by two doctors? (paired yes/no outcomes).

• Test: McNemar’s test.

• Why? Because it’s for paired categorical data (think 2x2 table with matched pairs).

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🧩 Step 4: Continuous vs Continuous

• Example: Is maternal weight correlated with infant birth weight?

• Test: Pearson’s correlation (parametric) or Spearman’s correlation (nonparametric).

• Why? Because both are continuous, and you’re looking for association.

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🧩 Step 5: Parametric vs. Nonparametric

• Parametric tests (t-test, ANOVA, z-test, Pearson) assume the data follow certain distributions (usually normal).

• Nonparametric tests (Wilcoxon, Mann–Whitney, Kruskal–Wallis, Spearman, Fisher’s exact) don’t rely on those assumptions.

• Rule of thumb: if sample size is small or data are skewed/outliers → go nonparametric.

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✅ In plain words:

• If your outcome is continuous → you’re comparing means → use t-test (2 groups), ANOVA (≥3 groups), or paired t-test.

• If your outcome is categorical → you’re comparing proportions → use chi-square, z-test, or Fisher’s exact.

• If both are continuous → use correlation (Pearson or Spearman).

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Two-sample z-test (with known variances)

Question: Do two independent populations have the same mean, assuming their variances are known?

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Step 1: Hypotheses

• Null hypothesis:

$$H_0:{\mu }_1={\mu }_2\text{or}{\mu }_1-{\mu }_2=0$$

• Alternative hypothesis (depends on question):

$$H_a:{\mu }_1\ne {\mu }_2\text{(two-sided)}$$

or

$$H_a:{\mu }_1>{\mu }_2\text{(one-sided)}$$

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Step 2: Sampling distribution of the difference

The sample means are:

$${\bar{X}}_1,{\bar{X}}_2$$

The difference has expectation:

$$E[{\bar{X}}_1-{\bar{X}}_2]={\mu }_1-{\mu }_2$$

And variance:

$$\text{Var}({\bar{X}}_1-{\bar{X}}_2)=\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}$$

So the standard error (SE) is:

$$SE=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$$

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Step 3: Test statistic

Under $H_0:{\mu }_1-{\mu }_2=0$:

$$Z=\frac{({\bar{X}}_1-{\bar{X}}_2)-0}{SE}=\frac{{\bar{X}}_1-{\bar{X}}_2}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$$

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Step 4: Decision rule

• Choose significance level $\alpha$ (e.g., 0.05).
• Two-sided test: reject $H_0$ if $|Z|>z_{\alpha /2}$.
• One-sided test: reject $H_0$ if $Z>z_{\alpha }$ or $Z<-z_{\alpha }$.
• Or compute the p-value and compare with $\alpha$.

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Step 5: Numerical Example

Suppose:

• Population 1: ${\sigma }_1^2=25$, $n_1=50$, ${\bar{X}}_1=100$
• Population 2: ${\sigma }_2^2=36$, $n_2=60$, ${\bar{X}}_2=95$

A. Compute SE

$$SE=\sqrt{\frac{25}{50}+\frac{36}{60}}=\sqrt{0.5+0.6}=\sqrt{1.1}\approx 1.048$$

B. Test statistic

$$Z=\frac{100-95}{1.048}\approx \frac{5}{1.048}\approx 4.77$$

C. Decision

• Degrees of freedom: not needed (Z test uses normal distribution).
• Critical value for two-sided $\alpha =0.05$: $z_{0.025}=1.96$.
• Our $Z=4.77>1.96$.

✅ Reject $H_0$.

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✅ Conclusion

There is a significant difference between the two population means (p < 0.01).