Large Sample Tests

Topic - When the sample size is large ($n \ge 30$), the Central Limit Theorem guarantees that the sampling distribution of the sample mean follows a normal distribution. In this regime, the Z-test provides the standard analytical tool to test hypotheses regarding single population means and differences between two independent population means.

---

1. Assumptions for Large Sample Z-Tests

To ensure that inference using the standard normal distribution is valid, the following conditions must be satisfied:

1. Continuous Data: The metric of interest must be continuous numerical data.
2. Simple Random Sampling: Observations are independent and identically distributed (i.i.d.).
3. Large Sample Size ($n \ge 30$): By the Central Limit Theorem, the distribution of $\bar{X}$ is normal. If the population standard deviation $\sigma$ is unknown, we substitute the sample standard deviation $s$ without needing Student's t-distribution.
4. Normality Verification: For moderate samples, normality of the data can be validated using the Shapiro-Wilk test (scipy.stats.shapiro).

---

2. One-Sample Z-Test for Population Mean

Tests whether a single population mean $\mu$ equals a hypothesized baseline value $\mu_0$.

Test Statistic Formulation

$Z_{\text{calc}} = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}$

If population standard deviation $\sigma$ is unknown and $n \ge 30$, we substitute sample standard deviation $s$:

$Z_{\text{calc}} = \frac{\bar{X} - \mu_0}{s / \sqrt{n}}, \quad \text{where } s = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2}$

Decision Rules across All Three Tails

Test Type	Hypotheses	Critical Value Rule	P-Value Rule	Confidence Interval Rule
Two-Tailed	$H_0: \mu = \mu_0$ $H_1: \mu \neq \mu_0$	Reject $H_0$ if $\lvert Z_{\text{calc}} \rvert > z_{\alpha/2}$	Reject $H_0$ if $p \le \alpha$ ($p = 2 \cdot P(Z > \lvert Z_{\text{calc}} \rvert)$)	Reject $H_0$ if $\mu_0 \notin \bar{X} \pm z_{\alpha/2}\frac{s}{\sqrt{n}}$
Left-Tailed	$H_0: \mu \ge \mu_0$ $H_1: \mu < \mu_0$	Reject $H_0$ if $Z_{\text{calc}} < -z_{\alpha}$	Reject $H_0$ if $p \le \alpha$ ($p = P(Z < Z_{\text{calc}})$)	N/A (One-sided bound)
Right-Tailed	$H_0: \mu \le \mu_0$ $H_1: \mu > \mu_0$	Reject $H_0$ if $Z_{\text{calc}} > z_{\alpha}$	Reject $H_0$ if $p \le \alpha$ ($p = P(Z > Z_{\text{calc}})$)	N/A (One-sided bound)

---

3. Two-Sample Z-Test for Independent Means

Compares the means of two independent populations ($\mu_1$ vs. $\mu_2$) from two samples ($n_1 \ge 30, n_2 \ge 30$).

Hypotheses

• Two-tailed: $H_0: \mu_1 = \mu_2 \iff \mu_1 - \mu_2 = 0$ against $H_1: \mu_1 \neq \mu_2$
• One-tailed: $H_0: \mu_1 \le \mu_2$ against $H_1: \mu_1 > \mu_2$

Test Statistic

$Z_{\text{calc}} = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}$

When $\sigma_1^2, \sigma_2^2$ are unknown:

$Z_{\text{calc}} = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Equal Variance Check (Levene's Test)

Before conducting two-sample tests, evaluate whether population variances are equal using Levene's test (scipy.stats.levene):

• $H_0: \sigma_1^2 = \sigma_2^2$ vs. $H_1: \sigma_1^2 \neq \sigma_2^2$
• If $p \ge 0.05$, fail to reject $H_0$ and assume equal variances.

---

4. End-to-End Problem Scenarios from Course Material

Scenario 1: Two-Tailed Test (Car Mileage Claim)

Problem: A car manufacturer claims the mileage of their new car is $\mu = 25\text{ km/L}$ with $\sigma = 2.5\text{ km/L}$. A sample of $n = 45$ cars gives a mean of $\bar{X} = 24\text{ km/L}$. Is there evidence to claim that the mean mileage differs from $25\text{ km/L}$? Use $\alpha = 0.01$.

1. Formulate Hypotheses: $H_0: \mu = 25 \quad \text{vs.} \quad H_1: \mu \neq 25 \quad (\text{Two-Tailed})$
2. Calculate Test Statistic: $Z_{\text{calc}} = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} = \frac{24 - 25}{2.5 / \sqrt{45}} = \frac{-1}{2.5 / 6.708} = \frac{-1}{0.3727} \approx -2.683$
3. Method A (Critical Value): For $\alpha = 0.01$, $z_{\alpha/2} = z_{0.005} = 2.575$. Since $|Z_{\text{calc}}| = |-2.683| = 2.683 > 2.575$, we Reject $H_0$.
4. Method B (P-Value): $p = 2 \cdot P(Z < -2.683) = 2 \cdot 0.00366 = 0.0073$ Since $p\text{-value} = 0.0073 < 0.01$, we Reject $H_0$.
5. Method C (Confidence Interval): $\text{CI}_{99\%} = 24 \pm 2.575 \cdot 0.3727 = 24 \pm 0.960 = [23.04, \; 24.96]$ Since hypothesized mean $\mu_0 = 25$ does not lie within $[23.04, 24.96]$, we Reject $H_0$.
6. Conclusion: There is statistically significant evidence at $\alpha = 0.01$ to conclude the mean mileage is different from $25\text{ km/L}$.

---

Scenario 2: Left-Tailed Test (PVC Pipe Thickness)

Problem: A sample of $n = 900$ PVC pipes has a mean thickness of $\bar{X} = 12.5\text{ mm}$. Test whether the sample comes from a population with mean $\mu \ge 13\text{ mm}$ against the claim that it is less than $13\text{ mm}$, given $\sigma = 1\text{ mm}$ at $\alpha = 0.05$.

1. Formulate Hypotheses: $H_0: \mu \ge 13 \quad \text{vs.} \quad H_1: \mu < 13 \quad (\text{Left-Tailed})$
2. Calculate Test Statistic: $Z_{\text{calc}} = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} = \frac{12.5 - 13}{1 / \sqrt{900}} = \frac{-0.5}{1 / 30} = -0.5 \cdot 30 = -15.0$
3. Decision: Critical value for left-tailed test at $\alpha = 0.05$ is $-z_{0.05} = -1.645$. Since $Z_{\text{calc}} = -15.0 < -1.645$ (and $p\text{-value} \approx 0.000$), we Reject $H_0$.
4. Conclusion: Significant evidence indicates average thickness is less than $13\text{ mm}$.

---

Scenario 3: Right-Tailed Test (Food Delivery Time)

Problem: An e-commerce food app claims delivery time is at most 60 minutes ($\mu \le 60$) with $\sigma = 30$ minutes. A sample of $n = 45$ orders yields a mean of $\bar{X} = 75$ minutes. Test the claim at $\alpha = 0.05$.

1. Formulate Hypotheses: $H_0: \mu \le 60 \quad \text{vs.} \quad H_1: \mu > 60 \quad (\text{Right-Tailed})$
2. Calculate Test Statistic: $Z_{\text{calc}} = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} = \frac{75 - 60}{30 / \sqrt{45}} = \frac{15}{4.472} \approx +3.35$
3. Decision: Critical value for right-tailed test at $\alpha = 0.05$ is $+z_{0.05} = +1.645$. $p\text{-value} = P(Z > 3.35) = 0.0004$. Since $p\text{-value} < 0.05$ and $Z_{\text{calc}} > 1.645$, we Reject $H_0$.
4. Conclusion: Strong evidence that the mean delivery time exceeds 60 minutes.

---

Scenario 4: Two-Sample Test (Hemoglobin Levels)

Problem: A blood study evaluates hemoglobin levels for males and females:

• Males: $n_1 = 160$, $\bar{X}_1 = 13\text{ g/dL}$, $s_1 = 4.1\text{ g/dL}$
• Females: $n_2 = 180$, $\bar{X}_2 = 15\text{ g/dL}$, $s_2 = 3.5\text{ g/dL}$ Test whether population mean hemoglobin levels differ between men and women at $\alpha = 0.01$.

1. Formulate Hypotheses: $H_0: \mu_1 = \mu_2 \quad \text{vs.} \quad H_1: \mu_1 \neq \mu_2 \quad (\text{Two-Tailed})$
2. Calculate Test Statistic: $\text{SE}_{\text{diff}} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{4.1^2}{160} + \frac{3.5^2}{180}} = \sqrt{\frac{16.81}{160} + \frac{12.25}{180}} = \sqrt{0.1051 + 0.0681} = \sqrt{0.1732} \approx 0.4161$ $Z_{\text{calc}} = \frac{\bar{X}_1 - \bar{X}_2}{\text{SE}_{\text{diff}}} = \frac{13 - 15}{0.4161} = \frac{-2}{0.4161} \approx -4.807$
3. Decision: Critical value at $\alpha = 0.01$ is $z_{0.005} = 2.575$. Since $|Z_{\text{calc}}| = |-4.807| = 4.807 > 2.575$ (and $p\text{-value} < 0.0001$), we Reject $H_0$.
4. Conclusion: There is strong empirical evidence that average hemoglobin levels differ significantly between men and women.

---

5. Implementation Lab

Execute large sample Z-tests in Python utilizing statsmodels and scipy.stats.

import numpy as np
from scipy import stats
from statsmodels.stats.weightstats import ztest

# 1. One-Sample Z-Test Function
def manual_z_test(samp_mean, pop_mean, std_dev, n, alternative='two-sided'):
    z_stat = (samp_mean - pop_mean) / (std_dev / np.sqrt(n))
    if alternative == 'two-sided':
        p_val = 2 * (1 - stats.norm.cdf(np.abs(z_stat)))
    elif alternative == 'smaller':
        p_val = stats.norm.cdf(z_stat)
    elif alternative == 'larger':
        p_val = 1 - stats.norm.cdf(z_stat)
    return z_stat, p_val

# Car Mileage Scenario (Two-Tailed)
z_calc, p_val = manual_z_test(samp_mean=24, pop_mean=25, std_dev=2.5, n=45, alternative='two-sided')
print(f"Car Mileage Z-Stat: {z_calc:.3f}, P-Value: {p_val:.5f}")

# 2. Two-Sample Z-Test Function (Hemoglobin Scenario)
def two_sample_z_test(m1, s1, n1, m2, s2, n2):
    se_diff = np.sqrt((s1**2 / n1) + (s2**2 / n2))
    z_stat = (m1 - m2) / se_diff
    p_val = 2 * (1 - stats.norm.cdf(np.abs(z_stat)))
    return z_stat, p_val

z_hemo, p_hemo = two_sample_z_test(13, 4.1, 160, 15, 3.5, 180)
print(f"Hemoglobin Z-Stat: {z_hemo:.3f}, P-Value: {p_hemo:.6f}")

info

Key Takeaways

• Z-Test Criterion: Appropriate whenever $n \ge 30$ due to asymptotic normality from the Central Limit Theorem.
• Three Equivalent Paths: For two-tailed tests, Critical Value ($|Z| > z_{\alpha/2}$), P-value ($p \le \alpha$), and Confidence Interval ($\mu_0 \notin \text{CI}$) always agree.
• Two-Sample Variance Pooling: If variances are unknown but sample sizes are large ($n_1, n_2 \ge 30$), substituting individual sample variances $s_1^2$ and $s_2^2$ is asymptotically exact.