Tags: #◇CSMODEL #◇Inference for Means #◇Computer Science #◇Data Science #◇Statistical Modeling #◇Python #◇Machine Learning

Module 08: Inference for Means

◇Key Concepts

1. Central Limit Theorem & The Student's t-Distribution

When estimating a population mean $\mu$ using a sample mean $\bar{x}$, the Central Limit Theorem (CLT) states that for independent observations from a population with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of $\bar{x}$ is approximately normal:
$\bar{x} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$

• The Unknown $\sigma$ Problem: In almost all real-world applications, the true population standard deviation $\sigma$ is unknown. We must estimate $\sigma$ using the sample standard deviation $s$, yielding an estimated standard error:
  $SE = \frac{s}{\sqrt{n}}$
• Why the t-Distribution? Substituting $s$ for $\sigma$ introduces additional variability into the test statistic, especially when sample size $n$ is small. Consequently, the standard normal distribution ($Z$) underestimates tail probabilities. To account for this extra uncertainty, we use the Student's t-distribution.
• Properties of the t-Distribution:
  Symmetric and bell-shaped, centered at 0.
  Has thicker (heavier) tails than the standard normal distribution ($Z$), meaning extreme values are more likely under a t-distribution.
  Parameterized by degrees of freedom ($df$).
  As $df$ increases ($df \to \infty$), the t-distribution approaches the standard normal distribution $N(0, 1)$.

2. Conditions for Inference Using the t-Distribution

Before performing any t-test or constructing t-confidence intervals, two core conditions must be verified:

1. Independence Condition:
   Observations in the sample must be independent of one another.
   Collected via random sampling or random assignment.
   If sampling without replacement, the sample size must not exceed 10% of the population ($n < 10\% N$).
2. Normality / Sample Size Condition:
   If the population distribution is normal, the sampling distribution of $\bar{x}$ is normal even for small $n$ ($n < 30$).
   If the population distribution is not normal or unknown:
   Small samples ($n < 30$): Data should be nearly normal with no severe skewness or prominent outliers.
   Large samples ($n \ge 30$): The Central Limit Theorem guarantees that the sampling distribution of $\bar{x}$ is nearly normal, even if the underlying population data exhibits moderate skewness.

3. One-Sample t-Test

• When to Use: Testing whether the sample mean $\bar{x}$ from a single numeric variable significantly differs from a hypothesized population value $\mu_0$.
• Hypotheses:
  Null Hypothesis ($H_0$): $\mu = \mu_0$ (No difference from hypothesized mean)
  Alternative Hypothesis ($H_A$): $\mu \neq \mu_0$ (Two-tailed) OR $\mu > \mu_0$ / $\mu < \mu_0$ (One-tailed)
• Test Statistic:
  $t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$
• Degrees of Freedom: $df = n - 1$.
• Confidence Interval:
  $\text{CI} = \bar{x} \pm t^*_{df, 1-\alpha/2} \times \frac{s}{\sqrt{n}}$

4. Paired Observations (Dependent Samples t-Test)

• When to Use: Comparing two sets of quantitative observations that are paired or matched (e.g., pre-test vs. post-test scores of the same individuals, textbook prices at UCLA Bookstore vs. Amazon for the exact same titles).
• Methodology: Paired data is reduced to a single vector of differences:
  $d_i = x_{1, i} - x_{2, i}$
  The inference is then conducted as a one-sample t-test on the differences $d_i$.
• Hypotheses:
  $H_0: \mu_d = 0$ (The true average difference is zero)
  $H_A: \mu_d \neq 0$ (The true average difference is non-zero)
• Test Statistic:
  $t = \frac{\bar{x}_d - 0}{\frac{s_d}{\sqrt{n_{\text{pairs}}}}}$
• Degrees of Freedom: $df = n_{\text{pairs}} - 1$.

5. Independent Two-Sample t-Test (Unpaired Observations)

• When to Use: Comparing the means of two distinct, independent groups (e.g., birth weight of babies born to smoker vs. non-smoker mothers). Sample sizes $n_1$ and $n_2$ do not need to be equal.
• Hypotheses:
  $H_0: \mu_1 - \mu_2 = 0 \iff \mu_1 = \mu_2$
  $H_A: \mu_1 - \mu_2 \neq 0 \iff \mu_1 \neq \mu_2$
• Point Estimate: Difference between sample means $\bar{x}_1 - \bar{x}_2$.
• Standard Error (Welch's / Unequal Variances):
  $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$
• Test Statistic:
  $t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$
• Degrees of Freedom ($df$):
  Conservative Manual Rule (Exam Standard): $df = \min(n_1 - 1, n_2 - 1)$.
  Welch-Satterthwaite Approximation: Used automatically by scipy.stats.ttest_ind(..., equal_var=False).

6. One-Tailed vs. Two-Tailed Tests & p-Value Adjustments

• Two-Tailed Test ($H_A: \mu \neq \mu_0$): Measures extreme values in both tails. $p = 2 \times P(T \ge |t|)$.
• One-Tailed Test ($H_A: \mu > \mu_0$ or $H_A: \mu < \mu_0$): Measures extreme values in a specific direction.
• Converting p-values:
  If test statistic $t$ aligns with the direction specified in $H_A$:
  $p_{\text{one-tailed}} = \frac{p_{\text{two-tailed}}}{2}$
  If test statistic $t$ goes in the direction opposite of $H_A$:
  $p_{\text{one-tailed}} = 1 - \frac{p_{\text{two-tailed}}}{2}$

7. Comparing Multiple Means: ANOVA & Bonferroni Correction

• When to Use: Comparing means across $k \ge 3$ independent groups simultaneously (e.g., batting averages among Outfielders, Infielders, and Catchers).
• Why Not Multiple Pairwise t-Tests? Running many pairwise t-tests causes Type I Error Inflation. If $\alpha = 0.05$ for a single test, running $K$ tests increases the overall probability of at least one false positive to $1 - (1 - \alpha)^K$.
• ANOVA Hypotheses:
  $H_0: \mu_1 = \mu_2 = \dots = \mu_k$ (All group means are equal)
  $H_A$: At least one group mean is different.
• ANOVA F-Statistic:
  $F = \frac{MSG}{MSE}$
  $MSG$ (Mean Square between Groups): Measures variability between group means ($\frac{SSG}{k-1}$).
  $MSE$ (Mean Square Error): Measures variability within groups ($\frac{SSE}{n_{\text{total}} - k}$).
• Post-Hoc Analysis & Bonferroni Correction:
  If ANOVA rejects $H_0$, conduct pairwise t-tests to identify specific group differences.
  Apply Bonferroni Correction to control family-wise error rate:
  $\alpha^* = \frac{\alpha}{K} \quad \text{where } K = \binom{k}{2} = \frac{k(k-1)}{2}$

8. Step-by-Step Manual Calculator Steps for t-Tests

1. State Hypotheses: Define $H_0$ and $H_A$ clearly with population parameters ($\mu, \mu_d, \mu_1 - \mu_2$).
2. Verify Assumptions: Check independence and normality / $n \ge 30$.
3. Compute Sample Statistics: Calculate $\bar{x}$, $s$, $n$ (or $\bar{x}_d, s_d, n_{\text{pairs}}$, or $\bar{x}_1, \bar{x}_2, s_1, s_2, n_1, n_2$).
4. Calculate Standard Error ($SE$):
   One-sample / Paired: $SE = \frac{s}{\sqrt{n}}$
   Independent Two-sample: $SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$
5. Compute Test Statistic ($t$):
   $t = \frac{\text{Point Estimate} - \text{Null Value}}{SE}$
6. Find Degrees of Freedom ($df$) & Critical Value ($t^*$):
   Look up $t^*$ in t-table for given $\alpha$ and $df$.
7. Make Decision & State Conclusion:
   If $|t| > t^*$ (or $p$-value $< \alpha$), reject $H_0$.
   If $|t| \le t^*$ (or $p$-value $\ge \alpha$), fail to reject $H_0$.
   State conclusion in plain language relative to the problem context.

◇Python Snippets

1. One-Sample t-Test (scipy.stats.ttest_1samp)

Tests if a sample vector data has a population mean equal to popmean.

import numpy as np
from scipy import stats

# Sample data: housing lot areas (in sq ft)
lot_area_sample = np.array([9500, 10200, 8900, 9800, 10500, 9100, 9400, 9900, 10100, 8800])
hypothesized_mean = 10000

# Perform one-sample t-test (two-sided by default)
res = stats.ttest_1samp(lot_area_sample, popmean=hypothesized_mean)

print(f"t-statistic: {res.statistic:.4f}")
print(f"p-value: {res.pvalue:.4f}")
print(f"Degrees of freedom: {res.df}")

# Decision rule at alpha = 0.05
alpha = 0.05
if res.pvalue < alpha:
    print("Reject H0: Sample mean differs significantly from 10,000.")
else:
    print("Fail to reject H0: Insufficient evidence of a difference.")

2. Independent Two-Sample t-Test (scipy.stats.ttest_ind)

Compares two independent samples. Use equal_var=False for Welch's t-test (standard best practice).

import pandas as pd
from scipy import stats

# Load birth records dataset
nc_df = pd.read_csv('nc.csv')

# Extract baby weights for smoker and non-smoker mothers
smoker_weights = nc_df[nc_df['habit'] == 'smoker']['weight'].dropna()
nonsmoker_weights = nc_df[nc_df['habit'] == 'nonsmoker']['weight'].dropna()

# Perform Welch's t-test (equal_var=False)
res_ind = stats.ttest_ind(smoker_weights, nonsmoker_weights, equal_var=False)

print(f"t-statistic: {res_ind.statistic:.4f}")
print(f"p-value: {res_ind.pvalue:.4f}")

# Interpretation:
# If p-value < 0.05, reject H0 (significant difference in newborn weight).

3. Paired Two-Sample t-Test (scipy.stats.ttest_rel)

Compares two dependent/paired samples (e.g., store price vs. Amazon price for the same items).

import numpy as np
from scipy import stats

# Prices of 6 textbooks at campus bookstore vs. Amazon
store_prices = np.array([30, 19, 19, 45, 25, 60])
amazon_prices = np.array([20, 15, 19, 38, 21, 52])

# Method A: Using scipy.stats.ttest_rel
res_paired = stats.ttest_rel(store_prices, amazon_prices)

# Method B: Equivalent one-sample t-test on differences
diffs = store_prices - amazon_prices
res_diff = stats.ttest_1samp(diffs, popmean=0)

print(f"Paired t-stat: {res_paired.statistic:.4f}, p-value: {res_paired.pvalue:.4f}")
print(f"Differences 1-sample t-stat: {res_diff.statistic:.4f}, p-value: {res_diff.pvalue:.4f}")
# Note: Both methods produce identical t-statistic and p-value!

4. Critical Values and Quantiles (scipy.stats.t.ppf)

Finds critical values $t^*$ for confidence intervals and hypothesis rejection regions.

from scipy import stats

n = 20
df = n - 1
alpha = 0.05

# Two-tailed critical t-value for 95% confidence level
t_star_95 = stats.t.ppf(1 - alpha / 2, df=df)
print(f"95% Two-tailed critical t-value (df={df}): {t_star_95:.4f}")

# One-tailed critical t-value for 95% confidence level (alpha = 0.05 in one tail)
t_star_one_tailed = stats.t.ppf(1 - alpha, df=df)
print(f"95% One-tailed critical t-value (df={df}): {t_star_one_tailed:.4f}")

5. Constructing Confidence Intervals for Means

Calculates margin of error and confidence bounds manually and using stats.t.interval.

import numpy as np
from scipy import stats

sample_data = np.array([4.4, 3.8, 5.1, 4.9, 4.2, 5.5, 3.9, 4.7, 4.1, 4.6])
n = len(sample_data)
sample_mean = np.mean(sample_data)
sample_std = np.std(sample_data, ddof=1)  # Sample std (ddof=1)
se = sample_std / np.sqrt(n)

alpha = 0.05
df = n - 1
t_crit = stats.t.ppf(1 - alpha / 2, df=df)
margin_of_error = t_crit * se

ci_lower = sample_mean - margin_of_error
ci_upper = sample_mean + margin_of_error

print(f"Sample Mean: {sample_mean:.2f}")
print(f"Margin of Error: {margin_of_error:.4f}")
print(f"Manual 95% CI: ({ci_lower:.4f}, {ci_upper:.4f})")

# Built-in SciPy method:
ci_scipy = stats.t.interval(confidence=0.95, df=df, loc=sample_mean, scale=se)
print(f"SciPy 95% CI: ({ci_scipy[0]:.4f}, {ci_scipy[1]:.4f})")

6. One-Way ANOVA & Bonferroni Post-Hoc Correction

Tests equality of means across multiple groups and applies Bonferroni correction.

import numpy as np
from scipy import stats

outfielders = np.array([0.32, 0.31, 0.35, 0.29, 0.33, 0.30])
infielders = np.array([0.318, 0.310, 0.325, 0.305, 0.312, 0.320])
catchers = np.array([0.302, 0.295, 0.288, 0.310, 0.290, 0.300])

# 1. Run One-Way ANOVA
f_stat, p_val = stats.f_oneway(outfielders, infielders, catchers)
print(f"ANOVA F-statistic: {f_stat:.4f}, p-value: {p_val:.4f}")

# 2. Bonferroni Correction for post-hoc pairwise tests
k = 3  # Number of groups
num_comparisons = k * (k - 1) // 2  # 3 comparisons
alpha_original = 0.05
alpha_bonferroni = alpha_original / num_comparisons

print(f"Original Alpha: {alpha_original}")
print(f"Bonferroni Adjusted Alpha*: {alpha_bonferroni:.4f}")

# Compare Outfielders vs Catchers post-hoc
res_posthoc = stats.ttest_ind(outfielders, catchers, equal_var=False)
print(f"Outfielder vs Catcher p-value: {res_posthoc.pvalue:.4f}")
if res_posthoc.pvalue < alpha_bonferroni:
    print("Significant difference between Outfielders and Catchers after Bonferroni correction.")

◇Formulas & Key Equations

◇Common Exam Traps

• Trap 1: Confusing Paired vs. Independent t-Tests

• Mistake: Running ttest_ind on data where measurements come from the same individuals (e.g., pre-treatment vs. post-treatment weights).

• Correction: If observations are linked/paired, use ttest_rel (or 1-sample t-test on differences). Using independent t-test ignores pairing and reduces test power.

• Trap 2: Forgetting equal_var=False in SciPy

• Mistake: Calling scipy.stats.ttest_ind(a, b) without specifying equal_var=False.

• Correction: By default, SciPy sets equal_var=True (Student's pooled t-test). In CSMODEL and modern statistics, Welch's t-test (equal_var=False) is required unless equal population variances are proven.

• Trap 3: Incorrect Degrees of Freedom Calculation

• Paired t-test: $df = n_{\text{pairs}} - 1$ (NOT $2n - 2$).

• Independent t-test (Manual Exam Rule): $df = \min(n_1 - 1, n_2 - 1)$ (NOT $n_1 + n_2 - 2$ unless pooled variance is assumed).

• ANOVA: Numerator $df_1 = k - 1$, Denominator $df_2 = n_{\text{total}} - k$.

• Trap 4: One-Tailed p-Value Division Error

• Mistake: Always dividing SciPy's pvalue by 2 to get a one-tailed p-value.

• Correction: Dividing by 2 is valid ONLY IF the sample statistic lies in the direction of the alternative hypothesis. If the sample statistic goes against $H_A$, the one-tailed p-value is $1 - \frac{p_{\text{two-tailed}}}{2}$. Alternatively, use alternative='greater' or alternative='less'.

• Trap 5: Misinterpreting $n \ge 30$ Normality Rule

• Mistake: Believing that $n \ge 30$ means the sample data itself becomes normally distributed.

• Correction: The sample data distribution does not change. $n \ge 30$ ensures that the sampling distribution of the mean $\bar{x}$ is approximately normal due to the Central Limit Theorem.

• Trap 6: Misinterpreting Confidence Intervals

• Mistake: Claiming "There is a 95% probability that the true mean $\mu$ falls inside the interval $(3.29, 5.51)$."

• Correction: In frequentist statistics, $\mu$ is a fixed unknown constant, not a random variable. The correct phrasing is: "We are 95% confident that the true population mean lies between 3.29 and 5.51", meaning 95% of intervals constructed using this procedure will capture $\mu$.

• Trap 7: Omitting Bonferroni Correction for Multiple Comparisons

• Mistake: Running multiple pairwise t-tests after ANOVA using standard $\alpha = 0.05$.

• Correction: Running $K$ tests inflates overall Type I error rate. You must compare p-values against $\alpha^* = \alpha / K$.

◇Practice Questions

Q1. A factory manufactures precision steel rods. The target mean diameter is $\mu_0 = 15.00$ mm. A quality control inspector takes a random sample of $n = 16$ rods and measures their diameters. The sample mean is $\bar{x} = 14.82$ mm and the sample standard deviation is $s = 0.36$ mm.
Perform a two-tailed hypothesis test at $\alpha = 0.05$ to determine if the manufacturing process is out of calibration. (Given: For $df = 15$, $t^*_{0.025} = 2.131$). A) $t = -2.000$; Fail to reject $H_0$ because $|-2.000| \le 2.131$. B) $t = -2.000$; Reject $H_0$ because $|-2.000| > 2.131$. C) $t = -0.500$; Fail to reject $H_0$ because $|-0.500| \le 2.131$. D) $t = -2.131$; Reject $H_0$ because $t = t^*$.

Q2. Inspect the following Python code snippet analyzing two independent website layout variations (A and B):

import numpy as np
from scipy import stats

layout_A = np.array([45, 52, 48, 60, 55, 50, 49, 53])
layout_B = np.array([38, 42, 40, 47, 44, 39, 41, 43])

stat, pval = stats.ttest_ind(layout_A, layout_B, equal_var=False)
print(f"Statistic: {stat:.3f}, p-value: {pval:.5f}")

Output:
Statistic: 3.842, p-value: 0.00179

What is the correct conclusion at a 5% significance level ($\alpha = 0.05$)? A) Fail to reject $H_0$ because the t-statistic (3.842) is greater than 0.05. B) Reject $H_0$ using Student's paired t-test; layout A and B have identical performance. C) Reject $H_0$ using Welch's t-test (equal_var=False); there is a statistically significant difference in user engagement between layout A and layout B ($p = 0.00179 < 0.05$). D) Accept $H_0$; layout B is significantly better than layout A.

Q3. A medical clinic tested a new cholesterol-lowering drug on $n = 25$ patients. The blood cholesterol level (in mg/dL) was measured before and 8 weeks after treatment for each patient. The average difference (Before - After) was $\bar{x}_d = 12.0$ mg/dL, and the sample standard deviation of differences was $s_d = 15.0$ mg/dL.
Construct the 95% confidence interval for the mean reduction in cholesterol. (Given: For $df = 24$, $t^*_{0.025} = 2.064$). A) $(5.808, 18.192)$ mg/dL B) $(0.000, 24.000)$ mg/dL C) $(9.523, 14.477)$ mg/dL D) $(6.192, 17.808)$ mg/dL

Q4. Consider the Python script below testing whether a new compiler produces faster execution times than a benchmark standard of $25.0$ seconds:

from scipy import stats

exec_times = [23.5, 24.1, 22.8, 24.8, 23.9, 24.2, 23.0] # n = 7
res = stats.ttest_1samp(exec_times, popmean=25.0, alternative='less')
t_crit = stats.t.ppf(0.05, df=len(exec_times)-1)

print(f"t_stat = {res.statistic:.3f}, p_val = {res.pvalue:.4f}")
print(f"t_crit = {t_crit:.3f}")

Output:
t_stat = -4.382, p_val = 0.0023
t_crit = -1.943

Which statement correctly interprets this output at $\alpha = 0.05$? A) Fail to reject $H_0$ because execution times cannot be negative. B) Reject $H_0$ because the observed t-statistic ($-4.382$) is more extreme than the critical value ($-1.943$) and $p = 0.0023 < 0.05$, proving the compiler significantly reduces execution time. C) Reject $H_0$ because $p = 0.0023 > 0.05$. D) Fail to reject $H_0$ because a two-sided test was required instead of alternative='less'.

Q5. A data scientist compares the processing time of 3 different algorithm implementations (Algorithm A, B, and C) across multiple benchmark tests ($k = 3$, $n_{\text{total}} = 33$).
An ANOVA test yields $SSG = 120.0$ and $SST = 420.0$.
Calculate the ANOVA $F$-statistic and determine the Bonferroni-adjusted significance level $\alpha^*$ for post-hoc pairwise comparisons at overall $\alpha = 0.05$. A) $F = 6.00$, $\alpha^* = 0.0167$ B) $F = 3.00$, $\alpha^* = 0.0500$ C) $F = 12.00$, $\alpha^* = 0.0250$ D) $F = 4.00$, $\alpha^* = 0.0167$

Q6. Match each research scenario (1, 2, 3) to its appropriate statistical test function in Python SciPy:

1. Comparing the exam scores of Section 1 vs. Section 2 (two separate groups of students).
2. Testing if the average test score of Section 1 differs from the historical departmental mean of 75.0.
3. Comparing midterm vs. final exam scores of the same group of students. A) 1: ttest_rel, 2: ttest_1samp, 3: ttest_ind B) 1: ttest_ind, 2: ttest_1samp, 3: ttest_rel C) 1: ttest_1samp, 2: ttest_ind, 3: ttest_rel D) 1: f_oneway, 2: ttest_ind, 3: ttest_1samp

Answer Key:

Q1: A — $t = -2.000$; Fail to reject $H_0$
Step-by-step solution:

1. $H_0: \mu = 15.00$, $H_A: \mu \neq 15.00$.
2. $SE = \frac{s}{\sqrt{n}} = \frac{0.36}{\sqrt{16}} = \frac{0.36}{4} = 0.09$ mm.
3. Test Statistic $t = \frac{\bar{x} - \mu_0}{SE} = \frac{14.82 - 15.00}{0.09} = \frac{-0.18}{0.09} = -2.000$.
4. Compare $|t| = 2.000$ to critical value $t^* = 2.131$ ($df = 15$).
5. Since $|-2.000| \le 2.131$, fail to reject $H_0$. There is insufficient evidence to conclude the process is out of calibration.
   Q2: C — Reject $H_0$ using Welch's t-test (equal_var=False)
   Explanation: The code calls ttest_ind(..., equal_var=False), which performs Welch's independent two-sample t-test. The resulting $p$-value ($0.00179$) is less than $\alpha = 0.05$, so we reject $H_0$ and conclude that user engagement differs significantly between layouts.
   Q3: A — $(5.808, 18.192)$ mg/dL
   Step-by-step solution:
6. $n = 25$, $df = 24$, $\bar{x}_d = 12.0$, $s_d = 15.0$.
7. $SE = \frac{s_d}{\sqrt{n}} = \frac{15.0}{\sqrt{25}} = \frac{15.0}{5} = 3.0$ mg/dL.
8. Margin of Error $ME = t^* \times SE = 2.064 \times 3.0 = 6.192$ mg/dL.
9. $\text{CI} = \bar{x}_d \pm ME = 12.0 \pm 6.192 = (5.808, 18.192)$ mg/dL.
   Q4: B — Reject $H_0$ because $t_{\text{stat}} (-4.382) < t_{\text{crit}} (-1.943)$ and $p = 0.0023 < 0.05$
   Explanation: The left-tailed test $H_A: \mu < 25.0$ yields a negative t-statistic. Since $t_{\text{stat}} = -4.382$ is less than the critical value $-1.943$ (falls deep into the rejection region) and $p = 0.0023 < 0.05$, we reject $H_0$.
   Q5: A — $F = 6.00$, $\alpha^* = 0.0167$
   Step-by-step solution:
10. $k = 3$ groups, $n_{\text{total}} = 33$.
11. $df_1 = k - 1 = 3 - 1 = 2$.
12. $df_2 = n_{\text{total}} - k = 33 - 3 = 30$.
13. $MSG = \frac{SSG}{df_1} = \frac{120.0}{2} = 60.0$.
14. $SSE = SST - SSG = 420.0 - 120.0 = 300.0$.
15. $MSE = \frac{SSE}{df_2} = \frac{300.0}{30} = 10.0$.
16. $F = \frac{MSG}{MSE} = \frac{60.0}{10.0} = 6.00$.
17. Pairwise comparisons $K = \frac{k(k-1)}{2} = \frac{3 \times 2}{2} = 3$.
18. Bonferroni adjusted $\alpha^* = \frac{0.05}{3} \approx 0.0167$.
    Q6: B — 1: ttest_ind, 2: ttest_1samp, 3: ttest_rel
    Explanation: Scenario 1 compares two separate/independent groups (ttest_ind). Scenario 2 compares a single group against a known standard (ttest_1samp). Scenario 3 compares paired measurements from the same students (ttest_rel).

◇References & Navigation

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• Related Notes: ◇Foundations for Inference, ◇Bayesian Inference