Interval Estimation & Hypothesis Test

What is Interval Estimation?

• A range of plausible values for a population parameter (e.g., mean $\mu$, proportion $p$) constructed from sample data.
• Contrasts with a point estimate (single value).
• Provides a measure of uncertainty or confidence.
• Often reported as Confidence Intervals (CIs).

Confidence Interval (CI): Basic Concept

• A CI for parameter $\theta$ is a random interval $[L(X),U(X)]$, where $X$ is sample data.
• For a $(1-\alpha )\times 100\%$ CI: $Pr(L(X)\le \theta \le U(X))=1-\alpha$
• This probability holds for repeated sampling.
• Commonly, $\alpha =0.05$ for a 95% CI.

Types of Confidence Intervals

• For Mean (population standard deviation $\sigma$ known): $\bar{X}\pm z^{\ast }\cdot \frac{\sigma }{\sqrt{n}}$
• For Mean (population standard deviation $\sigma$ unknown): $\bar{x}\pm t^{\ast }\cdot \frac{s}{\sqrt{n}}$ (where $t^{\ast }$ is the critical value from the t-distribution)
• For Proportion: $\hat{p}\pm z^{\ast }\cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ (where $\hat{p}$ is the sample proportion, $z^{\ast }$ is the critical value from the standard normal distribution)

Example: CI for Mean (Normal Case, Known ${\sigma }^2$)

• Assumptions: Sample $X=(X_1,...,X_n)$ from Normal $N(\mu ,{\sigma }^2)$, ${\sigma }^2$ is known.
• Steps:
  1. Compute sample mean: $\bar{X}=\frac{1}{n}{\sum }_{i=1}^n{X}_i$
  2. Standard Error (SE): $SE=\frac{\sigma }{\sqrt{n}}$
  3. $(1-\alpha )\times 100\%$ Z-interval: $[\bar{X}-z_{\alpha /2}\frac{\sigma }{\sqrt{n}},\bar{X}+z_{\alpha /2}\frac{\sigma }{\sqrt{n}}]$ (e.g., for 95% CI, $\alpha =0.05$, $z_{\alpha /2}=z_{0.025}\approx 1.96$)

Example Calculation (Unknown $\sigma$)

• Survey: $n=100$, $\bar{x}=170$ cm, $s=10$ cm. Find 95% CI for population mean $\mu$.
• Use t-distribution since $\sigma$ is unknown.
• $SE=\frac{s}{\sqrt{n}}=\frac{10}{\sqrt{100}}=1$
• Degrees of freedom $df=n-1=99$.
• For 95% CI, $df=99$, the critical $t$-value $t^{\ast }\approx 1.984$.
• CI = $\bar{x}\pm t^{\ast }\cdot SE=170\pm 1.984\cdot 1=170\pm 1.984$
• CI = $[168.016,171.984]$

Interpretation of CI

• If we repeat the experiment many times, about $(1-\alpha )\times 100\%$ (e.g., 95%) of the computed intervals will contain the true parameter $\mu$.
• The parameter $\mu$ is fixed; the interval varies from sample to sample.

Choosing Z-Score vs T-Distribution

• Use Z-score if: Sample size $n\ge 30$ OR population standard deviation $\sigma$ is known.
• Use T-distribution if: Sample size $n<30$ AND population standard deviation $\sigma$ is unknown.

Pivotal Quantity

• A function of sample data $X_1,...,X_n$ and unknown parameter(s) $\theta$, say $Q(X_1,...,X_n;\theta )$.
• Its probability distribution is known and does not depend on the unknown parameter(s) $\theta$.
• Crucial for constructing exact CIs and hypothesis tests.

Examples of Pivotal Quantities

• Normal Distribution, Known Variance ${\sigma }^2$: For estimating $\mu$, the pivotal quantity is: $Q=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}\sim N(0,1)$
• Normal Distribution, Unknown Variance ${\sigma }^2$: For estimating $\mu$, the pivotal quantity is: $Q=\frac{\bar{X}-\mu }{s/\sqrt{n}}\sim t_{n-1}$ (t-distribution with $n-1$ degrees of freedom)

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Hypothesis Test

Introduction & Role

• A fundamental statistical procedure to make inferences about population parameters based on sample data.
• Involves formulating hypotheses and using sample data to decide whether to reject the null hypothesis.
• Role: Provides a systematic, objective framework for making data-based decisions, quantifying evidence, controlling error probabilities, and standardizing scientific inquiry.

Basic Concepts

How Hypothesis Testing Works

• Null Hypothesis ($H_0$): The default assumption or statement being tested (e.g., “no effect”, “status quo”). Usually contains equality.
  • Formal: $H_0:\theta ={\theta }_0$ (Eq 1)
• Alternative Hypothesis ($H_a$ or $H_1$): The claim you suspect might be true, contradicting $H_0$.
  • Formal Types:
    • $H_a:\theta \ne {\theta }_0$ (two-sided/two-tailed) (Eq 2)
    • $H_a:\theta >{\theta }_0$ (right-sided/right-tailed) (Eq 3)
    • $H_a:\theta <{\theta }_0$ (left-sided/left-tailed) (Eq 4)

Testing Process Overview

1. Collect data.
2. Calculate a test statistic (summarizes data relative to $H_0$).
3. Make a decision: If the test statistic is “too unusual” assuming $H_0$ is true, reject $H_0$ in favor of $H_1$. Otherwise, fail to reject $H_0$.

Test Statistics

• Numerical summary of sample data used for decision making.
• Choice depends on parameter, assumed population distribution, and sample size.
• Z-statistic (Mean test, $\sigma$ known or large $n$): $Z=\frac{\bar{X}-{\mu }_0}{\sigma /\sqrt{n}}$ (Eq 5)
• T-statistic (Mean test, $\sigma$ unknown, small $n$): $t=\frac{\bar{X}-{\mu }_0}{s/\sqrt{n}}$ (Eq 6)

Tests for a Single Mean

Z-test (Known $\sigma$)

• Assumptions: $\sigma$ known; Population is Normal OR $n$ is large (CLT applies).
• Hypotheses: $H_0:\mu ={\mu }_0$ (Eq 7) $H_a:\mu \ne {\mu }_0$ (or $\mu >{\mu }_0$, or $\mu <{\mu }_0$) (Eq 8)
• Test Statistic: $Z=\frac{\bar{X}-{\mu }_0}{\sigma /\sqrt{n}}$ (Eq 9)
• Decision Rule (Example: Two-tailed): Reject $H_0$ if $|Z|>Z_{\alpha /2}$. Fail to reject $H_0$ if $|Z|\le Z_{\alpha /2}$.

Z-test Example (IQ Scores)

• Scenario: Claim $\mu >82$. $\sigma =20$. Sample: $n=81,\bar{X}=90$. Use $\alpha =0.05$.
• $H_0:\mu \le 82$, $H_1:\mu >82$ (Right-tailed).
• Significance Level: $\alpha =0.05$. Critical Z-value $Z_{0.05}=1.645$.
• Test Statistic: $Z=\frac{90-82}{20/\sqrt{81}}=\frac{8}{20/9}=\frac{8}{2.22}=3.60$.
• Decision: Since $3.60>1.645$, reject $H_0$.
• Conclusion: Sufficient evidence supports the claim that the mean IQ score is greater than 82.

T-test (Unknown $\sigma$)

• Assumptions: $\sigma$ unknown; Population is Normal OR $n$ is large.
• Hypotheses: $H_0:\mu ={\mu }_0$ (Eq 10) $H_a:\mu \ne {\mu }_0$ (or $\mu >{\mu }_0$, or $\mu <{\mu }_0$) (Eq 11)
• Test Statistic: $t=\frac{\bar{X}-{\mu }_0}{s/\sqrt{n}}$ (Eq 12) (follows t-distribution with $df=n-1$)
• Decision Rule (Example: Two-tailed): Reject $H_0$ if $|t|>t_{\alpha /2,n-1}$. Fail to reject $H_0$ if $|t|\le t_{\alpha /2,n-1}$.

T-test Example (Exam Scores)

• Scenario: Claim $\mu \ge 75$. Sample: $n=16,\bar{X}=71.5,s=8.5$. Use $\alpha =0.05$.
• $H_0:\mu \ge 75$, $H_1:\mu <75$ (Left-tailed). (Eq 13, 14)
• Significance Level: $\alpha =0.05$. Degrees of freedom $df=16-1=15$. Critical t-value $t_{0.05,15}=-1.753$.
• Test Statistic: $t=\frac{71.5-75}{8.5/\sqrt{16}}=\frac{-3.5}{8.5/4}=\frac{-3.5}{2.125}\approx -1.647$. (Eq 15-19)
• Decision: Since $-1.647>-1.753$, fail to reject $H_0$.
• Conclusion: Insufficient evidence to conclude the mean score is less than 75.

Tests for Two Means

Z-test for Two Means (Independent Samples, Known ${\sigma }_1,{\sigma }_2$)

• Use when: Comparing means from 2 independent populations; ${\sigma }_1,{\sigma }_2$ known; populations Normal OR sample sizes large.
• Hypotheses: (Often ${\Delta }_0=0$) $H_0:{\mu }_1-{\mu }_2={\Delta }_0$ (Eq 20) $H_a:{\mu }_1-{\mu }_2\ne {\Delta }_0$ (or $>,<$) (Eq 21)
• Test Statistic: $Z=\frac{({\bar{X}}_1-{\bar{X}}_2)-{\Delta }_0}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$ (Eq 22)
• Decision Rule (Example: Two-tailed): Reject $H_0$ if $|Z|>Z_{\alpha /2}$.

Z-test Two Means Example (Teaching Methods)

• Scenario: Compare Method A (${\mu }_1$) vs Method B (${\mu }_2$). Claim: Method B is more effective (${\mu }_2>{\mu }_1$).
• Data: A: $n_1=45,{\bar{X}}_1=72,{\sigma }_1=12$. B: $n_2=50,{\bar{X}}_2=78,{\sigma }_2=15$. Use $\alpha =0.05$.
• $H_0:{\mu }_1\ge {\mu }_2$ (or ${\mu }_1-{\mu }_2\ge 0$), $H_1:{\mu }_1<{\mu }_2$ (or ${\mu }_1-{\mu }_2<0$) (Left-tailed). (Eq 23, 24)
• Significance Level: $\alpha =0.05$. Critical Z-value $Z_{0.05}=-1.645$.
• Test Statistic: $Z=\frac{(72-78)-0}{\sqrt{\frac{12^2}{45}+\frac{15^2}{50}}}=\frac{-6}{\sqrt{3.2+4.5}}=\frac{-6}{\sqrt{7.7}}\approx \frac{-6}{2.77}\approx -2.17$. (Eq 25-30)
• Decision: Since $-2.17<-1.645$, reject $H_0$.
• Conclusion: Sufficient evidence that Method B is more effective than Method A.

T-test for Two Means (Independent Samples, Unknown but Equal $\sigma$)

• Use when: Comparing 2 independent means; ${\sigma }_1,{\sigma }_2$ unknown but assumed equal; populations Normal OR sample sizes large.
• Test Statistic: $t=\frac{({\bar{X}}_1-{\bar{X}}_2)-{\Delta }_0}{\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}}$ (Eq 31) (follows t-distribution with $df=n_1+n_2-2$)
• Pooled Variance ($s_p^2$): Estimate of the common variance ${\sigma }^2$. $s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$ (Eq 32)
• Degrees of Freedom: $df=n_1+n_2-2$.

T-test Two Means Example (Study Methods)

• Scenario: Compare Method 1 (${\mu }_1$) vs Method 2 (${\mu }_2$). Test if scores are different.
• Data: M1: $n_1=12,{\bar{X}}_1=76,s_1=8$. M2: $n_2=15,{\bar{X}}_2=82,s_2=7$. Use $\alpha =0.05$. Assume equal variances.
• $H_0:{\mu }_1={\mu }_2$ (or ${\mu }_1-{\mu }_2=0$), $H_1:{\mu }_1\ne {\mu }_2$ (or ${\mu }_1-{\mu }_2\ne 0$) (Two-tailed). (Eq 33, 34)
• Significance Level: $\alpha =0.05$. $df=12+15-2=25$. Critical t-values $t_{0.025,25}=\pm 2.060$.
• Calculate Pooled Variance: $s_p^2=\frac{(11)(8^2)+(14)(7^2)}{12+15-2}=\frac{704+686}{25}=\frac{1390}{25}=55.6$. (Eq 35-39)
• Test Statistic: $t=\frac{(76-82)-0}{\sqrt{55.6(\frac{1}{12}+\frac{1}{15})}}=\frac{-6}{\sqrt{55.6(0.0833+0.0667)}}=\frac{-6}{\sqrt{55.6(0.15)}}=\frac{-6}{\sqrt{8.34}}\approx \frac{-6}{2.89}\approx -2.08$. (Eq 40-43)
• Decision: Since $|-2.08|=2.08>2.060$, reject $H_0$.
• Conclusion: Sufficient evidence that the two study methods produce different test scores (Method 2 avg is higher).

Welch’s T-test (Independent Samples, Unknown and Unequal $\sigma$)

• Use when: Comparing 2 independent means; ${\sigma }_1,{\sigma }_2$ unknown and not assumed equal; populations Normal OR sample sizes large.
• Hypotheses: (Often ${\Delta }_0=0$) $H_0:{\mu }_1-{\mu }_2={\Delta }_0$ (Eq 44) $H_a:{\mu }_1-{\mu }_2\ne {\Delta }_0$ (or $>,<$) (Eq 45)
• Test Statistic: $t^{\prime }=\frac{({\bar{X}}_1-{\bar{X}}_2)-{\Delta }_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$ (Eq 46)
• Degrees of Freedom (Welch-Satterthwaite approximation): $df\approx \frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}{)}^2}{\frac{(s_1^2/n_1{)}^2}{n_1-1}+\frac{(s_2^2/n_2{)}^2}{n_2-1}}$ (Eq 47) (Often rounded down)
• Decision Rule (Example: Two-tailed): Reject $H_0$ if $|t^{\prime }|>t_{\alpha /2,df}$.

Welch’s T-test Example (Drug Recovery Time)

• Scenario: Compare New Drug (${\mu }_{new}$) vs Old Drug (${\mu }_{old}$). Test if new drug reduces recovery time (${\mu }_{new}<{\mu }_{old}$).
• Data: New: $n_N=10,{\bar{X}}_N=12.6,s_N=2.22$. Old: $n_O=10,{\bar{X}}_O=20.7,s_O=2.41$. Use $\alpha =0.05$. Variances not assumed equal.
• $H_0:{\mu }_{new}\ge {\mu }_{old}$ (or ${\mu }_{new}-{\mu }_{old}\ge 0$), $H_1:{\mu }_{new}<{\mu }_{old}$ (or ${\mu }_{new}-{\mu }_{old}<0$) (Left-tailed). (Eq 48, 49)
• Significance Level: $\alpha =0.05$.
• Calculate Test Statistic: $t^{\prime }=\frac{12.6-20.7}{\sqrt{\frac{2.22^2}{10}+\frac{2.41^2}{10}}}=\frac{-8.1}{\sqrt{\frac{4.9284}{10}+\frac{5.8081}{10}}}=\frac{-8.1}{\sqrt{0.493+0.581}}=\frac{-8.1}{\sqrt{1.074}}\approx \frac{-8.1}{1.036}\approx -7.82$. (Eq 50-53)
• Calculate Degrees of Freedom: Let $v_N=s_N^2/n_N\approx 0.493$, $v_O=s_O^2/n_O\approx 0.581$. $df\approx \frac{(v_N+v_O{)}^2}{\frac{v_N^2}{n_N-1}+\frac{v_O^2}{n_O-1}}=\frac{(0.493+0.581{)}^2}{\frac{0.493^2}{9}+\frac{0.581^2}{9}}=\frac{1.074^2}{\frac{0.243}{9}+\frac{0.338}{9}}=\frac{1.153}{0.027+0.0376}=\frac{1.153}{0.0646}\approx 17.85$. Use $df=17$. (Eq 54-56)
• Critical t-value: $t_{0.05,17}\approx -1.740$.
• Decision: Since $t^{\prime }=-7.82<-1.740$, reject $H_0$.
• Conclusion: Strong statistical evidence that the new drug reduces recovery time compared to the old drug. The difference (8.1 days) is also clinically significant. 95% CI for difference: [5.9, 10.3] days.