Two Sample Inference

Linear Combination of Random Variables

• Objective: Compare two proportions or two means by understanding how addition and multiplication affect the mean and variance.
• Set A Example: Given set A = {98, 99, 100, 101, 102}
  • Sample mean ($\bar{x}$): $\bar{x} = \frac{\Sigma x_i}{n} = 100$
  • Sample variance ($s^2$): $s^2 = \frac{\Sigma (x_i - \bar{x})^2}{n-1} = 2.5$
• Doubling Set A: 2A = {198, 199, 200, 201, 202}
  • New sample mean: $\bar{x} = 200$
  • New sample variance: $s^2 = 10 = 4 \times 2.5$
  • Doubling the values increases the sample variance by a factor of 4.
• Tripling Set A: 3A = {294, 297, 300, 303, 306}
  • New sample mean: $\bar{x} = 300$
  • New sample variance: $s^2 = 22.5 = 9 \times 2.5$
• Multiplying by -1: -A = {-98, -99, -100, -101, -102}
  • New sample mean: $\bar{x} = -100$
  • New sample variance: $s^2 = 2.5$
• Multiplying by -3:
  • New sample mean: $\bar{x} = -300$
  • New sample variance: $s^2 = 22.5 = 9 \times 2.5$
• Rule for Multiplication: Deriving a rule for calculating the sample mean and variance based on multiplication of the original set.

Effect of Addition on Sample Mean and Variance

• Adding 100 to all values in set A: Resulting set {198, 199, 200, 201, 202}
  • New sample mean: $\bar{x} = 200$
  • New sample variance: $s^2 = 2.5$
• Doubling and Subtracting: Doubling set A and subtracting 50 from each value.
  • New sample mean: $\bar{x} = 150$
• The sample variance would be four times the variance of your original set A. $s^2 = 10 = 4 \times 2.5$
• Rule for Addition: Establishing a rule for calculating the sample mean and variance based on the addition of values to the original set.

Expectation of a Linear Transformation (One Random Variable)

• Formula: $E[aX + c] = a \times E[X] + c$
  • Where:
    • $a$ and $c$ are constants.
    • $X$ is a random variable.
    • $E[\cdot]$ denotes the expected value (mean).

Examples of Linear Transformations (One Variable)

• Given: $E[X] = 2$
  • a. Calculate $E[5X]$
    • $E[5X] = 5 \times E[X] = 5 \times 2 = 10$
  • b. Calculate $E[-2X + 4]$
    • $E[-2X + 4] = -2 \times E[X] + 4 = -2 \times 2 + 4 = 0$
  • c. Calculate $E[5]$
    • $E[5] = 5$

Expectation of a Linear Combination (Two Random Variables)

• Formula: $E[aX + bY + c] = a \times E[X] + b \times E[Y] + c$
  • Where:
    • $a$, $b$, and $c$ are constants.
    • $X$ and $Y$ are random variables.
    • $E[\cdot]$ denotes the expected value (mean).

Examples of Linear Transformations (Two Variables)

• Given: $E[X] = 2$ and $E[Y] = 3$
  • a. Calculate $E[X - Y + 1]$
    • $E[X - Y + 1] = E[X] - E[Y] + 1 = 2 - 3 + 1 = 0$
  • b. Calculate $E[3X - 2Y + 5]$
    • $E[3X - 2Y + 5] = 3 \times E[X] - 2 \times E[Y] + 5 = 3 \times 2 - 2 \times 3 + 5 = 5$

Variance of a Linear Transformation (One Random Variable)

• Formula: $Var[aX + c] = a^2 \times Var[X]$
  • Where:
    • $a$ and $c$ are constants.
    • $X$ is a random variable.
    • $Var[\cdot]$ denotes the variance.

Examples of Variance Transformations (One Variable)

• Given: $Var[X] = 2$
  • a. Calculate $Var[2X + 4]$
    • $Var[2X + 4] = 2^2 \times Var[X] = 4 \times 2 = 8$
  • b. Calculate $Var[-X - 7]$
    • $Var[-X - 7] = (-1)^2 \times Var[X] = 1 \times 2 = 2$

Variance of a Linear Combination (Two Random Variables)

• Formula: $Var[aX + bY + c] = a^2 \times Var[X] + b^2 \times Var[Y] + 2abCov(X, Y)$
  • Where:
    • $a$, $b$, and $c$ are constants.
    • $X$ and $Y$ are random variables.
    • $Var[\cdot]$ denotes the variance.
    • $Cov(X, Y)$ is the covariance between $X$ and $Y$, measuring their joint variability.

Practice with Variance Calculations

• Given: $Var[X] = 2$, $Var[Y] = 3$, and $Cov(X, Y) = 0$
  • a. Calculate $Var[2X + 3Y + 4]$
  • b. Calculate $Var[X - Y]$
  • c. Calculate $Var[-3X - Y + 1]$

Applying Expectation Properties to Proportions

• Estimating $p<em>1 - p</em>2$ using $\hat{p}<em>1 - \hat{p}</em>2$
  • a. $E[\hat{p}<em>1] = p</em>1$
  • b. $E[\hat{p}<em>2] = p</em>2$
  • c. Expected value of the difference: $E[\hat{p}<em>1 - \hat{p}</em>2] = E[\hat{p}<em>1] - E[\hat{p}</em>2] = p<em>1 - p</em>2$
  • d. If \hat{p}1 - \hat{p}2 > 0, then $\hat{p}_1$ is larger.
  • e. If $\hat{p}<em>1 - \hat{p}</em>2 = 0$, the proportions are the same.

Applying Variance Properties to Proportions

• Estimating $p<em>1 - p</em>2$ using $\hat{p}<em>1 - \hat{p}</em>2$
  • a. Variance of $\hat{p}<em>1$: $\frac{\hat{p}</em>1(1 - \hat{p}<em>1)}{n</em>1}$
  • b. Variance of $\hat{p}<em>2$: $\frac{\hat{p}</em>2(1 - \hat{p}<em>2)}{n</em>2}$
  • c. Variance of the difference (assuming zero covariance):
    $Var(\hat{p}<em>1 - \hat{p}</em>2) = Var(\hat{p}<em>1) + Var(\hat{p}</em>2) \approx \frac{\hat{p}<em>1(1 - \hat{p}</em>1)}{n<em>1} + \frac{\hat{p}</em>2(1 - \hat{p}<em>2)}{n</em>2}$
  • d. Standard deviation of the difference (assuming zero covariance):
    $\sqrt{\frac{\hat{p}<em>1(1 - \hat{p}</em>1)}{n<em>1} + \frac{\hat{p}</em>2(1 - \hat{p}<em>2)}{n</em>2}}$

Confidence Interval for the Difference of Two Proportions

• Formula: $(\hat{p}<em>1 - \hat{p}</em>2) \pm z<em>{\alpha/2} \sqrt{\frac{\hat{p}</em>1(1 - \hat{p}<em>1)}{n</em>1} + \frac{\hat{p}<em>2(1 - \hat{p}</em>2)}{n_2}}$
  • Assumptions:
    • $n<em>1 \geq 30$ and $n</em>2 \geq 30$
    • $n<em>1\hat{p}</em>1 \geq 5$ and $n<em>2\hat{p}</em>2 \geq 5$
    • $n<em>1(1 - \hat{p}</em>1) \geq 5$ and $n<em>2(1 - \hat{p}</em>2) \geq 5$

Confidence Interval Example

• Scenario: Comparing the proportion of men ($p<em>1$) and women ($p</em>2$) taking vitamins.
  • Men: $n<em>1 = 100$, 72 take vitamins, $\hat{p}</em>1 = 0.72$
  • Women: $n<em>2 = 200$, 122 take vitamins, $\hat{p}</em>2 = 0.61$
  • a. 95% confidence interval for $p<em>1 - p</em>2$:
    $0.72 - 0.61 \pm 1.96 \sqrt{\frac{0.72(1 - 0.72)}{100} + \frac{0.61(1 - 0.61)}{200}} \approx [-0.0010, 0.2210]$
  • b. Interpretation: We are 95% confident that $p<em>1 - p</em>2$ is between -0.0010 and 0.2210.

Practice: Confidence Interval for Energy Drink vs. Coffee

• Scenario: Comparing high blood pressure rates between energy drink and coffee consumers.
  • Energy drinks: $n_1 = 120$, 40 have high blood pressure.
  • Coffee: $n_2 = 80$, 17 have high blood pressure.
  • Task:
    • a. Calculate the 95% confidence interval for the difference $p<em>{energy} - p</em>{coffee}$.
    • b. Interpret the confidence interval.

Applying Expectation Properties to Means

• Estimating $\mu<em>1 - \mu</em>2$ using $\bar{x}<em>1 - \bar{x}</em>2$
  • a. $E[\bar{x}<em>1] = \mu</em>1$
  • b. $E[\bar{x}<em>2] = \mu</em>2$
  • c. Expected value of the difference: $E[\bar{x}<em>1 - \bar{x}</em>2] = E[\bar{x}<em>1] - E[\bar{x}</em>2] = \mu<em>1 - \mu</em>2$
  • d. If \bar{x}1 - \bar{x}2 < 0, then $\bar{x}_2$ is larger.
  • e. If $\bar{x}<em>1 - \bar{x}</em>2 = 0$, the sample means are the same.

Applying Variance Properties to Means

• Estimating $\mu<em>1 - \mu</em>2$ using $\bar{x}<em>1 - \bar{x}</em>2$
  • a. Variance of $\bar{x}<em>1$ (approximated using $s</em>1$): $\frac{s<em>1^2}{n</em>1}$
  • b. Variance of $\bar{x}<em>2$ (approximated using $s</em>2$): $\frac{s<em>2^2}{n</em>2}$
  • c. Variance of the difference:
    $Var(\bar{x}<em>1 - \bar{x}</em>2) = Var(\bar{x}<em>1) + Var(\bar{x}</em>2) \approx \frac{s<em>1^2}{n</em>1} + \frac{s<em>2^2}{n</em>2}$
  • d. Standard deviation of the difference:
    $\sqrt{\frac{s<em>1^2}{n</em>1} + \frac{s<em>2^2}{n</em>2}}$

Confidence Interval for the Difference of Two Means

• Formula: $\bar{x}<em>1 - \bar{x}</em>2 \pm t<em>{\alpha/2, \hat{\nu}} \sqrt{\frac{s</em>1^2}{n<em>1} + \frac{s</em>2^2}{n_2}}$
  • Degrees of freedom: $\hat{\nu} = min(n<em>1 - 1, n</em>2 - 1)$

Discussion on Degrees of Freedom Calculation

• Conservative estimate: $\hat{\nu} = min(n<em>1 - 1, n</em>2 - 1)$
• Satterthwaite's degrees of freedom: $\hat{\nu}<em>{Satterthwaite} = \frac{\left(\frac{s</em>1^2}{n<em>1} + \frac{s</em>2^2}{n<em>2}\right)^2}{\frac{\left(\frac{s</em>1^2}{n<em>1}\right)^2}{n</em>1 - 1} + \frac{\left(\frac{s<em>2^2}{n</em>2}\right)^2}{n_2 - 1}}$
  • Justification: $min(n<em>1 - 1, n</em>2 - 1) \leq \hat{\nu}<em>{Satterthwaite} \leq n</em>1 + n_2 - 2$

Example: Hypertension Drug Trial

• Scenario: Evaluating the effectiveness of a hypertension drug in lowering systolic blood pressure (SBP).
  • Drug: $n<em>1 = 16$, $\bar{x}</em>1 = 125$, $s_1 = 10$
  • Placebo: $n<em>2 = 14$, $\bar{x}</em>2 = 138$, $s_2 = 20$
  • a. 95% confidence interval for $\mu<em>1 - \mu</em>2$:
    $125 - 138 \pm t_{\alpha/2, 13} \sqrt{\frac{10^2}{16} + \frac{20^2}{14}} \approx [-25.75, -0.25]$
  • b. Interpretation: We are 95% confident that the difference in population means $\mu<em>{Drug} - \mu</em>{Placebo}$ is between -25.75 and -0.25.

Practice: Corn Stalk Height Comparison

• Scenario: Examining the difference in corn stalk height between Iowa and Nebraska.
  • Iowa: $n<em>1 = 15$, $\bar{x}</em>1 = 155$ cm, $s_1 = 16$
  • Nebraska: $n<em>2 = 20$, $\bar{x}</em>2 = 145$ cm, $s_2 = 10$
  • Task:
    • a. Calculate the 95% confidence interval for $\mu<em>{Iowa} - \mu</em>{Nebraska}$.
    • b. Interpret the confidence interval.

How to Write a Hypothesis for 2-Samples

• Null Hypothesis: $H<em>0: \mu</em>1 - \mu<em>2 = 0$ or $H</em>0: \mu<em>1 = \mu</em>2$
• Possible Alternate Hypotheses:
  • $H<em>a: \mu</em>1 - \mu<em>2 > 0$ or $H</em>a: \mu<em>1 > \mu</em>2$
  • $H<em>a: \mu</em>1 - \mu<em>2 < 0$ or $H</em>a: \mu<em>1 < \mu</em>2$
  • $H<em>a: \mu</em>1 - \mu<em>2 \neq 0$ or $H</em>a: \mu<em>1 \neq \mu</em>2$
• Same approach applies to hypothesis testing about two proportions.

Hypothesis Testing Examples

• a. Claim: Mean workout time for men > women. P-value = 0.4358
  • $H<em>0: \mu</em>M = \mu_W$
  • Ha: \muM > \mu_W
  • Conclusion: Fail to reject $H_0$. Not enough evidence to suggest men work out more.
• b. Claim: Mean time to finish statistics homework = math homework. P-value = $10^{-7}$
  • $H<em>0: \mu</em>S = \mu_M$
  • $H<em>a: \mu</em>S \neq \mu_M$
  • Conclusion: Reject $H_0$. There is enough evidence to suggest the mean times are different.

Hypothesis Testing Practice

• a. Claim: Mean points in professional football = college football. P-value = 0.1438
• b. Claim: Mean fast food spending now > 10 years ago. P-value = 1.43 × $10^{-23}$

Test Statistic for 2-Sample Means

• Formula: t = \frac{\bar{x}1 - \bar{x}2 - (\mu1 - \mu2)0}{\sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n_2}}}}
  • $\bar{x}_i$ = Sample mean for i = 1,2
  • $s_i^2$ = Sample variance for i = 1,2
  • $n_i$ = Sample size for i = 1,2
  • $(μ<em>1 - μ</em>2)_0$ = Hypothesized difference of the two means

Example: Corn Stalk Height Hypothesis Test

• Scenario: Testing the claim that Iowa corn grows taller than Nebraska corn.
  • Iowa: $n<em>1 = 15$, $\bar{x}</em>1 = 155$ cm, $s_1 = 16$
  • Nebraska: $n<em>2 = 20$, $\bar{x}</em>2 = 145$ cm, $s_2 = 10$
  • a. Hypotheses: $H<em>0: \mu</em>I = \mu<em>N$, $H</em>a: \mu<em>I > \mu</em>N$
  • b. Test statistic: $t = \frac{155 - 145 - 0}{\sqrt{\frac{16^2}{15} + \frac{10^2}{20}}} \approx 2.13$
  • c. P-value: $1 - t.dist(2.13, 14, TRUE) \approx 0.0257$
  • d. Conclusion: Reject $H_0$. Enough evidence to suggest Iowa corn is taller.

Using Satterthwaite Degrees of Freedom

• Continuing the corn stalk height example.
  • a. Calculate the p-value using Satterthwaite's degrees of freedom:
    $\hat{\nu}_{Satterthwaite} = \frac{\left(\frac{16^2}{15} + \frac{10^2}{20}\right)^2}{\frac{\left(\frac{16^2}{15}\right)^2}{15 - 1} + \frac{\left(\frac{10^2}{20}\right)^2}{20 - 1}} \approx 22$
    $p \approx 1 - t.dist(2.13, 22, TRUE) \approx 0.0223$

Practice: Alex vs. Brian Game Scores

• Scenario: Testing if Alex's mean score equals Brian's mean score over 6 games each.
  • Task:
    • a. State the null and alternate hypotheses.
    • b. Calculate the test statistic.
    • c. Determine the p-value.
    • d. Draw a conclusion.

Installing Data Analysis ToolPak for Windows

• Step 1: Click on "File" followed by "Options"
• Step 2: Click "Add-ins"
• Step 3: Click "Go"
• Step 4: Check "Analysis ToolPak"
• Step 5: Click "OK." Double check it worked. Click "Data" followed by "Data Analysis"
• We can use this for "T-Test: Two Sample Assuming Unequal Variances"