Chapter 10: Comparing Two Populations or Treatments

Comparing Two Populations or Treatments

  • Chapter 10 Lessons Overview     * Lesson 10.1: Estimating a Difference Between Two Proportions (Page 590).     * Lesson 10.2: Testing a Claim about a Difference Between Two Proportions (Page 599).     * Lesson 10.3: Estimating a Difference Between Two Means (Page 610).     * Lesson 10.4: Testing a Claim about a Difference Between Two Means (Page 620).     * Lesson 10.5: Paired Data: Estimating a Mean Difference (Page 631).     * Lesson 10.6: Paired Data: Testing a Claim about a Mean Difference (Page 641).

Estimating a Difference Between Two Proportions

  • Goal of the Lesson     * Estimate the difference $p_1 - p_2$ between the proportions of successes in Population 1 and Population 2.     * The ideal strategy is to take a separate random sample from each population and use the difference between sample proportions p^1−p^2\hat{p}_1 - \hat{p}_2 as the point estimate.

  • Describing the Sampling Distribution of p^<em>1−p^2\hat{p}<em>1 - \hat{p}_2     * Let p^1\hat{p}_1 be the proportion of successes in an SRS of size $n_1$ from a large population with proportion $p_1$.     * Let p^2\hat{p}_2 be the proportion of successes in an SRS of size $n_2$ from a large population with proportion $p_2$.     * Shape: Approximately normal if $n_1p_1$, $n_1(1-p_1)$, $n_2p_2$, and $n_2(1-p_2)$ are all at least $10$.     * Mean: μ</em>p^1−p^2=p1−p2\mu</em>{\hat{p}_1 - \hat{p}_2} = p_1 - p_2.     * Standard Deviation:σp^1−p^2=p1(1−p1)n1+p2(1−p2)n2\sigma_{\hat{p}_1 - \hat{p}_2} = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}

  • Conditions for Constructing a Confidence Interval for $p_1 - p_2$     * Random: Data must come from separate random samples from the two populations of interest or from two groups in a randomized experiment.     * Large Counts: The counts of "successes" and "failures" in each sample or group (n1p^1n_1\hat{p}_1, n1(1−p^1)n_1(1-\hat{p}_1), n2p^2n_2\hat{p}_2, and n2(1−p^2)n_2(1-\hat{p}_2)) must all be at least $10$.     * Application Note: Because $p_1$ and $p_2$ are unknown when estimating, use observed sample proportions p^1\hat{p}_1 and p^2\hat{p}_2 to check the Large Counts condition. Observed counts should be rounded to the nearest integer.

  • Importance of Random Condition     * Random sampling allows for the generalization of results to the populations of interest.     * Random assignment in an experiment permits cause-and-effect conclusions.

Two-Sample z Interval for a Difference Between Two Proportions

  • General Formula for Confidence Intervals     * point estimate±margin of error\text{point estimate} \pm \text{margin of error}     * point estimate±(critical value)×(standard error of statistic)\text{point estimate} \pm (\text{critical value}) \times (\text{standard error of statistic})

  • Standard Error of p^1−p^2\hat{p}_1 - \hat{p}_2     * Since $p_1$ and $p_2$ are unknown parameters, they are replaced with p^1\hat{p}_1 and p^2\hat{p}_2 to calculate the standard error ($SE$):         SEp^1−p^2=p^1(1−p^1)n1+p^2(1−p^2)n2SE_{\hat{p}_1 - \hat{p}_2} = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}

  • Calculating the C% Confidence Interval     * If conditions are met, the interval is:         (p^1−p^2)±z∗p^1(1−p^1)n1+p^2(1−p^2)n2(\hat{p}_1 - \hat{p}_2) \pm z^* \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}     * z∗z^* is the critical value for the standard normal distribution with $C\%$ of its area between −z∗-z^* and z∗z^*.

  • Example: Facebook usage (Teens vs. Adults)     * Data: SRS of $1502$ U.S. adults (69% use Facebook); SRS of $743$ U.S. teens (51% use Facebook).     * Parameters: p1p_1 = proportion of all U.S. adults using Facebook; p2p_2 = proportion of all U.S. teens using Facebook.     * Large Counts Check:         * Adult successes: 1502×0.69=1036.4→10361502 \times 0.69 = 1036.4 \rightarrow 1036         * Adult failures: 1502×0.31=465.6→4661502 \times 0.31 = 465.6 \rightarrow 466         * Teen successes: 743×0.51=378.9→379743 \times 0.51 = 378.9 \rightarrow 379         * Teen failures: 743×0.49=364.1→364743 \times 0.49 = 364.1 \rightarrow 364         * All counts ≥10\ge 10.     * Calculation (95% CI):         * (0.69−0.51)±1.960.69(0.31)1502+0.51(0.49)743(0.69 - 0.51) \pm 1.96 \sqrt{\frac{0.69(0.31)}{1502} + \frac{0.51(0.49)}{743}}         * 0.18±0.043→0.137 to 0.2230.18 \pm 0.043 \rightarrow 0.137 \text{ to } 0.223

  • Interpreting the Confidence Interval     * Standard Sentence: "We are 95% confident that the interval from 0.137 to 0.223 captures the true difference in the proportion of all U.S. adults and teens who use Facebook."     * Plain Language Sentence: "The interval suggests that adult Facebook use is between 13.7 and 22.3 percentage points greater than teen Facebook use in the United States."     * Switching Order: If the order was switched (Teens - Adults), the interval would be −0.223 to −0.137-0.223 \text{ to } -0.137. Both are correct if the order of subtraction is defined.

The Four-Step Process for Inference

  • Step 1: STATE     * State the parameters you want to estimate and the confidence level.

  • Step 2: PLAN     * Identify the methodology (e.g., Two-sample z interval for p1−p2p_1 - p_2) and check the Random and Large Counts conditions.

  • Step 3: DO     * Perform calculations to find the point estimate, margin of error, and final interval.

  • Step 4: CONCLUDE     * Interpret the interval in context, including the descriptive second sentence regarding the size and direction of the difference.

  • Example: President Trump Approval Rating (Jan 2017 to Oct 2019)     * Problem: Poll of $1525$ adults in Jan 2017 ($686$ approved); poll of $1526$ adults in Oct 2019 ($595$ approved). Construct 90% CI for change ($p_{19} - p_{17}$).     * STATE: 90% CI for p19−p17p_{19} - p_{17}.     * PLAN: Two-sample z interval. Random: separate random samples. Large Counts: $595, 931, 686, 839$ are all ≥10\ge 10.     * DO: p^19=0.39,p^17=0.45\hat{p}_{19} = 0.39, \hat{p}_{17} = 0.45.         * (0.39−0.45)±1.6450.39(0.61)1526+0.45(0.55)1525(0.39 - 0.45) \pm 1.645 \sqrt{\frac{0.39(0.61)}{1526} + \frac{0.45(0.55)}{1525}}         * −0.06±0.029→−0.089 to −0.031-0.06 \pm 0.029 \rightarrow -0.089 \text{ to } -0.031     * CONCLUDE: We are 90% confident that the change in proportion of approval was between a decrease of 3.1 and 8.9 percentage points. Note: This differs from identifying a percentage decrease ($6/45 = 13.3\%$ decrease compared to 6 percentage points).

Testing a Claim about a Difference Between Two Proportions

  • Introduction to Significance Tests     * Tests determine if an observed difference reflects an actual population difference or chance variation.     * Null Hypothesis ($H_0$): $H_0: p_1 - p_2 = 0$ (or $H_0: p_1 = p_2$).     * Alternative Hypothesis ($H_a$): Can be $p_1 - p_2 > 0$, $p_1 - p_2 < 0$, or p1−p2≠0p_1 - p_2 \neq 0.

  • Conditions for Significance Tests     * Same as confidence intervals: Random and Large Counts.

  • Standardized Test Statistic (z)     * z = \frac{\text{statistic} - \text{null value}}{\text{standard deviation of statistic}}}     * In a significance test, we assume $H_0$ is true ($p_1 = p_2$). Therefore, we do not use separate sample proportions in the denominator for standard deviation.

  • Combined (Pooled) Sample Proportion ($p_c$)     * Combine data from both samples to estimate the common proportion ($p_c$):         pc=X1+X2n1+n2=total successestotal individualsp_c = \frac{X_1 + X_2}{n_1 + n_2} = \frac{\text{total successes}}{\text{total individuals}}     * Test Statistic Formula:z=(p^1−p^2)−0pc(1−pc)n1+pc(1−pc)n2z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\frac{p_c(1-p_c)}{n_1} + \frac{p_c(1-p_c)}{n_2}}}

  • Example: Children Eating Breakfast     * Data: School 1 (SRS $n=80$, $61$ ate); School 2 (SRS $n=150$, $124$ ate).     * Hypotheses: $H_0: p_1 - p_2 = 0$; Ha:p1−p2≠0H_a: p_1 - p_2 \neq 0.     * Calculations:         * p^1=0.7625,p^2=0.8267\hat{p}_1 = 0.7625, \hat{p}_2 = 0.8267.         * pc=61+12480+150=185230≈0.804p_c = \frac{61 + 124}{80 + 150} = \frac{185}{230} \approx 0.804.         * z=0.7625−0.82670.804(0.196)80+0.804(0.196)150=−1.17z = \frac{0.7625 - 0.8267}{\sqrt{\frac{0.804(0.196)}{80} + \frac{0.804(0.196)}{150}}} = -1.17     * P-value: Using Table A or technology for a two-sided test: 2×P(Z≤−1.17)=0.24202 \times P(Z \le -1.17) = 0.2420.     * Conclusion: With a P-value of 0.2420 > α=0.05\alpha = 0.05, fail to reject $H_0$. There is not convincing evidence of a difference.

Inference for Randomized Experiments

  • Comparison to Random Sampling     * Experiments often use recruited volunteers rather than random selection.     * Random assignment to treatments allows for cause-and-effect conclusions applicable to "people like those in the study."

  • Example: Financial Incentives to Quit Smoking     * Study: GM sponsored study; 878 volunteers.     * Groups: 439 assigned to $750 incentivized group ($15\%$ quit); 439 assigned to traditional group ($5\%$ quit).     * Hypotheses: $H_0: p_F - p_T = 0$; $H_a: p_F - p_T > 0$ at α=0.01\alpha = 0.01.     * Calculations:         * pc=66+22439+439=0.10p_c = \frac{66 + 22}{439 + 439} = 0.10         * z=0.15−0.050.10(0.90)439+0.10(0.90)439=4.94z = \frac{0.15 - 0.05}{\sqrt{\frac{0.10(0.90)}{439} + \frac{0.10(0.90)}{439}}} = 4.94     * P-value: Approximately 0.     * Conclusion: Reject $H_0$. There is convincing evidence that financial incentives help employees like these quit smoking.     * Type I Error Risk: Rejecting $H_0$ when it is true would mean spending money on an ineffective method.

  • Simulation-Based Inference     * Researchers can simulate repeating the random assignment many times under the assumption that the null hypothesis is true (the treatment has no effect).     * If the observed difference (e.g., 0.10) never or rarely occurs in simulations (e.g., 0 times in 500 repetitions), the result is statistically significant (Estimated P-value ≈0\approx 0).

Technology Corner: Calculations for Two Proportions

  • Applet Usage (One Categorical Variable, Multiple Groups)     1. Choose "Two-way table" input.     2. Enter variable and group names.     3. Input success and failure counts.     4. Select "2-sample z interval" or "2-sample z test".     5. Note: Set preferences to proportions if output is in percentages.

  • TI-83/84 Calculator Usage     * Confidence Intervals: STAT →\rightarrow TESTS →\rightarrow B: 2-PropZInt. Enter inputs (x1,n1,x2,n2,C-Levelx_1, n_1, x_2, n_2, C\text{-Level}). $x$ values must be integers.     * Significance Tests: STAT →\rightarrow TESTS →\rightarrow 6: 2-PropZTest. Enter inputs and select alternative hypothesis orientation (p1≠p2p_1 \neq p_2, $< p_2$, or $> p_2$). Highlight "Calculate" or "Draw".