Comprehensive Study Notes on Normal Distributions, Hypothesis Testing, and Proportion Inference

Density Curves and the Normal Distribution (Section P.5)

• Random Variables:

  • Definition: A random variable is a numeric quantity that changes from trial to trial in a random process. It is typically denoted by capital letters such as $X$ or $Y$.

  • Discrete Random Variable: A random variable that possesses a finite (or countably infinite) set of possible values.

  • Example: The numerical result obtained from rolling a fair six-sided die ($X \in \{1, 2, 3, 4, 5, 6\}$).

  • Example: The count of girls in a randomly selected family with 3 children ($X \in \{0, 1, 2, 3\}$).

  • Continuous Random Variable: A random variable that can assume any numerical value within a given continuous interval on the real line.

  • Example: The measured height of a randomly selected individual.

  • Example: The exact duration of time required to read a designated textbook section.

• Density Curves:

  • Definition: A density curve is a smooth, continuous theoretical mathematical model representing the probability distribution of a continuous random variable.

  • Fundamental Properties:

  • Total Area Property: The total area underneath the density curve above the horizontal axis is exactly equal to $1$ (corresponding to $100\%$ of the distribution).

  • Interval Proportion Property: The proportion of the population or the probability that the random variable falls within any designated interval $[a, b]$ is given precisely by the area under the curve over that interval.

  • Example: Commute Distances in Atlanta:

  • A right-skewed continuous density curve models commute distances in Atlanta.

  • Total area across all potential commute distances equals $1.0$.

  • Region for commute distances between $30$ and $40\,\text{miles}$ accounts for $0.15$ ($15\%$) of the total area.

  • Region for commute distances strictly less than $30\,\text{miles}$ represents $0.80$ ($80\%$) of the total area.

  • Region for commute distances strictly greater than $40\,\text{miles}$ represents $0.05$ ($5\%$) of the total area.

  • Sum of disjoint regions: $0.80 + 0.15 + 0.05 = 1.00$.

• Normal Distributions ($N(\mu, \sigma)$):

  • Definition: A symmetric, bell-shaped density curve uniquely defined by its population mean $\mu$ and population standard deviation $\sigma$.

  • Center: The peak and center of symmetry of the distribution is situated directly at the mean $\mu$.

  • Horizontal Spread: The dispersion is governed by $\sigma$. Approximately $95\%$ of the total area under the normal curve lies within two standard deviations of the mean: $(\mu - 2\sigma, \mu + 2\sigma)$.

  • Notation: Written as $X \sim N(\mu, \sigma)$, denoting that the random variable $X$ is distributed normally with mean $\mu$ and standard deviation $\sigma$.

• The Standard Normal Distribution ($N(0, 1)$) and Z-Scores:

  • Definition: The specific normal distribution that has a mean of $\mu = 0$ and a standard deviation of $\sigma = 1$. It is denoted as $Z \sim N(0, 1)$.

  • Standardizing (Forward Conversion): To convert an observation $X$ from any general normal distribution $N(\mu, \sigma)$ into its corresponding standardized value ($z$-score) in $N(0, 1)$, apply:     $Z = \frac{X - \mu}{\sigma}$

  • Unstandardizing (Reverse Conversion): To convert a standardized $z$-score back to the original measurement scale $X$ in $N(\mu, \sigma)$, apply:     $X = \mu + Z\sigma$

• Finding Probabilities, Percentiles, and Cutoffs:

  • Methodology: Because no elementary closed-form algebraic formula evaluates the integral of a normal density curve, probabilities and percentiles are found using standard normal cumulative probability tables (Table IV) or statistical software (StatKey).

  • Worked Example: General Normal Transformation ($N(68, 4)$):

  • Problem: Find the proportion of the distribution lying above $X = 72$ for $N(68, 4)$.

  • Standardization Step: Convert $X = 72$ to a $z$-score:       $Z = \frac{72 - 68}{4} = \frac{4}{4} = 1.0$

  • Standard Normal Area Lookup: Table IV yields the cumulative area to the left of $Z = 1.00$ as $0.8413$.

  • Upper Tail Area: Subtract the cumulative left area from $1$:       $\text{Area to the right of } 1.0 = 1 - 0.8413 = 0.1587$

  • Symmetry Verification: By symmetry, the right-tail area above $Z = 1.0$ equals the left-tail area below $Z = -1.0$, which reads directly from Table IV as $0.1587$.

  • Conclusion: The area above $72$ in an $N(68, 4)$ distribution is identical to the area above $1.0$ in an $N(0, 1)$ distribution: $0.1587$.

  • Worked Example: Exam Scores ($N(75, 10)$):

  • Exam scores follow an $N(\mu = 75, \sigma = 10)$ distribution.

  • Question 1: What proportion of students score above 90 points?

    • Standardize $X = 90$:         $Z = \frac{90 - 75}{10} = \frac{15}{10} = 1.50$

    • From Table IV or StatKey, the cumulative area to the left of $Z = 1.50$ is $0.9332$.

    • Proportion above $90$ points:         $1 - 0.9332 = 0.0668 \approx 0.067$

  • Question 2: What is the score cutoff for the lowest 20% of the distribution?

    • Identify the $20^{\text{th}}$ percentile in the standard normal distribution ($\text{Area} = 0.2000$ to the left).

    • In Table IV, locate the area closest to $0.2000$: $Z = -0.84$ yields an area of $0.2005$.

    • Unstandardize $Z = -0.84$ into the raw score metric:         $X = \mu + Z\sigma = 75 + (-0.84)(10) = 75 - 8.4 = 66.6$

    • High-precision software (StatKey) gives $X = 66.584$.

    • Students scoring below $66.6$ points fall into the bottom $20\%$ and are referred for tutoring.

Hypothesis Tests Using Normal Distributions (Section 5.1)

• Connection Between Simulation Distributions and the Normal Distribution:

  • Graphs of sampling distributions, bootstrap distributions, and randomization distributions across large samples consistently display a bell shape.

  • This behavior is formalized by the Central Limit Theorem.

• The Central Limit Theorem (CLT):

  • Formal Statement: For random samples drawn with a sufficiently large sample size, the distribution of many common sample statistics can be closely approximated by a normal distribution.

• Modeling Randomization Distributions:

  • A bell-shaped randomization distribution is centered at the value specified by the null hypothesis and dispersed according to the standard error ($SE$).

  • Best Approximating Model:     $N(\text{null parameter}, SE)$

  • Finding p-values:

  • The p-value is evaluated as the area or proportion under this approximating normal curve beyond the observed sample statistic in the direction dictated by the alternative hypothesis $H_a$.

  • Two-tailed tests: The p-value is obtained by taking twice the area of the smaller tail beyond the observed statistic.

  • Comparison of Methods: A p-value may be calculated either via simulated randomization distributions (simulation approach) or via the approximating normal distribution (theoretical approach).

• Standardized Test Statistics ($Z$):

  • Definition: When the distribution of a sample statistic under the null hypothesis $H_0$ is normally distributed, the test statistic is standardized into a $z$-score.

  • Formula:     $Z = \frac{\text{sample statistic} - \text{null parameter}}{SE}$

  • p-value Computation via Standard Normal Distribution ($N(0, 1)$):

  • Once the test statistic is converted to $Z$, the p-value is the proportion of the $N(0, 1)$ distribution falling beyond $Z$ in the direction of $H_a$.

• Comprehensive Worked Example: Organic Food and Longevity in Fruit Flies:

  • Context: A 16-year-old student conducted a controlled science experiment testing whether organic food confers health advantages over conventional food. Fruit flies were randomly divided into two nutritional groups: one fed conventional produce (potatoes, raisins, bananas, or soybeans) and the other fed organic versions of the same food. Biological parameters including longevity, fertility, stress resistance, and activity were tracked.

  • Data: After 13 days of feeding on potato diets:

  • Organic group: $318$ out of $n_O = 500$ flies were alive.

  • Conventional group: $285$ out of $n_C = 500$ flies were alive.

  • Step 1: Parameters, Hypotheses, and Observed Statistic:

  • Parameters: Let $p_O$ be the true proportion of organic-fed flies surviving $13\,\text{days}$, and let $p_C$ be the true proportion of conventional-fed flies surviving $13\,\text{days}$.

  • Hypotheses:       $H_0: p_O = p_C \quad \text{vs.} \quad H_a: p_O > p_C$       (Equivalently, $H_0: p_O - p_C = 0$ vs. $H_a: p_O - p_C > 0$)

  • Sample Proportions:       $\hat{p}_O = \frac{318}{500} = 0.636, \quad \hat{p}_C = \frac{285}{500} = 0.570$

  • Observed Sample Statistic:       $\hat{p}_O - \hat{p}_C = 0.636 - 0.570 = 0.066$

  • Step 2: Simulation Approach (Randomization Distribution):

  • From simulation, standard error is $SE = 0.031$.

  • Simulated p-value: $0.02$.

  • Decision: At a significance level of $\alpha = 0.05$, because $p\text{-value} = 0.02 \le 0.05$, reject $H_0$.

  • Conclusion: There is statistically significant evidence to conclude that fruit flies have a higher survival rate when fed an organic diet.

  • Step 3: Normal Approximation Method ($N(\text{null parameter}, SE)$):

  • Best approximating normal model: $N(0, 0.031)$.

  • The observed difference is $0.066$.

  • Proportion of $N(0, 0.031)$ lying to the right of $0.066$ yields a p-value of $0.017$.

  • Step 4: Standardized Test Statistic and Table IV Lookup:

  • Standardized test statistic calculation:       $Z = \frac{\text{sample statistic} - \text{null parameter}}{SE} = \frac{0.066 - 0}{0.031} = 2.129$

  • Because this is a right-tailed test, the p-value is the area to the right of $Z = 2.129$ in $N(0, 1)$.

  • Using Table IV (Negative $z$ values): By symmetry, the area to the right of $+2.129$ equals the area to the left of $-2.129$. In Table IV, under row $-2.1$ and column $.03$, the cumulative probability is $0.0166$.

  • Using Table IV (Positive $z$ values): Under row $2.1$ and column $.03$, the cumulative area to the left is $0.9834$. Therefore:       $\text{p-value} = 1 - 0.9834 = 0.0166$

  • Both table evaluations confirm $p\text{-value} = 0.0166$, matching the theoretical normal calculation.

Confidence Intervals Using Normal Distributions (Section 5.2)

• Modeling Bootstrap Distributions:

  • By the Central Limit Theorem, when the sample size is sufficiently large, a bell-shaped bootstrap distribution is centered at the original sample statistic with dispersion given by the bootstrap standard error ($SE$).

  • Best Approximating Model:     $N(\text{sample statistic}, SE)$

  • A $P\%$ confidence interval is obtained by identifying the interval containing the central $P\%$ of the area under this approximating normal distribution.

• Percentiles of Normal Distributions:

  • Example: Middle 95% of $N(40, 5)$:

  • To isolate the central $95\%$ of area, allocate $\frac{100\% - 95\%}{2} = 2.5\% = 0.025$ into each tail.

  • In the standard normal distribution $N(0, 1)$, the endpoints containing the middle $95\%$ are $-1.96$ and $+1.96$.

  • Converting these critical endpoints to $N(40, 5)$:       $\text{Lower Bound} = 5(-1.96) + 40 = -9.8 + 40 = 30.2$       $\text{Upper Bound} = 5(1.96) + 40 = 9.8 + 40 = 49.8$

  • Middle $95\%$ interval: $(30.2, 49.8)$.

  • Example: Middle 99% of Standard Normal Distribution:

  • To isolate the central $99\%$ of area, allocate $\frac{100\% - 99\%}{2} = 0.5\% = 0.005$ into each tail.

  • Looking up a left-tail cumulative area of $0.0050$ in Table IV, the value lies equidistant between $-2.57$ ($0.0051$) and $-2.58$ ($0.0049$), yielding $Z = -2.575$ (or approximately $\pm 2.576$).

  • The symmetric interval in $N(0, 1)$ containing the middle $99\%$ of the area is $(-2.576, 2.576)$.

• Constructing Confidence Intervals Using $z^*$:

  • When a bootstrap distribution is normal with standard error $SE$, the general confidence interval formula is:     $\text{Sample statistic} \pm z^* \cdot SE$

  • The critical multiplier $z^*$ is chosen such that the area under the standard normal curve between $-z^*$ and $+z^*$ equals the confidence level $P\%$.

  • Standard Critical Values ($z^*$):

  • 90% Confidence Level: Excludes $5\% = 0.05$ in each tail; $z^* = 1.645$.

  • 95% Confidence Level: Excludes $2.5\% = 0.025$ in each tail; $z^* = 1.960$.

  • 99% Confidence Level: Excludes $0.5\% = 0.005$ in each tail; $z^* = 2.576$.

• Comprehensive Worked Example: Online Dating Survey:

  • Context: A survey conducted in July 2015 sampled $n = 2000$ American adults at random. Across the sample, $15\%$ ($\hat{p} = 0.15$) reported having used online dating services.

  • Method 1: Bootstrap Distribution (StatKey):

  • Bootstrap resampling yields an empirical standard error of $SE = 0.008$.

  • The resulting 90% confidence interval spanning the central $90\%$ of bootstrap statistics is $(0.137, 0.163)$.

  • Method 2: Approximating Normal Distribution:

  • The bootstrap distribution is modeled as $N(0.15, 0.008)$.

  • The central $90\%$ of this normal distribution spans from $0.137$ to $0.163$.

  • Method 3: Standard Normal Formula ($\text{Statistic} \pm z^* \cdot SE$):

  • Original sample statistic: $\hat{p} = 0.15$.

  • Standard error: $SE = 0.008$.

  • Desired confidence: $90\%$, requiring $z^* = 1.645$.

  • Margin of error calculation:       $ME = z^* \cdot SE = 1.645 \cdot 0.008 = 0.01316$

  • Interval calculation:       $0.15 \pm 1.645(0.008) = 0.15 \pm 0.01316 = (0.13684, 0.16316) \approx (0.137, 0.163)$

  • Interpretation: One can be $90\%$ confident that the true proportion of all American adults who have used online dating is between $13.7\%$ and $16.3\%$.

Inference for a Single Proportion (Section 6.1)

• Sampling Distribution of a Sample Proportion ($\hat{p}$):

  • When taking random samples of size $n$ from a population with true proportion $p$, the theoretical sampling distribution of $\hat{p}$ has:

  • Center: Centered at the population parameter $p$.

  • Standard Error Formula:       $SE = \sqrt{\frac{p(1 - p)}{n}}$

  • Normality Conditions (Sample Size Check): The sampling distribution is reasonably modeled by a normal distribution provided that:       $np \ge 10 \quad \text{and} \quad n(1 - p) \ge 10$

• Standard Error Calculations for Varying Sample Sizes:

  • Case 1: $p = 0.25$ and $n = 50$     $SE = \sqrt{\frac{0.25(1 - 0.25)}{50}} = \sqrt{\frac{0.25(0.75)}{50}} = \sqrt{\frac{0.1875}{50}} = \sqrt{0.00375} \approx 0.061$

  • Case 2: $p = 0.25$ and $n = 200$     $SE = \sqrt{\frac{0.25(1 - 0.25)}{200}} = \sqrt{\frac{0.1875}{200}} = \sqrt{0.0009375} \approx 0.031$

  • Observation: Quadrupling the sample size from $50$ to $200$ cuts the standard error in half (from $0.061$ to $0.031$), illustrating the inverse-square-root relationship between sample size and standard error.

• Example: On-Time Flight Arrivals:

  • The Bureau of Transportation Statistics states that $80\%$ ($p = 0.80$) of an airline's flights arrive on time (defined as within $15\,\text{minutes}$ of scheduled arrival).

  • A random sample of $n = 400$ flights is examined.

  • Verification of Normality Conditions:     $np = 400(0.80) = 320 \ge 10$     $n(1 - p) = 400(1 - 0.80) = 400(0.20) = 80 \ge 10$     Both criteria are met; the distribution is approximately normal.

  • Standard Error Calculation:     $SE = \sqrt{\frac{0.80(1 - 0.80)}{400}} = \sqrt{\frac{0.80(0.20)}{400}} = \sqrt{\frac{0.16}{400}} = \sqrt{0.0004} = 0.02$

• Confidence Intervals for a Single Population Proportion:

  • Sample Statistic: Observed sample proportion $\hat{p}$.

  • Standard Error of the Estimate:     $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

  • Normality Conditions: The sample size must satisfy:     $n\hat{p} \ge 10 \quad \text{and} \quad n(1 - \hat{p}) \ge 10$

  • Formula for a $P\%$ Confidence Interval:     $\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

  • Step-by-Step Procedure:

  1. Verify that $n\hat{p} \ge 10$ and $n(1 - \hat{p}) \ge 10$.

  2. Identify the critical value $z^*$ in $N(0, 1)$ matching the required confidence level.

  3. Calculate the standard error and margin of error, then compute the confidence interval endpoints.

  • Worked Example: Peanut Proportion in Mixed Nuts:

  • In a sample of $n = 100$ mixed nuts, $52$ are peanuts.

  • Sample proportion: $\hat{p} = \frac{52}{100} = 0.52$.

  • Critical value for a $95\%$ confidence interval: $z^* = 1.96$.

  • Check conditions:       $n\hat{p} = 100(0.52) = 52 \ge 10$       $n(1 - \hat{p}) = 100(1 - 0.52) = 100(0.48) = 48 \ge 10$

  • Standard error calculation:       $SE = \sqrt{\frac{0.52(1 - 0.52)}{100}} = \sqrt{\frac{0.52(0.48)}{100}} = \sqrt{\frac{0.2496}{100}} = \sqrt{0.002496} \approx 0.04996$

  • Confidence interval calculation:       $0.52 \pm 1.96 \cdot 0.04996 = 0.52 \pm 0.0979 = (0.4221, 0.6179) \approx (0.42, 0.62)$

  • Contextual Interpretation: We are $95\%$ confident that the true proportion of peanuts in this brand of mixed nuts is between $0.42$ and $0.62$ (or between $42\%$ and $62\%$).

  • Comparison: This theoretical interval is identical to the interval of $0.42$ to $0.62$ derived via bootstrap resampling.

• Determining Sample Size for a Desired Margin of Error:

  • Margin of Error Formula:     $ME = z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

  • As sample size $n$ increases, the margin of error $ME$ decreases.

  • Sample Size Formula:     $n = \left(\frac{z^*}{ME}\right)^2 \tilde{p}(1 - \tilde{p})$

  • Selection of $\tilde{p}$:

  • Use an available prior estimate or sample proportion (such as $\hat{p}$) if known.

  • If no prior estimate is available, adopt the conservative estimate $\tilde{p} = 0.5$, which maximizes the product $\tilde{p}(1 - \tilde{p}) = 0.25$ and guarantees the resulting sample size achieves the desired margin of error.

  • Rounding rule: Always round non-integer calculations up to the next highest integer to guarantee the target precision.

  • Worked Example: Quebec Sovereignty Poll:

  • In a prior poll of $800$ Quebec residents, $28\%$ favored secession from Canada ($\tilde{p} = 0.28$).

  • Target precision: Estimate the true proportion to within $\pm 1\%$ ($ME = 0.01$) with $99\%$ confidence ($z^* = 2.576$).

  • Calculation:       $n = \left(\frac{2.576}{0.01}\right)^2 (0.28)(1 - 0.28)$       $n = (257.6)^2 \cdot (0.28)(0.72) = 66357.76 \cdot 0.2016 = 13377.72$

  • Rounding up yields a required sample size of $n = 13,378\,\text{residents}$.

• Hypothesis Testing for a Single Proportion:

  • Null and Alternative Hypotheses:     $H_0: p = p_0 \quad \text{vs.} \quad H_a: p \ne p_0 \quad (\text{or } H_a: p > p_0, \; H_a: p < p_0)$     where $p_0$ is the hypothesized null parameter value.

  • Null Standard Error:

  • Because hypothesis testing assumes the null hypothesis $H_0$ is true, the standard error must be computed using the null parameter $p_0$ rather than the sample statistic $\hat{p}$:       $SE_0 = \sqrt{\frac{p_0(1 - p_0)}{n}}$

  • Standardized Test Statistic:     $Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$

  • Sample Size Conditions: To ensure validity of the normal distribution, check:     $n p_0 \ge 10 \quad \text{and} \quad n(1 - p_0) \ge 10$

  • Step-by-Step Procedure:

  1. State $H_0$ and $H_a$, define the population parameter $p$, and identify $\hat{p}$.

  2. Verify that $n p_0 \ge 10$ and $n(1 - p_0) \ge 10$.

  3. Calculate the test statistic $Z$.

  4. Find the p-value using the standard normal distribution ($N(0, 1)$) based on $H_a$.

  5. Compare the p-value to the significance level $\alpha$ and state the formal conclusion in context.

  • Worked Example: Autonomous vs. Human-Controlled Vehicles:

  • Context: In an October 2015 Nielsen survey, a random sample of $n = 340$ US middle school students were asked whether they would prefer a self-driving car or a car they control themselves. In the sample, $59\%$ ($\hat{p} = 0.59$) preferred driving themselves, while $41\%$ preferred self-driving vehicles. Test at the $\alpha = 0.05$ significance level whether more than $50\%$ of all US middle school students prefer driving themselves.

  • Hypotheses:       $H_0: p = 0.50 \quad \text{vs.} \quad H_a: p > 0.50$       where $p$ represents the true proportion of all US middle school students who prefer controlling their own car.

  • Condition Verification:       $n p_0 = 340(0.50) = 170 \ge 10$       $n(1 - p_0) = 340(1 - 0.50) = 170 \ge 10$       Conditions are verified.

  • Standardized Test Statistic Calculation:       $SE_0 = \sqrt{\frac{0.50(1 - 0.50)}{340}} = \sqrt{\frac{0.25}{340}} = \sqrt{0.00073529} \approx 0.027116$       $Z = \frac{\hat{p} - p_0}{SE_0} = \frac{0.59 - 0.50}{0.027116} = \frac{0.09}{0.027116} \approx 3.32$

  • p-value Determination:

    • For a right-tailed test, the p-value is the area to the right of $Z = 3.32$ in $N(0, 1)$.

    • From Table IV, the cumulative area to the left of $Z = 3.32$ is $0.9995$.       $\text{p-value} = 1 - 0.9995 = 0.0005$

  • Conclusion:

    • Because $p\text{-value} = 0.0005 < \alpha = 0.05$, reject $H_0$.

    • We have sufficient evidence to conclude that more than $50\%$ of US middle school students prefer a car that they control themselves.

Inference for a Difference in Proportions (Section 6.3)

• Context: Comparing Proportions from Two Distinct Groups:

  • Focuses on comparing proportions between two distinct, independent populations or treatment groups ($p_1$ versus $p_2$).

  • Distinction: This does not apply to two categories within the same single categorical sample (e.g., comparing the proportion voting for Candidate A versus Candidate B within a single poll).

• Sampling Distribution of the Difference in Proportions ($\hat{p}_1 - \hat{p}_2$):

  • When taking independent random samples of size $n_1$ and $n_2$ from populations with proportions $p_1$ and $p_2$:

  • Center: The distribution is centered at $p_1 - p_2$.

  • Standard Error Formula:       $SE = \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}}$

  • Normality Conditions: The sampling distribution is approximately normal if all four conditions hold:       $n_1 p_1 \ge 10, \quad n_1(1 - p_1) \ge 10, \quad n_2 p_2 \ge 10, \quad n_2(1 - p_2) \ge 10$

• Worked Example: Belief in "One True Love":

  • A study asked respondents whether they agree with the statement: "There is only one true love for each person."

  • Results by gender ($2625$ total participants):

  • Males: $372$ of $n_M = 1213$ agree.

  • Females: $363$ of $n_F = 1412$ agree.

  • Sample proportions:     $\hat{p}_M = \frac{372}{1213} \approx 0.307, \quad \hat{p}_F = \frac{363}{1412} \approx 0.257$

  • Point estimate of the difference:     $\hat{p}_M - \hat{p}_F = 0.307 - 0.257 = 0.05$

  • Condition checks:     $n_M \hat{p}_M = 1213(0.307) = 372 \ge 10$     $n_M(1 - \hat{p}_M) = 1213(1 - 0.307) = 841 \ge 10$     $n_F \hat{p}_F = 1412(0.257) = 363 \ge 10$     $n_F(1 - \hat{p}_F) = 1412(1 - 0.257) = 1049 \ge 10$

  • Standard error calculation:     $SE = \sqrt{\frac{0.307(1 - 0.307)}{1213} + \frac{0.257(1 - 0.257)}{1412}} = \sqrt{\frac{0.21275}{1213} + \frac{0.19095}{1412}}$     $SE = \sqrt{0.0001754 + 0.0001352} = \sqrt{0.0003106} \approx 0.018$

• Confidence Intervals for a Difference in Proportions ($p_1 - p_2$):

  • Standard Error using Sample Estimates:     $SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$

  • Normality Conditions: Verify that all four counts are at least $10$:     $n_1 \hat{p}_1 \ge 10, \quad n_1(1 - \hat{p}_1) \ge 10, \quad n_2 \hat{p}_2 \ge 10, \quad n_2(1 - \hat{p}_2) \ge 10$

  • Formula for Confidence Interval:     $(\hat{p}_1 - \hat{p}_2) \pm z^* \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$

• Comprehensive Worked Example: Crows Holding a Grudge:

  • Context: Biologists studying American crows trapped and tagged seven individual crows while wearing a caveman mask. A control group wore a neutral mask and did not trap any crows. More than two years later, at sites exceeding one kilometer from the original trapping location, crow reactions were recorded for both mask types. Scolding behaviors were directed at the caveman mask not only by the original tagged birds, but also by untagged crows and crows born after the trapping took place.

  • Data:

  • Caveman mask: Scolded in $158$ out of $n_C = 444$ encounters ($\hat{p}_C = \frac{158}{444} \approx 0.356$).

  • Neutral mask: Scolded in $109$ out of $n_N = 922$ encounters ($\hat{p}_N = \frac{109}{922} \approx 0.118$).

  • Condition Checks:     $n_C \hat{p}_C = 444(0.356) = 158 \ge 10$     $n_C(1 - \hat{p}_C) = 444(1 - 0.356) = 286 \ge 10$     $n_N \hat{p}_N = 922(0.118) = 109 \ge 10$     $n_N(1 - \hat{p}_N) = 922(1 - 0.118) = 813 \ge 10$     All four sample size criteria are satisfied.

  • Confidence Level and Critical Value: For a $90\%$ confidence level, $z^* = 1.645$.

  • Standard Error Calculation:     $SE = \sqrt{\frac{0.356(1 - 0.356)}{444} + \frac{0.118(1 - 0.118)}{922}} = \sqrt{\frac{0.356(0.644)}{444} + \frac{0.118(0.882)}{922}}$     $SE = \sqrt{\frac{0.22926}{444} + \frac{0.10408}{922}} = \sqrt{0.0005164 + 0.0001129} = \sqrt{0.0006293} \approx 0.025086$

  • Confidence Interval Calculation:     $\text{Point Estimate} = \hat{p}_C - \hat{p}_N = 0.356 - 0.118 = 0.238$     $ME = 1.645 \cdot 0.025086 \approx 0.04127$     $0.238 \pm 0.04127 = (0.19673, 0.27927) \approx (0.197, 0.279)$

  • Contextual Interpretation: We are $90\%$ confident that the true difference in the proportion of encounters resulting in scoldings between the caveman mask and the neutral mask ($p_C - p_N$) is between $0.197$ and $0.279$ (or between $19.7\%$ and $27.9\%$).

• Hypothesis Testing for a Difference in Proportions:

  • Hypotheses:     $H_0: p_1 - p_2 = 0 \quad (\text{i.e., } p_1 = p_2) \quad \text{vs.} \quad H_a: p_1 - p_2 \ne 0 \quad (\text{or } H_a: p_1 - p_2 > 0, \; H_a: p_1 - p_2 < 0)$

  • Null Parameter: $0$.

  • The Pooled Proportion ($\hat{p}$):

  • Definition: The combined sample proportion obtained by pooling the successes of both groups across the combined total sample size $n_1 + n_2$.

  • Rationale: Under the null hypothesis $H_0: p_1 = p_2$, both populations share an identical underlying proportion $p$. Pooling provides the best, most efficient estimate of this shared parameter.

  • Formula:       $\hat{p} = \frac{\text{total successes}}{\text{total sample size}} = \frac{x_1 + x_2}{n_1 + n_2}$

  • Introductory Example: If group 1 has $\hat{p}_1 = \frac{38}{50}$ and group 2 has $\hat{p}_2 = \frac{41}{50}$, the pooled proportion is:       $\hat{p} = \frac{38 + 41}{50 + 50} = \frac{79}{100} = 0.79$

  • Standard Error Under the Null Hypothesis ($SE_{\text{pool}}$):

  • Using the pooled estimate $\hat{p}$ for both groups:       $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n_1} + \frac{\hat{p}(1 - \hat{p})}{n_2}}$

  • Standardized Test Statistic:     $Z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_1} + \frac{\hat{p}(1 - \hat{p})}{n_2}}}$

  • p-value Computation: Provided the sample size conditions hold (either checking that observed counts in each cell exceed $10$, or that $n_1\hat{p}, n_1(1-\hat{p}), n_2\hat{p}, n_2(1-\hat{p}) \ge 10$), the p-value is calculated from $N(0, 1)$.

• Comprehensive Worked Example: "Split or Steal" Game Show Decision Analysis:

  • Context: On the British television show Golden Balls, the final round features two contestants deciding whether to "split" or "steal" a jackpot. Payoff matrix: if both split, they divide the money equally; if one splits and the other steals, the stealer takes the entire jackpot; if both steal, both leave with nothing. Researchers analyzed $287$ episodes, recording $574$ total individual decisions classified by contestant age.

  • Contingency Data Table:

  • Under 40: $187$ Split, $195$ Steal, Total $n_Y = 382$

  • Over 40: $116$ Split, $76$ Steal, Total $n_O = 192$

  • Combined Column Totals: $303$ Split, $271$ Steal, Total $n = 574$

  • Step 1: Hypotheses and Sample Proportions:

  • Let $p_Y$ be the proportion of younger players (under 40) who split, and $p_O$ be the proportion of older players (over 40) who split.

  • Hypotheses:       $H_0: p_Y = p_O \quad \text{vs.} \quad H_a: p_Y \ne p_O$

  • Individual sample proportions:       $\hat{p}_Y = \frac{187}{382} \approx 0.4895 \approx 0.490$       $\hat{p}_O = \frac{116}{192} \approx 0.6042 \approx 0.604$

  • Observed difference in sample proportions:       $\hat{p}_Y - \hat{p}_O = 0.4895 - 0.6042 = -0.1147$

  • Step 2: Pooled Proportion Calculation:     $\hat{p} = \frac{187 + 116}{382 + 192} = \frac{303}{574} \approx 0.52788 \approx 0.528$     $1 - \hat{p} = 1 - 0.52788 = 0.47212 \approx 0.472$

  • Step 3: Verification of Conditions:

  • The observed cell counts are $187, 195, 116, 76$, all of which substantially exceed $10$.

  • Expected counts under the pooled model:       $n_Y \hat{p} = 382(0.52788) \approx 201.7 \ge 10$       $n_Y(1 - \hat{p}) = 382(0.47212) \approx 180.3 \ge 10$       $n_O \hat{p} = 192(0.52788) \approx 101.4 \ge 10$       $n_O(1 - \hat{p}) = 192(0.47212) \approx 90.6 \ge 10$

  • Conditions for the normal approximation are satisfied.

  • Step 4: Standard Error and Test Statistic Calculation:

  • Pooled standard error calculation:       $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n_Y} + \frac{\hat{p}(1 - \hat{p})}{n_O}} = \sqrt{\frac{0.52788(0.47212)}{382} + \frac{0.52788(0.47212)}{192}}$       $SE = \sqrt{\frac{0.24922}{382} + \frac{0.24922}{192}} = \sqrt{0.00065241 + 0.00129803} = \sqrt{0.00195044} \approx 0.044164$

  • Standardized test statistic:       $Z = \frac{\hat{p}_Y - \hat{p}_O - 0}{SE} = \frac{-0.1147}{0.044164} \approx -2.60 \quad (\text{or } -2.58 \text{ using rounded inputs})$

  • Step 5: p-value Calculation and Decision:

  • For a two-tailed test, $\text{p-value} = 2 \times P(Z \le -2.58)$.

  • From Table IV, the area to the left of $Z = -2.58$ is $0.0049$.

  • Two-tailed p-value:       $\text{p-value} = 2 \times 0.0049 = 0.0098$       (Using $Z = -2.60$ gives $2 \times 0.0047 = 0.0094$)

  • Decision: Because $p\text{-value} \approx 0.010 < \alpha = 0.05$, reject $H_0$.

  • Contextual Conclusion: The sample evidence indicates a statistically significant difference between the proportion of younger contestants and older contestants who choose to split the jackpot on Golden Balls (older contestants split significantly more often than younger contestants).

Exam 3 Master Formula Sheet, Terminology, and Review Guide

• Comprehensive Definitions Checklist:

  • Density Curve: A theoretical model for the distribution of a continuous random variable having total area $1$ and where area over any interval equals the proportion of data in that interval.

  • Normal Distribution ($N(\mu, \sigma)$): A symmetric, bell-shaped continuous probability distribution with center $\mu$ and scale $\sigma$, where approximately $95\%$ of values fall within $(\mu - 2\sigma, \mu + 2\sigma)$.

  • Standard Normal Distribution ($N(0, 1)$): The standardized normal distribution having mean $\mu = 0$ and standard deviation $\sigma = 1$.

  • Standardized Test Statistic ($Z$):     $Z = \frac{\text{sample statistic} - \text{null parameter}}{SE}$     Measures the distance between the observed statistic and the hypothesized null value in units of standard error.

  • p-value: The probability, assuming the null hypothesis is true, of obtaining a sample statistic at least as extreme as the observed statistic, calculated as tail area(s) from a randomization or standard normal distribution.

  • Standard Error ($SE$): The standard deviation of a sampling distribution of a statistic.

  • Confidence Interval: An interval estimate for an unknown population parameter constructed as $\text{sample statistic} \pm z^* \cdot SE$.

  • Pooled Proportion ($\hat{p}$): The overall proportion of successes across two samples combined into a single pooled group of size $n_1 + n_2$.

• Master Formula Comparison Table:

• Sample Size Determination for a Desired Margin of Error:   $n = \left(\frac{z^*}{ME}\right)^2 \tilde{p}(1 - \tilde{p})$

  • Use prior estimate $\tilde{p} = \hat{p}$ if known; otherwise use conservative default $\tilde{p} = 0.5$.

  • Always round up to the next integer.

• Standard Critical Values Reference Table:

• Key Skills and Problem-Solving Strategy Checklist:

  • Estimate areas underneath density curves ensuring sum of all regions equals $1.0$.

  • Standardize and unstandardize normal random variables via $Z = \frac{X - \mu}{\sigma}$ and $X = \mu + Z\sigma$.

  • Calculate probabilities and percentiles utilizing cumulative standard normal tables (Table IV).

  • Identify the approximating normal model for a randomization distribution: $N(\text{null parameter}, SE)$.

  • Identify the approximating normal model for a bootstrap distribution: $N(\text{sample statistic}, SE)$.

  • Recognize when to substitute $\hat{p}$ versus the null value $p_0$ or pooled value $\hat{p}$ into the standard error formula.

  • Verify sample size criteria ($\ge 10$ successes and failures) before applying normal distribution procedures.