Statistical Inference About Means and Proportions for Two Populations

Learning Objectives

• Element 1 (LO 1): Describe and explain the distribution form and parameters of the sampling distribution of the difference between the sample means of two different populations.
• LO 2: Calculate and interpret the interval estimate at a given level of confidence for a difference between two population means when the standard deviations of the two populations are known.
• LO 3: Conduct a hypothesis test about the difference between two population means when the standard deviations of the two populations are known.
• LO 4: Calculate and interpret the interval estimate at a given level of confidence for a difference between two population means when the standard deviations of the two populations are unknown.
• LO 5: Conduct a hypothesis test about the difference between two population means when the standard deviations of the two populations are unknown.
• LO 6: Conduct a hypothesis test about the difference between two population means when using a matched sample design.
• LO 7: Calculate and interpret the interval estimate at a given level of confidence for a difference between two population means when using a matched sample design.
• LO 8: Describe and explain the distribution form and parameters of the sampling distribution of the difference between the sample proportions of two different populations.
• LO 9: Calculate and interpret the interval estimate at a given level of confidence for a difference between two population proportions.
• LO 10: Conduct a hypothesis test about the difference between two population proportions.

Overview of Inferences About Two Populations

• Statistical inference extends beyond single-sample analyses of a single population mean or proportion to comparative studies involving two distinct populations.
• The prime parameter of interest is either the difference between two population means, denoted as $\mu_1 - \mu_2$, or the difference between two population proportions, denoted as $p_1 - p_2$.
• Practical scenarios include:
  • Estimating the difference between the mean starting salary for a population of males and the mean starting salary for a population of females.
  • Testing for differences in quality between two suppliers by evaluating the proportion of defective parts produced by Supplier A versus Supplier B.
• Independent random samples are selected separately from each population: a random sample of size $n_1$ from Population 1 and an independent random sample of size $n_2$ from Population 2.

Inferences About Two Population Means: $\sigma_1$ and $\sigma_2$ Known

• Parameters and Independent Samples:
  • Let $\mu_1$ denote the mean of Population 1 and $\mu_2$ denote the mean of Population 2.
  • The population standard deviations $\sigma_1$ and $\sigma_2$ are assumed to be known prior to collecting the sample data.
  • Independent random samples of sizes $n_1$ and $n_2$ yield sample means $\bar{x}_1$ and $\bar{x}_2$.
• Point Estimator:
  • The point estimator of the difference between the two population means is the difference between the sample means:   $\bar{x}_1 - \bar{x}_2$
• Sampling Distribution of $\bar{x}_1 - \bar{x}_2$:
  • Mean: The expected value of the sampling distribution is:   $E(\bar{x}_1 - \bar{x}_2) = \mu_1 - \mu_2$
  • Standard Error (Formula 10.2):$\sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$
  • Distribution Form: The sampling distribution of $\bar{x}_1 - \bar{x}_2$ is normally distributed if both underlying populations follow a normal distribution, or if the sample sizes are sufficiently large ($n_1 \ge 30$ and $n_2 \ge 30$) that the Central Limit Theorem applies.
• Interval Estimation Form:
  • General interval estimation formula: $\text{Point Estimate} \pm \text{Margin of Error}$.
  • Margin of Error (Formula 10.3):$\text{Margin of Error} = z_{\alpha/2}\sigma_{\bar{x}_1 - \bar{x}_2} = z_{\alpha/2}\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$
  • Interval Estimate (Formula 10.4):$(\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$
• Home Styles Furniture Store Application (Interval Estimation):
  • Home Styles operates two furniture retail stores in Buffalo, New York: one located in the inner city (Population 1) and one located in a suburban shopping center (Population 2).
  • Management seeks to analyze demographic differences in customer age.
  • Historical demographic data establish known standard deviations: $\sigma_1 = 9\,\text{years}$ and $\sigma_2 = 10\,\text{years}$.
  • Sample data collected:
  • Inner city: $n_1 = 36$, $\bar{x}_1 = 40\,\text{years}$
  • Suburban: $n_2 = 49$, $\bar{x}_2 = 35\,\text{years}$
  • Point estimate computation:   $\bar{x}_1 - \bar{x}_2 = 40 - 35 = 5\,\text{years}$
  • At a $95\%$ confidence level ($\alpha = 0.05$), the critical value is $z_{\alpha/2} = z_{0.025} = 1.96$.
  • Margin of error calculation:   $1.96\sqrt{\frac{9^2}{36} + \frac{10^2}{49}} = 1.96\sqrt{\frac{81}{36} + \frac{100}{49}} = 4.06\,\text{years}$
  • The $95\%$ confidence interval is:   $5 \pm 4.06\,\text{years}$
  • Lower limit: $5 - 4.06 = 0.94\,\text{years}$
  • Upper limit: $5 + 4.06 = 9.06\,\text{years}$
• Excel Worksheet Implementation for Interval Estimation (homestyle_mod):
  • Open data file homestyle_mod. Column A stores the age data for $36$ inner city customers; Column B stores the age data for $49$ suburban customers.
  • Descriptive statistics: entered into cells E5:F6.
  • Known population standard deviations: entered into cells E8 and F8.
  • Cell E9 standard error formula:   =SQRT(E8^2/E5 + F8^2/F5)
  • Confidence coefficient of $0.95$ is entered into cell E11; level of significance $\alpha = 1 - \text{confidence coefficient}$ is computed in cell E12.
  • Cell E13 computes the critical value using the function NORMSINV.
  • Cell E14 calculates the margin of error by multiplying cell E13 by cell E9.
  • Cell E16 calculates the point estimate ($\bar{x}_1 - \bar{x}_2$).
  • Cell E17 computes the lower limit ($0.94$); cell E18 computes the upper limit ($9.06$).
  • Template generalization formulas: dynamic ranges in cells E5:F6 replace fixed ranges:
  • Cell E5: =COUNT(A:A)
  • Cell F5: =COUNT(B:B)
  • Cell E6: =AVERAGE(A:A)
  • Cell F6: =AVERAGE(B:B)

Hypothesis Tests About $\mu_1 - \mu_2$: $\sigma_1$ and $\sigma_2$ Known

• Hypothesis Testing Formulations:
  • Let $d_0$ represent the hypothesized difference between $\mu_1$ and $\mu_2$.
  • Three structural forms are available:
  • Upper-tail test:     $H_0: \mu_1 - \mu_2 \le d_0 \quad \text{versus} \quad H_a: \mu_1 - \mu_2 > d_0$
  • Lower-tail test:     $H_0: \mu_1 - \mu_2 \ge d_0 \quad \text{versus} \quad H_a: \mu_1 - \mu_2 < d_0$
  • Two-tail test:     $H_0: \mu_1 - \mu_2 = d_0 \quad \text{versus} \quad H_a: \mu_1 - \mu_2 \ne d_0$
  • When testing for equality, $d_0 = 0$, simplifying the null hypothesis to $H_0: \mu_1 - \mu_2 = 0$ ($\mu_1 = \mu_2$).
• Test Statistic (Formula 10.5):$z = \frac{(\bar{x}_1 - \bar{x}_2) - d_0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}$
• Training Center Quality Evaluation Application:
  • Individuals trained at Center A (Population 1) and Center B (Population 2) take a standardized exam to determine training quality differences.
  • Parameter definition: $\mu_1$ is the mean exam score for Center A; $\mu_2$ is the mean exam score for Center B.
  • Hypotheses:   $H_0: \mu_1 - \mu_2 = 0 \quad \text{versus} \quad H_a: \mu_1 - \mu_2 \ne 0$
  • Historical standardized exams demonstrate score standard deviations close to $10\,\text{points}$, setting known parameters: $\sigma_1 = 10$ and $\sigma_2 = 10$.
  • Specified level of significance: $\alpha = 0.05$.
  • Sample data:
  • Center A: $n_1 = 30$, $\bar{x}_1 = 82$
  • Center B: $n_2 = 40$, $\bar{x}_2 = 78$
  • Test statistic calculation:   $z = \frac{(82 - 78) - 0}{\sqrt{\frac{10^2}{30} + \frac{10^2}{40}}} = \frac{4}{\sqrt{\frac{100}{30} + \frac{100}{40}}} = 1.66$
  • $p$-Value Approach:
  • Standard normal distribution cumulative area for $z = 1.66$ is $0.9515$.
  • Upper tail area: $1.0000 - 0.9515 = 0.0485$.
  • Two-tailed $p$-value:     $p\text{-value} = 2(0.0485) = 0.0970$
  • Rejection rule: Reject $H_0$ if $p\text{-value} \le \alpha$.
  • Decision: Because $p\text{-value} = 0.0970 > 0.05$, do not reject $H_0$. There is insufficient evidence to conclude the training centers differ in quality.
  • Critical Value Approach:
  • For $\alpha = 0.05$, $z_{\alpha/2} = z_{0.025} = 1.96$.
  • Rejection rule: Reject $H_0$ if $z \le -1.96$ or $z \ge 1.96$.
  • Decision: Since $z = 1.66$ falls inside $(-1.96, 1.96)$, do not reject $H_0$.
• Excel Data Analysis Tool for Hypothesis Testing (examscores):
  • Tool: z-Test: Two-Sample for Means.
  • Data file: examscores. Column A contains $30$ scores for Center A; Column B contains $40$ scores for Center B.
  • Variance inputs: $\sigma_1^2 = 10^2 = 100$ and $\sigma_2^2 = 10^2 = 100$.
  • Procedure:
  1. Click Data tab on the ribbon.
  2. Click Data Analysis in the Analyze group.
  3. Select z-Test: Two-Sample for Means.
  4. Enter A1:A31 in Variable 1 Range.
  5. Enter B1:B41 in Variable 2 Range.
  6. Enter 0 in Hypothesized Mean Difference.
  7. Enter 100 in Variable 1 Variance (known).
  8. Enter 100 in Variable 2 Variance (known).
  9. Check Labels.
  10. Enter 0.05 in Alpha.
  11. Select Output Range and specify D4.
  12. Click OK.
  • Output interpretation:
  • Cell E11: Test statistic $z = 1.6562$.
  • Cell E12: One-tail $p$-value labeled $P(Z \le z)\text{ one-tail}$.
  • Cell E14: Two-tail $p$-value labeled $P(Z \le z)\text{ two-tail} = 0.0977$.
  • Since $0.0977 > 0.05$, do not reject $H_0$.
• Sample Size Considerations:
  • Random samples with $n_1 \ge 30$ and $n_2 \ge 30$ are adequate in most practical applications.
  • When either sample size is less than $30$, the underlying population distributions must be reasonably assumed to be approximately normal.

Inferences About Two Population Means: $\sigma_1$ and $\sigma_2$ Unknown

• Foundational Conditions:
  • When $\sigma_1$ and $\sigma_2$ are unknown, the sample standard deviations $s_1$ and $s_2$ serve as point estimators.
  • The interval estimation and hypothesis testing procedures are based on the $t$-distribution rather than the standard normal distribution.
• Interval Estimation Form (Formula 10.6):$(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$
• Degrees of Freedom Formula (Formula 10.7):$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1 - 1}\left(\frac{s_1^2}{n_1}\right)^2 + \frac{1}{n_2 - 1}\left(\frac{s_2^2}{n_2}\right)^2}$
  • When $df$ is non-integer, round down to the nearest integer to ensure a larger critical value and a more conservative estimate.
• Clearwater National Bank Application (Interval Estimation):
  • The bank investigates checking account balances across two branches: Cherry Grove (Population 1) and Beechmont (Population 2).
  • Sample parameters:
  • Cherry Grove: $n_1 = 28$, $\bar{x}_1 = \$1025$, $s_1 = 150$
  • Beechmont: $n_2 = 22$, $\bar{x}_2 = \$910$, $s_2 = 125$
  • Degrees of freedom calculation:   $df = \frac{\left(\frac{150^2}{28} + \frac{125^2}{22}\right)^2}{\frac{1}{28 - 1}\left(\frac{150^2}{28}\right)^2 + \frac{1}{22 - 1}\left(\frac{125^2}{22}\right)^2} = 47.8$
  • Rounded down to integer: $df = 47$.
  • At $95\%$ confidence, the critical value for $47\,df$ is $t_{0.025} = 2.012$.
  • Point estimate:   $\bar{x}_1 - \bar{x}_2 = 1025 - 910 = \$115$
  • Margin of error calculation:   $2.012\sqrt{\frac{150^2}{28} + \frac{125^2}{22}} = 2.012(38.96) = \$78$
  • Interval bounds: $115 \pm 78$
  • Lower limit: $115 - 78 = \$37$
  • Upper limit: $115 + 78 = \$193$
• Excel Worksheet Implementation (Clearwater National Bank / Figure 10.5):
  • Raw data entered in Column A (Cherry Grove) and Column B (Beechmont).
  • Dynamic formulas for summary statistics:
  • Sample sizes: E5 = COUNT(A:A), F5 = COUNT(B:B)
  • Sample means: E6 = AVERAGE(A:A), F6 = AVERAGE(B:B)
  • Sample standard deviations: E7 = STDEV.S(A:A), F7 = STDEV.S(B:B)
  • Confidence coefficient of $0.95$ is entered into cell E12.
  • Point estimate in cell E18: $115$.
  • Lower limit in cell E19: $37$.
  • Upper limit in cell E20: $193$.
• Hypothesis Testing Formulations ($\sigma_1, \sigma_2$ Unknown):
  • Test Statistic (Formula 10.8):$t = \frac{(\bar{x}_1 - \bar{x}_2) - d_0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$
  • Degrees of freedom determined via Formula 10.7.
• Software Evaluation Study Application (Hypothesis Testing):
  • A study evaluates whether a new software package reduces information systems project completion time relative to current technology.
  • Populations: Systems analysts using current technology (Population 1) versus analysts using the new software package (Population 2).
  • Means: $\mu_1$ is the mean completion time for current technology; $\mu_2$ is the mean completion time for the new package.
  • Research hypothesis posits that the new software will yield a shorter completion time ($\mu_2 < \mu_1$, which means $\mu_1 - \mu_2 > 0$).
  • Stated hypotheses:   $H_0: \mu_1 - \mu_2 \le 0 \quad \text{versus} \quad H_a: \mu_1 - \mu_2 > 0$
  • Significance level: $\alpha = 0.05$.
  • Sample data (from $24$ analysts, $12$ per group):
  • Current technology: $n_1 = 12$, $\bar{x}_1 = 325\,\text{hours}$, $s_1 = 40$
  • New software: $n_2 = 12$, $\bar{x}_2 = 286\,\text{hours}$, $s_2 = 44$
  • Test statistic computation:   $t = \frac{(325 - 286) - 0}{\sqrt{\frac{40^2}{12} + \frac{44^2}{12}}} = \frac{39}{\sqrt{\frac{1600}{12} + \frac{1936}{12}}} = 2.27$
  • Degrees of freedom computation:   $df = \frac{\left(\frac{40^2}{12} + \frac{44^2}{12}\right)^2}{\frac{1}{12 - 1}\left(\frac{40^2}{12}\right)^2 + \frac{1}{12 - 1}\left(\frac{44^2}{12}\right)^2} = 21.8$
  • Rounding down gives $df = 21$.
  • From the $t$-distribution table with $21\,df$, the value $t = 2.27$ places the upper-tail area ($p$-value) between $0.01$ and $0.025$.
  • Because the $p\text{-value} < 0.05$, reject $H_0$.
  • Conclusion: The research supports the finding that the new software package provides a shorter mean completion time (noted in text phrasing as supporting a slower completion time differential relative to the null baseline).
• Excel Data Analysis Tool (softwareredist / Figures 10.6 and 10.7):
  • Tool: t-Test: Two-Sample Assuming Unequal Variances.
  • File: softwareredist.
  • Setup: Variable 1 Range = A1:A13, Variable 2 Range = B1:B13, Hypothesized Mean Difference = 0, Labels checked, Alpha = 0.05, Output Range = D1.
  • Tool output:
  • Cell E9: Test statistic $t = 2.2721$.
  • Cell E10: One-tail $p$-value $P(T \le t)\text{ one-tail} = 0.0166$.
  • Cell E12: Two-tail $p$-value $P(T \le t)\text{ two-tail}$.
  • Decision: Because $p\text{-value} = 0.0166 < 0.05$, reject $H_0$.
• Robustness and Sample Size Guidelines:
  • The $t$-distribution procedure is robust to non-normality.
  • Total sample size $n_1 + n_2 \ge 20$ with equal or nearly equal sample sizes provides reliable results even with non-normal populations.
  • Larger sample sizes are required if data distributions exhibit substantial skewness or outliers.
  • Sample sizes smaller than $20$ require verified approximate normality of the parent populations.

Inferences About Two Populations: Matched Samples

• Study Design Comparison:
  • Independent Sample Design: Separate, unrelated random samples are assigned to each treatment/population.
  • Matched Sample Design: Each sampled element provides a paired pair of observations, receiving both treatments in a randomized order.
  • Primary Advantage: Testing treatments under identical conditions (using the exact same subjects) eliminates between-subject variability, reducing sampling error and increasing the precision of the estimator.
• Hypothesis Formulation for Matched Samples:
  • Let $d_i$ denote the observed difference for pair $i$.
  • Let $\mu_d$ denote the mean difference for the underlying population.
  • Null and alternative hypotheses:   $H_0: \mu_d = 0 \quad \text{versus} \quad H_a: \mu_d \ne 0$
• Manufacturing Task Completion Application:
  • A manufacturing company tests two production methods across a random sample of $n = 6$ workers.
  • Observed paired differences ($d_i = \text{Method 1} - \text{Method 2}$): $0.6$, $-0.2$, $0.5$, $0.3$, $0.0$, $0.6$.
  • Sample mean difference:   $\bar{d} = \frac{\sum d_i}{n} = \frac{1.8}{6} = 0.30$
  • Sample standard deviation of differences:   $s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}} = \sqrt{\frac{0.56}{5}} = 0.335$
  • Assumption: The population of differences is normally distributed.
  • Test Statistic (Formula 10.9):$t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}$
  • Degrees of freedom: $df = n - 1 = 6 - 1 = 5$.
  • Calculation:   $t = \frac{0.30 - 0}{0.335 / \sqrt{6}} = \frac{0.30}{0.1368} = 2.20$
  • Table lookup for $5\,df$ shows upper-tail area between $0.025$ and $0.05$.
  • For a two-tailed test, doubling the area yields a $p$-value between $0.05$ and $0.10$.
  • Decision: Since $p\text{-value} > 0.05$, do not reject $H_0$. Exact Excel $p\text{-value} = 0.0795$.
• Interval Estimation for Matched Samples:
  • General form: $\bar{d} \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}$.
  • Critical value for $95\%$ confidence with $5\,df$ is $t_{0.025} = 2.571$.
  • Calculation:   $0.30 \pm 2.571\left(\frac{0.335}{\sqrt{6}}\right) = 0.30 \pm 0.35\,\text{minutes}$
  • Lower limit: $0.30 - 0.35 = -0.05\,\text{minutes}$
  • Upper limit: $0.30 + 0.35 = 0.65\,\text{minutes}$
• Excel Data Analysis Tool (match / Figures 10.8 and 10.9):
  • Tool: t-Test: Paired Two Sample for Means.
  • File: match. Column A contains worker IDs, Column B contains Method 1 times, Column C contains Method 2 times.
  • Setup: Variable 1 Range = B1:B7, Variable 2 Range = C1:C7, Hypothesized Mean Difference = 0, Labels checked, Alpha = 0.05, Output Range = E1.
  • Output interpretation:
  • Cell F11: One-tail $p$-value $P(T \le t)\text{ one-tail}$.
  • Cell F13: Two-tail $p$-value $P(T \le t)\text{ two-tail} = 0.0795$.
  • Decision: Because $p\text{-value} = 0.0795 > 0.05$, do not reject $H_0$.

Inferences About Two Population Proportions: $p_1 - p_2$

• Parameters and Estimators:
  • Let $p_1$ denote the proportion in Population 1 and $p_2$ denote the proportion in Population 2.
  • Samples of size $n_1$ and $n_2$ yield sample proportions $\bar{p}_1$ and $\bar{p}_2$.
  • Point Estimator (Formula 10.10):$\bar{p}_1 - \bar{p}_2$
  • Standard Error (Formula 10.11):$\sigma_{\bar{p}_1 - \bar{p}_2} = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$
  • Normality Assumption: The sampling distribution of $\bar{p}_1 - \bar{p}_2$ is approximately normal if all four sample condition thresholds are met:   $n_1 p_1 \ge 5, \quad n_1(1-p_1) \ge 5, \quad n_2 p_2 \ge 5, \quad n_2(1-p_2) \ge 5$
• Interval Estimation:
  • Population proportions are unknown; sample proportions are substituted into the standard error formula.
  • Margin of Error (Formula 10.12):$\text{Margin of Error} = z_{\alpha/2}\sqrt{\frac{\bar{p}_1(1-\bar{p}_1)}{n_1} + \frac{\bar{p}_2(1-\bar{p}_2)}{n_2}}$
  • Confidence Interval (Formula 10.13):$(\bar{p}_1 - \bar{p}_2) \pm z_{\alpha/2}\sqrt{\frac{\bar{p}_1(1-\bar{p}_1)}{n_1} + \frac{\bar{p}_2(1-\bar{p}_2)}{n_2}}$
• Tax Preparation Firm Application (Interval Estimation):
  • The firm evaluates erroneous returns across Office 1 (Population 1) and Office 2 (Population 2).
  • Sample data:
  • Office 1: $n_1 = 250$, erroneous returns = $35$$\bar{p}_1 = \frac{35}{250} = 0.14$
  • Office 2: $n_2 = 300$, erroneous returns = $27$$\bar{p}_2 = \frac{27}{300} = 0.09$
  • Point estimate:   $\bar{p}_1 - \bar{p}_2 = 0.14 - 0.09 = 0.05$
  • At $90\%$ confidence ($\alpha = 0.10$), critical value $z_{\alpha/2} = z_{0.05} = 1.645$.
  • Margin of error calculation:   $1.645\sqrt{\frac{0.14(1 - 0.14)}{250} + \frac{0.09(1 - 0.09)}{300}} = 1.645\sqrt{\frac{0.1204}{250} + \frac{0.0819}{300}} = 0.045$
  • Interval estimate bounds: $0.05 \pm 0.045$
  • Lower limit: $0.05 - 0.045 = 0.005$
  • Upper limit: $0.05 + 0.045 = 0.095$
• Excel Worksheet Implementation for Interval Estimation (tax_mod / Figure 10.10):
  • File: tax_mod. Columns A and B contain Yes or No error indicators.
  • Sample size: Cells E5 and F5 use COUNT ($250$ and $300$).
  • Response of interest: Type Yes into cells E6 and F6.
  • Error counts: Cells E7 and F7 use COUNTIF.
  • Sample proportions: Computed in cells E8 and F8.
  • Confidence coefficient of $0.90$ entered into cell E10; significance level $\alpha = 0.10$ computed in cell E11.
  • Critical value $z$ computed in cell E12 using NORMSINV.
  • Estimated standard error computed in cell E14.
  • Margin of error computed in cell E15.
  • Point estimate computed in cell E17 ($0.05$).
  • Limits: Lower limit in cell E18 ($0.0048$); Upper limit in cell E19 ($0.0952$).

Hypothesis Tests About $p_1 - p_2$

• Formulation of Hypotheses (Testing No Difference):
  • Upper-tail test:   $H_0: p_1 - p_2 \le 0 \quad \text{versus} \quad H_a: p_1 - p_2 > 0$
  • Lower-tail test:   $H_0: p_1 - p_2 \ge 0 \quad \text{versus} \quad H_a: p_1 - p_2 < 0$
  • Two-tail test:   $H_0: p_1 - p_2 = 0 \quad \text{versus} \quad H_a: p_1 - p_2 \ne 0$
• Pooled Estimator of $p$:
  • Under the assumption that $H_0$ is true as an equality, $p_1 = p_2 = p$.
  • Standard error under $H_0$ (Formula 10.14):   $\sigma_{\bar{p}_1 - \bar{p}_2} = \sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}$
  • Because $p$ is unknown, sample proportions are combined into a pooled estimator (Formula 10.15):   $\bar{p} = \frac{n_1\bar{p}_1 + n_2\bar{p}_2}{n_1 + n_2}$
• Test Statistic (Formula 10.16):$z = \frac{\bar{p}_1 - \bar{p}_2}{\sqrt{\bar{p}(1-\bar{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$
  • Applicable when sample sizes satisfy: $n_1 p_1 \ge 5$, $n_1(1-p_1) \ge 5$, $n_2 p_2 \ge 5$, and $n_2(1-p_2) \ge 5$.
• Tax Preparation Firm Application (Hypothesis Testing):
  • Test whether error rates differ between Office 1 and Office 2 at $\alpha = 0.10$.
  • Hypotheses:   $H_0: p_1 - p_2 = 0 \quad \text{versus} \quad H_a: p_1 - p_2 \ne 0$
  • Data: $n_1 = 250$, $\bar{p}_1 = 0.14$; $n_2 = 300$, $\bar{p}_2 = 0.09$.
  • Pooled estimator computation:   $\bar{p} = \frac{250(0.14) + 300(0.09)}{250 + 300} = \frac{35 + 27}{550} = \frac{62}{550} = 0.1127$
  • Test statistic calculation:   $z = \frac{0.14 - 0.09}{\sqrt{0.1127(1 - 0.1127)\left(\frac{1}{250} + \frac{1}{300}\right)}} = \frac{0.05}{\sqrt{0.1000(0.004 + 0.003333)}} = 1.85$
  • $p$-Value Calculation:
  • Cumulative normal area for $z = 1.85$ is $0.9678$.
  • Upper tail area: $1.0000 - 0.9678 = 0.0322$.
  • Two-tail $p$-value:     $p\text{-value} = 2(0.0322) = 0.0644$
  • Decision: Because $p\text{-value} = 0.0644 < 0.10$, reject $H_0$. Error rates differ significantly between the two offices.
• Excel Worksheet Implementation for Hypothesis Testing (tax_mod / Figure 10.11):
  • Hypothesized difference entered in cell E10 ($0$).
  • Point estimate in cell E11 ($0.05$).
  • Pooled estimate $\bar{p}$ computed in cell E13 ($0.1127$).
  • Standard error estimate computed in cell E14 via Formula 10.14.
  • Test statistic computed in cell E15 via formula:   =(E11 - E10)/E14   (yields $z = 1.8462$).
  • Cumulative probabilities via NORM.S.DIST:
  • Cell E17: Lower-tail $p$-value.
  • Cell E18: Upper-tail $p$-value.
  • Cell E19: Two-tail $p$-value computed as twice the minimum of the one-tailed values ($0.0649$).
  • Decision: Reject $H_0$ since $0.0649 < 0.10$.

Case Problem: PAR Incorporated

• Business Context:
  • PAR Incorporated is a manufacturer of golf equipment seeking to increase market share by developing a cut-resistant, durable golf ball coating.
  • Research concerns center on whether the new coating adversely affects driving distance.
  • Driving distance for the new cut-resistant ball must remain comparable to the current ball model.
• Testing Methodology:
  • A mechanical hitting machine was used to hit balls of both models, eliminating human swinging variation so that distance differences reflect true product differences.
  • Driving distances were measured to the nearest yard and cataloged in file LO four five.
• Managerial Report Requirements:
  1. Formulate and justify a hypothesis test comparing the driving distances of current and new balls.
  2. Analyze data, report the test statistic, compute the $p$-value, state the statistical conclusion, and provide managerial recommendations.
  3. Provide descriptive statistical summaries of driving distance data for each ball model.
  4. Calculate the $95\%$ confidence interval for the population mean driving distance of each model, as well as the $95\%$ confidence interval for the difference between the two population means.
  5. Evaluate and discuss whether larger sample sizes and additional testing are required.

Appendix: Inferences About Two Populations with R

• Difference Between Two Population Means: $\sigma_1$ and $\sigma_2$ Known:
  • Script: Dipmaid's nodes.R.
  • Context: Evaluating Center A ($\mu_1$) versus Center B ($\mu_2$) standardized exam scores with known parameters $\sigma_1 = 10$ and $\sigma_2 = 10$.
  • Working file: examscores_r.csv.
  • Procedural steps in RStudio console pane:
  • Step 1: Set the working directory to the folder containing examscores_r.csv.
  • Step 2: Read the CSV file into a data frame.
  • Step 3: Use subset to create a data frame containing only the variable Center.A.
  • Step 4: Use na.omit to remove missing values from center_a_df.
  • Step 5: Use nrow to calculate the sample size for Center A.
  • Step 6: Create a numeric vector of scores for Center A.
  • Step 7: Use mean to compute the sample mean for Center A.
  • Step 8: Repeat Steps 3 through 7 for Center B.
  • Step 9: Calculate the standard error of the difference between sample means.
  • Step 10: Calculate the test statistic $z$.
  • Step 11: Use pnorm to calculate the $p$-value.
  • Statistical results: Test statistic $z = 1.656$, two-tailed $p\text{-value} = 0.097$.
  • One-tailed modifications:
  • Lower-tail ($H_a: \mu_1 - \mu_2 < 0$): adjust Step 11 to calculate lower-tail area directly.
  • Upper-tail ($H_a: \mu_1 - \mu_2 > 0$): adjust Step 11 to calculate upper-tail area directly.
• Difference Between Two Population Means: $\sigma_1$ and $\sigma_2$ Unknown:
  • Script: Difmeid's unknowns.R.
  • Context: Clearwater National Bank checking balances comparing Cherry Grove ($n_1 = 28$) and Beechmont ($n_2 = 22$).
  • Working file: checkscached_r.csv.
  • Procedural steps:
  • Step 1: Set working directory to the location of checkscached_r.csv.
  • Step 2: Read the CSV file into a data frame.
  • Step 3: Create a vector of balances for Beechmont.
  • Step 4: Create a vector of balances for Cherry Grove.
  • Step 5: Use the R function t.test to execute the two-sample hypothesis test.
  • Statistical results: Test statistic $t = 2.9546$, two-tailed $p\text{-value} = 0.0048$.
  • Alternative directions (lower-tail or upper-tail) are implemented by modifying the alternative argument in t.test in Step 5.
• Difference Between Two Population Means: Matched Samples:
  • Script: DIPMANES match.R.
  • Context: Production completion times across $n = 6$ workers for Method 1 and Method 2.
  • Working file: matched_r.csv.
  • Procedural steps:
  • Step 1: Set working directory to the location of matched_r.csv.
  • Step 2: Read the CSV file into a data frame.
  • Step 3: Create a vector of Method 1 completion times.
  • Step 4: Create a vector of Method 2 completion times.
  • Step 5: Use the R function t.test with the argument paired = TRUE to perform the matched-sample hypothesis test.

Summary of Formulas

• Formula 10.1 — Point Estimator of the Difference Between Two Population Means:$\bar{x}_1 - \bar{x}_2$
• Formula 10.2 — Standard Error of $\bar{x}_1 - \bar{x}_2$ ($\sigma_1, \sigma_2$ Known):$\sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$
• Formula 10.3 — Margin of Error ($\sigma_1, \sigma_2$ Known):$z_{\alpha/2}\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$
• Formula 10.4 — Interval Estimate of $\mu_1 - \mu_2$ ($\sigma_1, \sigma_2$ Known):$(\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$
• Formula 10.5 — Test Statistic for $\mu_1 - \mu_2$ ($\sigma_1, \sigma_2$ Known):$z = \frac{(\bar{x}_1 - \bar{x}_2) - d_0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}$
• Formula 10.6 — Interval Estimate of $\mu_1 - \mu_2$ ($\sigma_1, \sigma_2$ Unknown):$(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$
• Formula 10.7 — Degrees of Freedom for Two Independent Random Samples ($t$-Distribution):$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1 - 1}\left(\frac{s_1^2}{n_1}\right)^2 + \frac{1}{n_2 - 1}\left(\frac{s_2^2}{n_2}\right)^2}$
• Formula 10.8 — Test Statistic for $\mu_1 - \mu_2$ ($\sigma_1, \sigma_2$ Unknown):$t = \frac{(\bar{x}_1 - \bar{x}_2) - d_0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$
• Formula 10.9 — Test Statistic for Matched Samples:$t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}}$
• Formula 10.10 — Point Estimator of the Difference Between Two Population Proportions:$\bar{p}_1 - \bar{p}_2$
• Formula 10.11 — Standard Error of $\bar{p}_1 - \bar{p}_2$:$\sigma_{\bar{p}_1 - \bar{p}_2} = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$
• Formula 10.12 — Margin of Error for $p_1 - p_2$:$z_{\alpha/2}\sqrt{\frac{\bar{p}_1(1-\bar{p}_1)}{n_1} + \frac{\bar{p}_2(1-\bar{p}_2)}{n_2}}$
• Formula 10.13 — Interval Estimate of $p_1 - p_2$:$(\bar{p}_1 - \bar{p}_2) \pm z_{\alpha/2}\sqrt{\frac{\bar{p}_1(1-\bar{p}_1)}{n_1} + \frac{\bar{p}_2(1-\bar{p}_2)}{n_2}}$
• Formula 10.14 — Standard Error of $\bar{p}_1 - \bar{p}_2$ When $p_1 = p_2 = p$:$\sigma_{\bar{p}_1 - \bar{p}_2} = \sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}$
• Formula 10.15 — Pooled Estimator of $p$:$\bar{p} = \frac{n_1\bar{p}_1 + n_2\bar{p}_2}{n_1 + n_2}$
• Formula 10.16 — Test Statistic for $p_1 - p_2$:$z = \frac{\bar{p}_1 - \bar{p}_2}{\sqrt{\bar{p}(1-\bar{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$