One-Sample Hypothesis Testing: Means, Proportions, Error Rates, and Practical Decision-Making

Fundamentals of One-Sample Hypothesis Testing

• Core Concepts and Purpose:

  • Hypothesis testing provides a structured statistical framework to evaluate whether empirical sample data provides sufficient evidence to reject a predetermined baseline claim about a population parameter.
  • Unlike confidence intervals—which estimate a range of plausible values for a population parameter (such as the population mean $\mu$) at a specified confidence level percentage—hypothesis tests formally evaluate the viability of a specific hypothesized parameter value.
  • Hypothesis testing and confidence intervals rely on the same foundational probability distributions and test statistics ($z$-statistics and $t$-statistics).

• The Three Primary Cases:

  • Case 1 (Mean with Known Population Standard Deviation): Used when testing a population mean $\mu$ and the population standard deviation $\sigma$ is known. Employs the standard normal distribution ($z$-test statistic).
  • Case 2 (Mean with Unknown Population Standard Deviation): Used when testing a population mean $\mu$ and $\sigma$ is unknown. The sample standard deviation $s$ is substituted, requiring the Student's $t$-distribution ($t$-test statistic) with degrees of freedom $df = n - 1$.
  • Case 3 (Proportions): Used when analyzing categorical, binary data (success/failure or yes/no outcomes) to test a population proportion $p$. Evaluated using a $z$-test statistic.

• Formulating Hypotheses:

  • Hypothesis tests always operate on a paired set of competing statements:
    • Null Hypothesis ($H_0$): The formal baseline claim asserting that no effect, no change, or no difference exists. It anchors the distribution to a specific hypothesized parameter value (e.g., $H_0: \mu = \mu_0$ or $H_0: p = p_0$).
    • Alternative Hypothesis ($H_A$ or $H_a$): The investigative claim representing the specific direction or departure being tested based on the research objective.
  • Directionality of Tests:
    • Left-Tailed Test (Less-than): The alternative hypothesis specifies that the parameter is strictly less than the null value ($H_A: \mu < \mu_0$). The critical region is in the extreme left tail.
    • Right-Tailed Test (Greater-than): The alternative hypothesis specifies that the parameter is strictly greater than the null value ($H_A: \mu > \mu_0$). The critical region is in the extreme right tail.
    • Two-Tailed Test (Not-equal): The alternative hypothesis specifies that the parameter is simply not equal to the null value ($H_A: \mu \neq \mu_0$). The critical region is split equally between both extremes of the distribution.

• Significance Level ($\alpha$) and Error Control:

  • The significance level $\alpha$ represents the threshold probability of rejecting the null hypothesis when it is actually true (a Type I error). It must be selected a priori (before examining the data or performing calculations).
  • $\alpha = 0.05$ serves as the conventional rule-of-thumb default across general business and academic research.
  • Selecting $\alpha$ is inherently context-dependent and dictated by the consequences of making a false alarm:
    • A higher $\alpha$ (e.g., $\alpha = 0.10$ or greater) may be tolerated in low-stakes exploratory analyses, such as testing the behavioral probability of users logging into a social media platform on a Sunday afternoon.
    • A substantially lower $\alpha$ (e.g., $\alpha = 0.01$, $\alpha = 0.001$, or lower) is mandatory in high-stakes, safety-critical domains, such as testing the structural integrity or failure rates of an O-ring on a human-crewed rocket booster.
  • StatCrunch sets $\alpha = 0.05$ as its automatic default significance level.

One-Sample $t$-Test for the Mean

• Theoretical Framework and Test Statistic:

  • When the population standard deviation $\sigma$ is unknown, the sample standard deviation $s$ is used to estimate the dispersion of the sampling distribution.
  • The test statistic is calculated as a standardized ratio measuring the distance between the observed sample mean $\bar{x}$ and the hypothesized population mean $\mu_0$, expressed in units of estimated standard errors:         $t = \frac{\bar{x} - \mu_0}{SE}$
  • The standard error of the mean ($SE$) occupies the denominator of the test statistic:         $SE = \frac{s}{\sqrt{n}}$
  • The degrees of freedom ($df$) required to parameterize the Student's $t$-distribution is defined as:         $df = n - 1$
  • Real-World Origin of $\mu_0$: The hypothesized baseline $\mu_0$ is derived from historical benchmarks or control environments (e.g., historical crash data measuring fatal crashes per month in automobiles without airbags, against which modern samples from airbag-equipped vehicles are tested to evaluate reductions in fatal crashes).

• Detailed Empirical Example: Support Resolution Times:

  • Context: A technical support operations center monitors resolution times (in minutes) across customer interactions. Historical benchmark performance establishes a baseline mean resolution time of $45\,\text{minutes}$.
  • Hypotheses Formulation:
    • Null hypothesis: $H_0: \mu = 45\,\text{minutes}$
    • Alternative hypothesis: $H_A: \mu < 45\,\text{minutes}$ (Left-tailed test)
  • Sample Data Parameters:
    • Sample size: $n = 100$
    • Sample mean: $\bar{x} = 44.4\,\text{minutes}$
    • Sample standard deviation: $s = 2.799684\,\text{minutes}$ (reported approximately as $2.8\,\text{minutes}$)
    • Significance level: $\alpha = 0.05$
  • Step-by-Step Execution:
    1. Calculate the estimated difference from the baseline:             $\bar{x} - \mu_0 = 44.4 - 45 = -0.6\,\text{minutes}$
    2. Calculate the standard error ($SE$):             $SE = \frac{s}{\sqrt{n}} = \frac{2.799684}{\sqrt{100}} = \frac{2.799684}{10} \approx 0.2799684\,\text{minutes}$             (Approximately $0.28\,\text{minutes}$).
    3. Calculate the observed $t$-test statistic:             $t = \frac{-0.6}{0.2799684} \approx -2.143$             This indicates the sample mean is $2.143$ standard errors below the hypothesized population mean.
    4. Determine degrees of freedom:             $df = 100 - 1 = 99$
    5. Calculate the $p$-value:             Using the cumulative distribution function for a left-tailed test: $P(T \le -2.143)$ with $99$ degrees of freedom yields $p\text{-value} \approx 0.0173$.
    6. Decision Rule Application:             Compare the $p$-value to $\alpha$:             $0.0173 < 0.05$             Because the $p$-value is strictly less than $\alpha$, reject the null hypothesis ($H_0$).
    7. Statistical Interpretation:             There is sufficient, statistically significant evidence at the $\alpha = 0.05$ level to support the alternative hypothesis that the population mean resolution time is less than $45\,\text{minutes}$.

• Computational Implementation:

  • Microsoft Excel:
    • Excel's Data Analysis ToolPak does not provide a direct one-sample $t$-test module.
    • The left-tailed $p$-value is generated using the formula =T.DIST(x, deg_freedom, cumulative):             =T.DIST(-2.143, 99, TRUE)
    • For a right-tailed test, the upper tail probability is obtained by taking =1 - T.DIST(t, df, TRUE).
  • StatCrunch Navigation:
    1. Navigate to Stat > T Stats > One Sample > With Data (or With Summary if raw data is unavailable).
    2. Select the column containing the resolution time data.
    3. Under the Perform section, ensure the radio button for Hypothesis test for μ is selected.
    4. Enter the hypothesized null value: set H_0: μ = 45 (this automatically updates the alternative value).
    5. Select the appropriate inequality sign for the alternative hypothesis: < for a left-tailed test.
    6. Retain the default significance level of $0.05$.
    7. Click Compute to verify output: $t\text{-stat} = -2.14$, $p\text{-value} = 0.0173$, and standard error $SE \approx 0.28$.

One-Sample $z$-Test for Proportions

• Theoretical Framework and Mathematical Structure:

  • Used to analyze categorical variables with binary outcomes (e.g., success vs. failure, yes vs. no, resolution achieved on first contact vs. not achieved).
  • Let $X$ represent the total number of observed successes and $n$ represent the total sample size. The point estimate (sample proportion $\hat{p}$) is defined as:         $\hat{p} = \frac{X}{n}$
  • Tests regarding population proportions always rely on the standard normal distribution ($z$-statistic).
  • Critical Formula Distinction (Hypothesis Tests vs. Confidence Intervals):
    • In confidence interval construction, the standard error is estimated using the sample proportion $\hat{p}$:             $SE_{CI} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$
    • In hypothesis testing, the null hypothesis assumes that the true population proportion equals the specified null baseline $p_0$. Consequently, the standard error must be computed using the hypothesized null proportion $p_0$:             $SE_{HT} = \sqrt{\frac{p_0(1 - p_0)}{n}}$
  • The test statistic is formulated as:         $z = \frac{\hat{p} - p_0}{SE_{HT}} = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$

• Detailed Empirical Example: First-Contact Call Resolution Rates:

  • Context: A contact center measures first-contact customer resolution. Historical operations dictate a baseline success rate of $70\%$ ($p_0 = 0.70$). The center investigates whether a new protocol yields a statistically significant increase in the first-contact resolution rate.
  • Hypotheses Formulation:
    • Null hypothesis: $H_0: p = 0.70$
    • Alternative hypothesis: $H_A: p > 0.70$ (Right-tailed test)
  • Sample Data Parameters:
    • Sample size: $n = 100$
    • Observed successes: $77\,\text{yes}$ responses, $23\,\text{no}$ responses ($X = 77$
    • Significance level: $\alpha = 0.05$
  • Step-by-Step Execution:
    1. Calculate the sample proportion ($\hat{p}$):             $\hat{p} = \frac{77}{100} = 0.77$
    2. Calculate the hypothesis test standard error ($SE$) using $p_0$:             $SE = \sqrt{\frac{0.70 \times (1 - 0.70)}{100}} = \sqrt{\frac{0.70 \times 0.30}{100}} = \sqrt{\frac{0.21}{100}} = \sqrt{0.0021} \approx 0.04582575$             (Reported approximately as $0.04583$).
    3. Calculate the observed $z$-test statistic:             $z = \frac{\hat{p} - p_0}{SE} = \frac{0.77 - 0.70}{0.04583} = \frac{0.07}{0.04583} \approx 1.528$
    4. Calculate the $p$-value (Right-tailed area):             Using Excel's cumulative standard normal function =NORM.S.DIST(z, TRUE):             =NORM.S.DIST(1.528, TRUE) evaluates the cumulative area from $-\infty$ to $1.528$, yielding approximately $0.9367$.             To obtain the upper/right-tail area for a greater-than test:             $p\text{-value} = 1 - P(Z \le 1.528) = 1 - 0.9367 = 0.0633$
    5. Decision Rule Application:             Compare the $p$-value to $\alpha$:             $0.0633 > 0.05$             Because the $p$-value is greater than $\alpha$, fail to reject the null hypothesis ($H_0$).
    6. Statistical Interpretation:             There is insufficient statistical evidence at the $\alpha = 0.05$ level to conclude that the population resolution rate exceeds $0.70$.

• Computational Implementation:

  • StatCrunch Navigation:
    1. Navigate to Stat > Proportion Stats > One Sample > With Summary.
    2. Enter Number of successes: $77$.
    3. Enter Number of observations: $100$.
    4. Under Perform, select the Hypothesis test for p radio button.
    5. Enter the hypothesized value: set H_0: p = 0.7.
    6. Set the alternative inequality sign to >.
    7. Retain the default significance level of $0.05$.
    8. Click Compute to verify: sample proportion $\hat{p} = 0.77$, standard error $SE \approx 0.04583$, $z\text{-stat} = 1.53$, and $p\text{-value} = 0.0633$.

Rigorous Interpretation of $p$-Values and Decision Logic

• Formal Definition of the $p$-Value:

  • Assuming that the null hypothesis ($H_0$) and all underlying distributional assumptions are strictly true, the $p$-value is the exact probability of obtaining a test statistic at least as extreme as the observed value, in the direction specified by the alternative hypothesis ($H_A$).
  • It quantifies how surprising or incompatible the empirical sample evidence is with the baseline claim stated in $H_0$.
  • A smaller $p$-value indicates stronger empirical evidence against the null hypothesis.

• Critical Misconceptions and Fallacies:

  • Fallacy 1: The $p$-value is not the probability that the null hypothesis is true.
  • Fallacy 2: The $p$-value is not the probability that the alternative hypothesis is true.
  • Fallacy 3: The $p$-value is not the probability that the observed results occurred purely by chance, random accident, or luck in a casual sense.
  • Audit Misinterpretation Example: In an operational AI system audit yielding $p = 0.0173$, it is completely false to assert that "there is a $1.73\%$ chance that the null hypothesis is true." It merely reflects that if the null were true, an outcome this far from the benchmark would occur in only $1.73\%$ of repeated random samples.

• The Binary Decision Rule and Linguistic Precision:

  • Condition 1: If $p\text{-value} < \alpha$:
    • Formal action: Reject $H_0$.
    • Meaning: The finding is statistically significant; there is sufficient empirical evidence supporting $H_A$.
    • Caveat: Rejecting $H_0$ does not mathematically prove that $H_A$ is true.
  • Condition 2: If $p\text{-value} \ge \alpha$:
    • Formal action: Fail to reject $H_0$.
    • Meaning: The finding is not statistically significant; there is insufficient empirical evidence to conclude that $H_A$ is true.
    • Philosophical rule: Never "accept" or "approve" the null hypothesis. Failing to reject $H_0$ does not prove that $H_0$ is true, nor does it establish that the parameter exactly equals the null value. It indicates only an absence of contradictory evidence.

Decision Errors and Statistical Power

• Classification of Inferential Errors:

  • Type I Error (False Positive / False Alarm):
    • Occurs when a test rejects the null hypothesis $H_0$ when $H_0$ is, in truth, correct.
    • The probability of committing a Type I error is exactly equal to the chosen significance level: $P(\text{Type I Error}) = \alpha$.
    • Operational Example: In a cold storage facility where temperatures must not exceed $4^\circ\text{C}$, committing a Type I error means concluding that the mean temperature exceeds $4^\circ\text{C}$ when it is actually running within target, thereby triggering an expensive, unneeded operational investigation.
  • Type II Error (False Negative / Missed Detection):
    • Occurs when a test fails to reject the null hypothesis $H_0$ when $H_0$ is, in truth, false (i.e., a real effect or parameter shift exists).
    • The probability of committing a Type II error at a specified true alternative parameter value is denoted by $\beta$: $P(\text{Type II Error}) = \beta$.
    • Operational Example: In the same cold storage monitoring system, committing a Type II error means failing to detect that the true mean temperature has risen above $4^\circ\text{C}$, allowing potential product spoilage to continue undetected.

• Statistical Power:

  • Statistical power is the probability that a test correctly rejects a false null hypothesis at a specified true alternative parameter value:         $\text{Power} = 1 - \beta$
  • Interpretation via Repeated Sampling: If a test possesses a statistical power of $80\%$ ($\text{Power} = 0.80$, implying $\beta = 0.20$) for a specified true reduction below $45\,\text{minutes}$, conducting the test across $100$ distinct random samples would successfully detect the true reduction in approximately $80$ instances and fail to detect it in approximately $20$ instances.
  • Power analyses are vital in business and organizational reporting to verify whether an evaluation used an adequate sample size or if an insignificance result was merely an artifact of an underpowered test.

• Factors Governing Statistical Power:

  • Sample Size ($n$): Increasing $n$ reduces standard error, narrowing the sampling distribution and increasing power.
  • Magnitude of the True Effect: A larger true difference between the actual parameter and the hypothesized value $\mu_0$ makes the departure easier to detect, thereby increasing power.
  • Data Variability ($s$ or $\sigma$): Lower internal variability in the population or sample produces less statistical noise, increasing power.
  • Error Rate Interplay Discussed in Lecture: Power and error rates are inherently linked. The lecture notes that artificially inflating power or altering testing thresholds can elevate the risk of false alarms (Type I errors), creating an inevitable trade-off between sensitivity and false alarm rates.

Statistical Significance vs. Practical Significance

• Conceptual Distinction:

  • Statistical Significance: Confirms that an observed sample difference is unlikely to have arisen solely from random sampling variability under the null hypothesis ($p\text{-value} < \alpha$).
  • Practical Significance: Assesses whether the actual physical magnitude of the observed effect is large enough to matter in real-world application, business operations, or decision-making.

• The Materiality Threshold:

  • Practical decisions are constrained by real-world contexts, including capital expenditures, operational risks, implementation disruption, and organizational feasibility.
  • A materiality threshold represents the minimum physical or economic effect size required to justify taking action or investing capital.
  • Example 1 (Department Redesign): An operational sample reveals a statistically significant reduction in processing time ($p = 0.0173$), but the point estimate indicates a reduction of only $0.6\,\text{minutes}$ ($44.4\,\text{minutes}$ vs. $45.0\,\text{minutes}$). If managerial cost-benefit models dictate that a facility layout redesign requires at least a $3.0\,\text{minute}$ reduction to offset capital costs, the finding possesses statistical significance but entirely lacks practical significance.
  • Example 2 (Large Sample Power Inflation): By collecting very large sample sizes, analysts can drive standard errors close to zero, rendering even trivial parameter differences statistically significant. Reports claiming statistical significance must always be paired with point estimates and confidence intervals to evaluate materiality.

Midterm Exam Preparation and Worked Practice Problems

• Exam Structure and Testing Guidelines:

  • The midterm examination consists entirely of multiple-choice questions.
  • Many questions feature two-part answers requiring both a numeric calculation and the corresponding formal statistical interpretation.
  • Students are provided an official reference sheet containing mathematical formulas, notation, and standard tables.
  • Software calculations (such as executing =T.DIST or =NORM.S.DIST in Excel) will not be tested on the midterm; required $p$-values will be supplied in tabular form. Manual calculations focus on standard errors, test statistics, and parameter differences.
  • Conceptual questions will address variable types (numeric vs. categorical) and measurement scales (nominal, ordinal, interval, and ratio).

• Problem 1: One-Tailed Mean Test on Inventory Picking Time:

  • Problem Statement: A warehouse manager asks whether the population mean inventory picking time is below $18\,\text{minutes}$. The following $p$-values are supplied from software output:
    • Left-tail $p\text{-value} = 0.0296$
    • Right-tail $p\text{-value} = 0.9704$
    • Two-tail $p\text{-value} = 0.0592$         Evaluate at $\alpha = 0.05$.
  • Hypotheses Setup: The manager's objective is to determine if times are "below $18\,\text{minutes}$".
    • $H_0: \mu = 18\,\text{minutes}$
    • $H_A: \mu < 18\,\text{minutes}$
  • $p$-Value Selection: Because the alternative hypothesis is a less-than test (left-tailed), select the left-tail probability: $p\text{-value} = 0.0296$.
  • Decision and Interpretation: Compare $p\text{-value}$ to $\alpha$: $0.0296 < 0.05$. Reject $H_0$. There is statistically significant evidence supporting the claim that the true population mean picking time is below $18\,\text{minutes}$.

• Problem 2: $t$-Test Statistic and Directional Sign for Processing Costs:

  • Problem Statement: A purchasing team evaluates invoice processing costs against a null hypothesis $H_0: \mu = \$48.00$. The sample yields a mean processing cost of $\bar{x} = \$49.20$ with a provided standard error $SE = \$0.40$. Calculate the estimated change, determine the $t$-test statistic, and interpret its sign.
  • Step 1: Calculate the estimated change (numerator):$\bar{x} - \mu_0 = 49.20 - 48.00 = 1.20\,\text{dollars}$
  • Step 2: Calculate the $t$-test statistic:$t = \frac{\bar{x} - \mu_0}{SE} = \frac{1.20}{0.40} = +3.0$
  • Interpretation of Sign and Magnitude: The test statistic is strictly positive ($+3.0$) because the observed sample mean ($\$49.20$) exceeds the hypothesized baseline ($\$48.00$). It indicates that the sample mean is exactly $3.0$ standard errors above the hypothesized null mean.

• Problem 3: Flawed Interpretation of Non-Significant Results:

  • Problem Statement: A test evaluates $H_0: \mu = 10\,\text{minutes}$ versus $H_A: \mu > 10\,\text{minutes}$. The resulting $p$-value is $0.083$ at $\alpha = 0.05$. An analyst concludes: "The population mean is definitely $10\,\text{minutes}$ because the result is not statistically significant." Evaluate this conclusion.
  • Evaluation: Compare $p\text{-value}$ to $\alpha$: $0.083 > 0.05$. The correct decision is to fail to reject $H_0$. While the analyst correctly identifies that the result is not statistically significant, the conclusion that the mean is *"definitely $10\,\text{minutes}"* is fundamentally false. Failing to reject the null hypothesis merely indicates an absence of sufficient evidence to claim$\mu > 10\,\text{minutes}; it never proves that the null hypothesis is true.\n\n* **Problem 4: Statistical Significance vs. Materiality in Order Fulfillment:**\n * *Problem Statement:* A two-sided mean test assesses order throughput with null baseline \mu_0 = 320\,\text{orders/hour}$. A sample of$n = 64$yields a sample mean$\bar{x} = 321.2\,\text{orders/hour}$and a standard error$SE = 0.3\,\text{orders/hour}$. The resulting test yields an observed$t\text{-stat} = 4.0$and a$p\text{-value} = 0.0002$. Management requires an operational gain of at least$6.0\,\text{orders/hour}$to justify capital expenditure on new sorting equipment. At$\alpha = 0.05, does this evidence justify purchasing the equipment?\n * *Statistical Evaluation:* p\text{-value} = 0.0002 < 0.05$. The test is highly statistically significant, providing strong evidence to reject$H_0$and conclude that the true mean throughput exceeds$320\,\text{orders/hour}.\n * *Practical/Materiality Evaluation:* The observed point estimate gain is:\n        \bar{x} - \mu_0 = 321.2 - 320.0 = 1.2\,\text{orders/hour}\n        The required materiality threshold is 6.0\,\text{orders/hour}$. Because the observed gain of$1.2\,\text{orders/hour}$falls far short of the required threshold of$6.0\,\text{orders/hour}, the statistically significant finding lacks practical significance. The purchase of new equipment is not justified.\n\n* **Problem 5: Cold Storage Temperature Monitoring and Decision Errors:**\n * *Problem Statement:* An automated monitoring system tests H_0: \mu = 4^\circ\text{C}$against$H_A: \mu > 4^\circ\text{C}$. Rejection of$H_0$triggers a full engineering investigation. Define the Type I and Type II errors for this system, and state the effect of lowering$\alpha$from$0.10$to$0.02.\n * *Type I Error:* Rejecting H_0$when$\mu = 4^\circ\text{C}. The system concludes the temperature is running high when it is actually at target, triggering an unnecessary, costly investigation.\n * *Type II Error:* Failing to reject H_0$when$\mu > 4^\circ\text{C}. The system fails to detect an actual elevated temperature condition, leading to unaddressed product risk.\n * *Impact of Lowering \alpha$:* Lowering the significance threshold makes the criteria for rejection stricter, thereby reducing the probability of committing a Type I error ($\alpha$decreases from$0.10$to$0.02). As discussed in lecture, modifying threshold values involves trade-offs between detection power and false alarms.\n\n* **Problem 6: Two-Sided Test of Customer Renewal Proportions:**\n * *Problem Statement:* In a random sample of 240$customers,$126$renewed their subscriptions. A two-sided test evaluates$H_0: p = 0.55$against$H_A: p eq 0.55$, yielding a$p\text{-value} = 0.4363$. Evaluate at$\alpha = 0.05.\n* *Step 1: Calculate sample proportion \hat{p}:*\n    \hat{p} = \frac{126}{240} = 0.525\n* *Step 2: Decision and Evaluation:*\n    Compare p\text{-value}$to$\alpha$:$0.4363 > 0.05$. Fail to reject$H_0.\n* *Interpretation:* There is insufficient evidence to conclude that the population renewal rate differs from 0.55$. This does not prove that the true renewal rate is exactly$0.55.\n\n# Questions and Discussion\n\n* **Question on Paper-and-Pencil Midterm Problem Solving:**\n * *Prompt:* A student inquired whether future classes would demonstrate how to solve exam problems manually on paper, noting potential anxiety about the mechanics of solving written exam problems.\n * *Response:* Exam preparation problems with full worked-out solutions are integrated directly into lecture slide decks and class reviews. Physical handwriting on screen is intentionally avoided to prevent readability issues. Students must start each word problem by methodically writing down all known parameters (e.g., recognizing that a stated sample mean corresponds to \bar{x}), locating the relevant formulas on the official reference sheet, and determining whether the context requires a mean test, proportion test, or confidence interval.\n\n* **Question on Exam Coverage of Software Functions:**\n * *Prompt:* A student asked whether memorizing Microsoft Excel syntax (such as `=T.DIST` or `=NORM.S.DIST`) is required for the midterm exam or if it is restricted to homework assignments.\n * *Response:* Software formulas and syntax are strictly for homework assignments. The midterm exam will not require students to write or set up Excel formulas. Any necessary p-values will be directly supplied in tabular formats on the exam.\n\n* **Question on Null Hypothesis Notation (\le$vs.$=):**\n * *Prompt:* A student asked whether the null hypothesis for a test where the manager investigates a value below 18$should be written as$\mu \le 18$or$\mu = 18.\n * *Response:* Both conventions appear across academic literature. Pearson course materials and StatCrunch standardize the null hypothesis using strict equality (H_0: \mu = 18$). Mathematically, for any continuous distribution, the probability of obtaining an exact discrete single-point value is zero ($P(X = c) = 0$), making the operational boundary point the reference of interest. On the examination, null hypotheses will typically be written using equality ($=.\n\n* **Question on Accepting vs. Failing to Reject the Null:**\n * *Prompt:* A student asked whether obtaining a large p$-value (such as$0.51) means that analysts "approve" or "accept" the null hypothesis.\n * *Response:* In rigorous statistical practice, analysts never say they "accept" or "approve" the null hypothesis. The correct linguistic phrasing is exclusively to "fail to reject the null hypothesis." Rejecting the null provides affirmative evidence for the alternative hypothesis; failing to reject provides no positive evidence of anything, reflecting only that the data lacks sufficient evidence to overturn the baseline.\n\n* **Question on Alpha Allocation in One-Tailed vs. Two-Tailed Tests:**\n * *Prompt:* A student asked whether \alpha$remains at$0.05$for a one-tailed test or whether it is divided into$0.025.\n * *Response:* In a one-tailed test, the entire significance level (\alpha = 0.05$) is placed into the single relevant tail. In a two-tailed test, the significance level is split evenly between both tails, allocating$\frac{\alpha}{2} = 0.025$to the left tail and$0.025 to the right tail.\n\n* **Question on Deriving Hypotheses from Problem Narrative:**\n * *Prompt:* A student expressed confusion regarding why an investigation asking whether picking times are below 18\,\text{minutes}$does not place "below$18" into the null hypothesis.\n * *Response:* The null hypothesis always represents the baseline status quo or starting condition, anchoring the distribution to the benchmark value (\mu = 18$). The specific question or directional change investigated in the problem narrative defines the alternative hypothesis ($H_A$). Because the inquiry asks whether times are below$18$, the alternative must be$\mu < 18$. If management had asked whether times exceeded$18$, the alternative would be$\mu > 18$$.

• Question on Exam Schedule:

  • Prompt: A student asked whether the midterm exam is administered on the upcoming Monday.
  • Response: The midterm exam is not scheduled for Monday. Monday's class will introduce new material from Unit 4 (which will not appear on the midterm) and dedicate at least half of the lecture time to further midterm review.