Discussion 5.1: Z vs. T Distributions, Confidence Intervals, and One-Sample Hypothesis Testing

Overview of Statistical Inference for Population Means

• Statistical inference uses sample data to draw conclusions about unknown population parameters.
• Core mechanisms covered in this module:
  • Distinguishing between the standard normal distribution ($Z$ distribution) and Student's $t$-distribution ($T$ distribution).
  • Constructing and interpreting confidence intervals for both known population parameters and unknown parameters estimated via sample statistics.
  • Conducting and interpreting one-sample hypothesis tests for single population means.

Properties of the Z Distribution and T Distribution

• The $Z$ Distribution:
  • Represents the standard normal distribution, possessing a mean of $\mu = 0$ and a standard deviation of $\sigma = 1$.
  • Role of the Central Limit Theorem (CLT): The threshold of sample size $n \ge 30$ relies on the Central Limit Theorem, which establishes that the sampling distribution of the sample mean approaches normality as sample size increases, regardless of the underlying population distribution's shape.
  • Criteria for application:
  • Used for confidence intervals and hypothesis tests regarding means when the population is known to follow a normal distribution.
  • Used when the sample size is large ($n \ge 30$) AND the population variance ($\sigma^2$) or population standard deviation ($\sigma$) is known.
• The $T$ Distribution:
  • Structurally represents a "looser" version of the standard normal distribution, characterized by a lower peak at the center and heavier, fatter tails.
  • The spread and tail thickness depend on degrees of freedom ($df$), defined as $df = n - 1$.
  • As sample size $n$ (and consequently degrees of freedom) increases, the $T$ distribution converges toward the standard normal $Z$ distribution.
  • Criteria for application:
  • Used for confidence intervals and hypothesis tests about means when the sample size is small ($n < 30$).
  • Used whenever the population variance ($\sigma^2$) or population standard deviation ($\sigma$) is unknown, regardless of how large the sample size is.
• Comparison of critical cutoff values at a $95\%$ confidence level:
  • For the standard normal $Z$ distribution, the critical values encompassing the central $95\%$ of area are $-1.96$ and $1.96$.
  • For a $T$ distribution with sample size $n = 12$ ($df = 11$), the critical values expand outward to $-2.201$ and $2.201$ to capture the central $95\%$, reflecting increased uncertainty due to estimating standard deviation from a sample.

Decision Criteria: Selecting Between Z and T Distributions

• Step 1: Assess knowledge of the population dispersion:
  • Ask: "Do you know the POPULATION standard deviation ($\sigma$) or variance ($\sigma^2$)?"
• Step 2: Follow the branch based on the standard deviation:
  • If NO (population standard deviation/variance is unknown):
  • Use the $T$ distribution immediately, regardless of sample size.
  • If YES (population standard deviation/variance is known):
  • Ask: "Is sample size $n \ge 30$?"
  • If YES ($n \ge 30$): Use the $Z$ distribution.
  • If NO ($n < 30$): Use the $T$ distribution.

Fundamentals of Confidence Intervals

• Definition: A confidence interval estimates an unknown population parameter based on an observed sample statistic (also referred to as a point estimator).
• Structure of the interval:
  • Adds and subtracts a margin of error around the point estimate (such as the sample mean $\bar{x}$ or population mean $\mu$).
  • The margin of error accounts for sampling variability and quantifies the reality that a single sample estimate will rarely equal the true population parameter exactly.

Confidence Intervals for Known Population Parameters (Z Distribution)

• Standard Intervals for Known Population Means and Standard Deviations:
  • Applied when a population is normally distributed, and both the population mean ($\mu$) and population standard deviation ($\sigma$) are known.
  • Benchmark confidence interval calculations:
  • $90\%$ Confidence Interval: $\mu \pm 1.65\,\sigma$
  • $95\%$ Confidence Interval: $\mu \pm 1.96\,\sigma$ (certain practical sources use $\pm 2\,\sigma$ as a common approximation shortcut)
  • $99\%$ Confidence Interval: $\mu \pm 2.58\,\sigma$
• Empirical Rule (Normal Distribution Percentages):
  • Within $\mu \pm 1\,\sigma$ lies approximately $68.27\%$ of the distribution.
  • Within $\mu \pm 2\,\sigma$ lies approximately $95.45\%$ of the distribution.
  • Within $\mu \pm 3\,\sigma$ lies approximately $99.73\%$ of the distribution.

• Exact or Custom Intervals Formula:
  • To compute an exact answer or construct an interval for confidence levels other than $90\%$, $95\%$, or $99\%$:
  • General formula:     $\mu \pm Z_{\alpha/2}\,\sigma$
  • Margin of error expression:     $\text{Margin of Error} = Z_{\alpha/2}\,\sigma$
  • Variable definitions:
  • $\mu$ = population mean
  • $\sigma$ = population standard deviation
  • $Z_{\alpha/2}$ = critical value derived from the standard normal ($Z$) distribution
• Spreadsheet Implementation:
  • Determine $Z_{\alpha/2}$ using the Excel function NORM.S.INV.
  • Input formula for probability: (1 - confidence_level) / 2.
  • Multiply the resulting critical value by $\sigma$ to determine the margin of error.
  • Add and subtract this margin of error from $\mu$.
• Applied Scenario:
  • Assume SAT scores follow a normal distribution with a known population mean $\mu = 1100$ and a known population standard deviation $\sigma = 200$.
  • Objective: Construct a $95\%$ confidence interval around the population mean.
  • Calculation:
  • Critical $Z$ value for $95\%$ is $Z_{0.025} = 1.96$.
  • $\text{Margin of Error} = 1.96 \times 200 = 392$.
  • Lower limit: $1100 - 392 = 708$.
  • Upper limit: $1100 + 392 = 1492$.
  • Resulting interval: $[708, 1492]$.

Confidence Intervals for Sample Means Using the Z Distribution

• Application Criteria:
  • Sample size is large ($n \ge 30$).
  • Population standard deviation ($\sigma$) is known.
  • Population mean ($\mu$) is unknown and estimated by the sample mean ($\bar{x}$).
• Mathematical Formulation:   $\bar{x} \pm Z_{\alpha/2}\,\frac{\sigma}{\sqrt{n}}$
• Margin of Error Component:   $\text{Margin of Error} = Z_{\alpha/2}\,\frac{\sigma}{\sqrt{n}}$
• Component Definitions:
  • $\bar{x}$ = sample mean
  • $\sigma$ = population standard deviation
  • $n$ = sample size
  • $Z_{\alpha/2}$ = critical cutoff value from the $Z$ distribution
  • $\frac{\sigma}{\sqrt{n}}$ = standard error of the mean
• Software Methods:
  • Method 1: Use NORM.S.INV to derive the critical $Z$ value (or utilize hard-coded values: $1.65$ for $90\%$, $1.96$ for $95\%$, $2.58$ for $99\%$), and multiply by $\frac{\sigma}{\sqrt{n}}$ to calculate the margin of error.
  • Method 2: Calculate the margin of error directly via the Excel function CONFIDENCE.NORM(alpha, standard_dev, size).
• Applied Scenario:
  • A dataset named TestScores contains a sample of $n = 70$ students' performance on a standardized test.
  • Nationwide population standard deviation is known to be $\sigma = 10$.
  • Task: Construct a $98\%$ confidence interval around the sample mean.
  • Setup:
  • $\alpha = 1 - 0.98 = 0.02$
  • Critical value: $Z_{\alpha/2} = Z_{0.01} = \text{NORM.S.INV}(0.01) \approx 2.326$
  • Standard error: $\frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{70}} \approx 1.1952$
  • Margin of error: $2.326 \times 1.1952 \approx 2.78$
  • Interval: $\bar{x} \pm 2.78$

Confidence Intervals for Sample Means Using the T Distribution

• Application Criteria:
  • Sample size is small ($n < 30$), OR
  • Population standard deviation ($\sigma$) is unknown (sample standard deviation $s$ is used as the point estimate), regardless of sample size.
• Mathematical Formulation:   $\bar{x} \pm t_{\alpha/2}\,\frac{s}{\sqrt{n}}$
• Degrees of Freedom:   $df = n - 1$
• Margin of Error Component:   $\text{Margin of Error} = t_{\alpha/2}\,\frac{s}{\sqrt{n}}$
• Component Definitions:
  • $\bar{x}$ = sample mean
  • $s$ = sample standard deviation
  • $n$ = sample size
  • $t_{\alpha/2}$ = critical value from the $T$ distribution with $df = n - 1$
  • $df$ = degrees of freedom
  • $\frac{s}{\sqrt{n}}$ = estimated standard error of the mean
• Software Methods:
  • Method 1: Compute the critical $t$ value using T.INV.2T(probability, deg_freedom) where probability equals $\alpha$, and then compute the margin of error by multiplying by $\frac{s}{\sqrt{n}}$.
  • Method 2: Compute margin of error directly using CONFIDENCE.T(alpha, standard_dev, size) where standard_dev is the sample standard deviation $s$.
  • Method 3: Use the Descriptive Statistics tool located in the Excel Analysis Toolpak with the Confidence Level for Mean option selected (requires raw data).
• Applied Scenario:
  • A dataset named TestScores contains a sample of $n = 70$ students' performance on a standardized test.
  • Population standard deviation is unknown.
  • Task: Construct a $92\%$ confidence interval around the sample mean.
  • Setup:
  • $\alpha = 1 - 0.92 = 0.08$
  • $df = 70 - 1 = 69$
  • Critical value: $t_{0.04, 69} = \text{T.INV.2T}(0.08, 69)$
  • Margin of error: $\text{CONFIDENCE.T}(0.08, s, 70)$
  • Interval: $\bar{x} \pm t_{0.04, 69}\,\frac{s}{\sqrt{70}}$

One-Sample Hypothesis Testing Framework

• Definition: A one-sample hypothesis test evaluates whether an assumption, claim, or hypothesized parameter about a single population mean is plausible based on empirical sample data.
• Selection of Test Statistic:
  • One-Sample $Z$ Test: Employed when sample size is large ($n \ge 30$) AND the population standard deviation ($\sigma$) is known.
  • One-Sample $T$ Test: Employed when sample size is small ($n < 30$) OR the population standard deviation ($\sigma$) is unknown.

One-Sample Z Tests

• Application Criteria:
  • Sample size is large ($n \ge 30$) AND population standard deviation ($\sigma$) is known.
• Test Statistic Formula:   $Z = \frac{\bar{x} - \mu - d}{\frac{\sigma}{\sqrt{n}}}$
• Component Definitions:
  • $\bar{x}$ = sample mean
  • $\mu$ = hypothetical population mean specified under the null hypothesis
  • $d$ = hypothesized difference (typically $d = 0$ in standard tests)
  • $\sigma$ = population standard deviation
  • $n$ = sample size
  • $\frac{\sigma}{\sqrt{n}}$ = standard error of the mean
• Software Methods:
  • Method 1: Calculate the $Z$ test statistic manually and use NORM.S.DIST(z, TRUE) to determine the cumulative lower-tail p-value.
  • Method 2: Use the Excel function Z.TEST(array, x, [sigma]) to compute the p-value directly from raw sample data (returns the one-tailed upper-tail p-value).
• Applied Scenario:
  • Problem: Test designers intended for test takers nationwide to achieve an average score of $75$. Did they achieve this goal or not?
  • Hypotheses:
  • Null Hypothesis ($H_0$): The average score is $75$$H_0: \mu = 75$
  • Alternative Hypothesis ($H_A$): The average score is not $75$$H_A: \mu \neq 75$
  • Testing Parameters:
  • Sample size: $n = 70$ from TestScores dataset
  • Population standard deviation: $\sigma = 10$
  • Significance level: $\alpha = 0.05$
  • Test type: Two-tailed one-sample $Z$ test

One-Sample T Tests

• Application Criteria:
  • Sample size is small ($n < 30$) OR population standard deviation ($\sigma$) is unknown.
• Test Statistic Formula:   $t = \frac{\bar{x} - \mu - d}{\frac{s}{\sqrt{n}}}$
• Degrees of Freedom:   $df = n - 1$
• Component Definitions:
  • $\bar{x}$ = sample mean
  • $\mu$ = hypothetical population mean specified under the null hypothesis
  • $d$ = hypothesized difference (typically $d = 0$)
  • $s$ = sample standard deviation
  • $n$ = sample size
  • $df$ = degrees of freedom
  • $\frac{s}{\sqrt{n}}$ = estimated standard error of the mean
• Software Methods:
  • Method 1: Compute the $t$ test statistic manually and calculate p-values using the Excel function T.DIST (or T.DIST.RT, T.DIST.2T based on tail direction).
  • Method 2: Use the Excel Analysis Toolpak tool t-Test: Two-Sample Assuming Unequal Variances (requires raw data and a secondary dummy column of zeros/constants to adapt it for a one-sample test).
• Applied Scenario:
  • Problem: A score of $70$ is considered passing. Was the nationwide average significantly higher than that?
  • Hypotheses:
  • Null Hypothesis ($H_0$): The average score is less than or equal to $70$$H_0: \mu \le 70$
  • Alternative Hypothesis ($H_A$): The average score is greater than $70$$H_A: \mu > 70$
  • Testing Parameters:
  • Sample size: $n = 70$ from TestScores dataset
  • Population standard deviation: Unknown
  • Significance level: $\alpha = 0.05$
  • Test type: Right-tailed one-sample $T$ test with $df = 69$

Reporting and Interpreting Hypothesis Test Results

• Standard Interpretation Format:
  • Every test must be stated in the canonical form:     P = [p-value]. At alpha = [significance level], we [fail to reject / reject] the null hypothesis.
• Decision Rules:
  • If $p \ge \alpha$: Fail to reject the null hypothesis ($H_0$).
  • If $p < \alpha$: Reject the null hypothesis ($H_0$).
• Strict Academic Phrasing Conventions:
  • Permitted terminology is restricted solely to "reject" or "fail to reject".
  • Never use the words "accept," "prove," or "disprove" in reference to a statistical hypothesis.

Applied Practice Exercises

• Exercise 1: Interval with Known Population Standard Deviation
  • Prompt: Calculate a $90\%$ confidence interval around the average exam score in the sample. The population standard deviation ($\sigma$) is $10$.
  • Strategy: Since $\sigma$ is known (and assuming large sample criteria $n \ge 30$), use the $Z$ distribution.
  • Margin of error setup: $1.645 \times \frac{10}{\sqrt{n}}$.
• Exercise 2: Directional One-Sample Test with Known Dispersion
  • Prompt: Perform a one-sample test based on the following hypotheses, given $\sigma = 10$ and $\alpha = 0.01$:
  • Null Hypothesis ($H_0$): Average exam score is greater than or equal to $82$$H_0: \mu \ge 82$
  • Alternative Hypothesis ($H_A$): Average exam score is less than $82$$H_A: \mu < 82$
  • Strategy: Perform a left-tailed one-sample $Z$ test using $\sigma = 10$.
• Exercise 3: Interval with Unknown Population Standard Deviation
  • Prompt: Calculate a $95\%$ confidence interval around the average exam score in the sample. Assume the population standard deviation is unknown.
  • Strategy: Because $\sigma$ is unknown, use the $T$ distribution with degrees of freedom $df = n - 1$ and sample standard deviation $s$.
  • Margin of error setup: $t_{0.025, n-1} \times \frac{s}{\sqrt{n}}$.
• Exercise 4: Test Involving a Hypothesized Difference
  • Prompt: Perform a one-sample test with $\alpha = 0.05$ assuming population standard deviation is unknown:
  • Null Hypothesis ($H_0$): Average exam score is at least $5$ points higher than $75$$H_0: \mu \ge 75 + 5$
  • Alternative Hypothesis ($H_A$): Average exam score is less than $5$ points higher than $75$$H_A: \mu < 75 + 5$
  • Procedural Note: The phrasing "at least 5 points higher" indicates a hypothesized mean difference parameter $d = 5$ relative to base benchmark $75$ (equating to a comparison threshold of $80$). Use a left-tailed one-sample $T$ test.
• Exercise 5: Sample SAT Score Interval Calculation
  • Prompt: Assume that the average SAT score for a sample of $n = 50$ students was $\bar{x} = 1020$, and the population standard deviation is $\sigma = 200$. Calculate a $99\%$ confidence interval around the sample average.
  • Determination of Distribution: The sample size is large ($n = 50 \ge 30$) and the population standard deviation is known ($\sigma = 200$), requiring the $Z$ distribution.
  • Numerical Evaluation:
  • Sample mean: $\bar{x} = 1020$
  • Population standard deviation: $\sigma = 200$
  • Sample size: $n = 50$
  • Confidence level: $0.99$, yielding $\alpha = 0.01$
  • Critical value: $Z_{\alpha/2} = Z_{0.005} \approx 2.576$ (or standard shortcut $2.58$)
  • Standard error of the mean:       $\frac{\sigma}{\sqrt{n}} = \frac{200}{\sqrt{50}} = \frac{200}{7.07107} \approx 28.284$
  • Margin of error calculation:       $\text{Margin of Error} = 2.576 \times 28.284 \approx 72.86$       (Using $2.58$: $2.58 \times 28.284 \approx 72.97$)
  • Interval limits (using $Z = 2.576$):       $\text{Lower Limit} = 1020 - 72.86 = 947.14$$\text{Upper Limit} = 1020 + 72.86 = 1092.86$
  • Resulting $99\%$ Confidence Interval: $[947.14, 1092.86]$.