STAT 200: Normal Distributions and Standardized Test Statistics

Introduction to Normal Distributions

• The normal distribution is a specific type of symmetrical distribution, characterized by a bell shape. It was first introduced to the course in Lesson 2.
• In recent lessons, bootstrapping and randomization methods were used to approximate sampling distributions. Many of these constructed distributions share similar bell-shaped characteristics, indicating they are approximately normal.
• The primary utility of the normal distribution this week is to approximate the sampling distribution when conducting hypothesis tests or constructing confidence intervals.

The Standard Normal Distribution and Z-Scores

• The Standard Normal Distribution is defined as a normal distribution with a mean of zero and a standard deviation of one.
• Notation used for the normal distribution is generally represented as:   $N(\mu, \sigma)$
• The specific notation for a standard normal distribution is:   $N(0, 1)$
  • The first value ($0$) represents the mean ($\mu$).
  • The second value ($1$) represents the standard deviation ($\sigma$).
• The normal distribution is a continuous distribution. This implies that the probability of any single, specific point occurring (e.g., $z = 2.0$) cannot be computed. Instead, probabilities are found for intervals of values.
• The Empirical Rule (also known as the 68-95-99.7 rule) provides specific interval probabilities for a normal distribution:
  • Approximately $68\%$ of the distribution falls within one standard deviation of the mean.
  • Approximately $95\%$ of the distribution falls within two standard deviations of the mean.
  • Approximately $99.7\%$ of the distribution falls within three standard deviations of the mean.
• Z-scores are used to describe an individual observation relative to the entire distribution in units of standard deviations. The formulas for calculating z-scores are:
  • Using sample statistic symbols:     $z = \frac{x - \bar{x}}{s}$
  • Using population parameter symbols:     $z = \frac{x - \mu}{\sigma}$
• Computing standardized test statistics in hypothesis testing relies heavily on this concept of the z-score.

Determining Area Under a Normal Distribution using Minitab Express

• The process for creating probability distribution plots is similar for both Windows and Mac versions of Minitab Express. In the Mac version, the probability distribution plot function is located in the top menu.
• General procedure in Minitab Express (Windows):
  1. Go to Statistics.
  2. Select Distribution Plot.
  3. Choose Display Probability.
  4. Ensure the default distribution is set to Normal, with Mean = $0$ and Standard Deviation = $1$ (for a Standard Normal Distribution).
• Example calculations for a Standard Normal Distribution ($N(0,1)$):
  • Area less than a z-score of 1: Select Specified x value, then Left Tail. Enter $x = 1$. The area is $0.841345$.
  • Area greater than a z-score of 1: Select Right Tail. Enter $x = 1$. The area is $0.158655$.
  • Area between z-scores of -1 and 1: Select Middle. Enter lower bound $-1$ and upper bound $1$. The area is $0.682689$ (consistent with the empirical rule's $68\%$ estimate).
  • Area in both tails (more extreme than -1 and 1): Select Equal Tails. Enter x value as $1$ (or $-1$); Minitab Express mirrors the symmetry. The total area is the sum of both tails.
• Non-Standard Example (SAT subscales):
  • SAT subscales are normed to a mean of $500$ and a standard deviation of $100$. This is expressed as $N(500, 100)$.
  • To find the area greater than a score of $600$: Change the Minitab parameters to Mean = $500$ and Standard Deviation = $100$. Select Right Tail and enter $x = 600$. The area is $0.158655$.

Determining Points from a Given Proportion

• This procedure is Used to find values on the x-axis that offset a specific proportion, such as finding z-multipliers for confidence intervals.
• Minitab Express requires proportions, not percentages. For example, a 99% interval is entered as $0.99$.
• Example: Finding z-scores that separate the middle $99\%$ of a distribution:
  1. Go to Display Probability for the Standard Normal Distribution.
  2. Select Specified Probability.
  3. Choose Equal Tails.
  4. If $99\%$ is in the middle, then $1\%$ ($0.01$) is split between the two tails ($0.005$ per tail). Minitab allows you to specify the area in the tails.
  5. The resulting z-scores are $\pm 2.57583$. These serve as the z-multipliers for a $99\%$ confidence interval.

The Central Limit Theorem (CLT)

• The Central Limit Theorem states that if the sample size is sufficiently large, the sampling distribution of many statistics will be approximately normally distributed.
• "Sufficiently large" is determined by the point at which the sampling distribution attains a normal shape. Specific thresholds for sample sizes will be discussed in Lessons 8 and 9.
• This theorem applies to all statistics previously covered: proportions, means, differences in proportions/means, correlations, and linear regression slopes.
• Researchers prefer using the normal distribution over bootstrapping or randomization because:
  • Z-scores use a standard scale (Mean = $0$, SD = $1$), making results easier to interpret.
  • They are easier to compute mathematically.
  • The methodology historically predates current high-power computing tools like StatKey.
• Conceptually, this method involves layering a standard normal distribution over the results obtained from randomization or bootstrapping.

Conducting a Hypothesis Test using Standardized Test Statistics

• This Five-Step Procedure is an evolution of the method learned in Lesson 5:
  • Step 1: Check assumptions and write hypotheses. The primary assumption is that the sampling distribution is approximately normal. For now, this is determined by visual inspection.
  • Step 2: Calculate the test statistic. This is the general form for all parameters:     $z = \frac{\text{Sample Statistic} - \text{Null Parameter}}{\text{Standard Error}}$     This formula converts the sample statistic into standard error units (a z-score) to show how far it lies from the hypothesized null parameter.
  • Step 3: Determine the P-value. The p-value is the area under the standard normal distribution that is more extreme than the test statistic in the direction indicated by the alternative hypothesis ($H_a$):
  • Right-tailed test: Area to the right of $z$.
  • Left-tailed test: Area to the left of $z$.
  • Two-tailed test: Sum of the areas in both tails (using $\pm z$).
  • Step 4: Make a decision. Comparison to the alpha level ($\alpha$):
  • If $p \leq \alpha$ (usually $0.05$), reject the null hypothesis ($H_0$).
  • If $p > \alpha$, fail to reject the null hypothesis ($H_0$).
  • Step 5: State a real-world conclusion. Interpret the significance in the context of the original research question. If $H_0$ is rejected, results are statistically significant, providing evidence of a population difference.

Hypotheses Testing Case Study: World Campus Adult Learners

• Research Question: Are more than $60\%$ of all World Campus STAT 200 students adult learners?
• Data: Sample size ($n = 100$), count of adult learners = $66$.
• Step 1: The sampling distribution is approximately normal. Hypotheses are:   $H_0: p = 0.60$$H_a: p > 0.60$
• Step 2: The sample statistic ($\hat{p}$) is $66 / 100 = 0.66$. The null parameter is $0.6$. The standard error (SE) from StatKey is $0.049$.   $z = \frac{0.66 - 0.6}{0.049} = 1.224$
• Step 3: Using Minitab Express ($N(0, 1)$, Right tail, $x = 1.224$), the p-value is $0.110476$.
• Step 4: Since $0.110476 > 0.05$, fail to reject the null hypothesis.
• Step 5: There is not sufficient evidence that more than $60\%$ of all World Campus STAT 200 students are adult learners.

Hypothesis Testing Case Study: Flight Arrival Times

• Research Question: Do the mean arrival times differ for Delta and United flights? (Arrival time of $0$ is on time, negative is early, positive is late).
• Original Sample: United Mean ($\bar{x}_1$) = $9.756\text{ minutes}$. Delta Mean ($\bar{x}_2$) = $-2.623\text{ minutes}$. Difference = $-12.38$.
• Step 1: Sample size is large, so sampling distribution is approximately normal. Hypotheses are:   $H_0: \mu_1 - \mu_2 = 0$$H_a: \mu_1 - \mu_2 \neq 0$
• Step 2: Null parameter = $0$. Standard error (SE) from StatKey = $1.751$. Difference in sample means = $-12.38$.   $z = \frac{-12.38 - 0}{1.751} = -7.07$
• Step 3: Two-tailed test. Find area less than $-7.07$ and greater than $7.07$. In Minitab Express, the p-value is extremely small. The p-value is effectively reported as $p < 0.0001$.
• Step 4: Since $p < 0.05$, reject the null hypothesis.
• Step 5: There is statistically significant evidence that the mean arrival times in the population differ for Delta and United flights.

Constructing Confidence Intervals using the Standard Form

• The general standard form of a confidence interval is:   $\text{Sample Statistic} \pm (z \times \text{Standard Error})$
• The $z$ value in this formula is the z-multiplier, which corresponds to the desired level of confidence. For example, for a $95\%$ confidence level, a multiplier of approximately $2$ was used previously (based on the empirical rule).
• Example: 90% Confidence Interval for Adult Learners
  • Sample statistic ($\hat{p}$) = $0.66$.
  • SE from bootstrap distribution = $0.048$.
  • To find $z$ for $90\%$: Middle = $0.90$, total tails = $0.10$, individual tail = $0.05$. Multiplier = $1.64485$.
  • Calculation: $0.66 \pm (1.64485 \times 0.048)$.
  • Interval: $0.581$ to $0.739$.
  • Interpretation: We are $90\%$ confident that the proportion of adult learners in the population is between $58.1\%$ and $73.9\%$.
• Example: 99% Confidence Interval for Commute Distances (Atlanta vs St. Louis)
  • St. Louis Mean = $14.156\text{ miles}$. Atlanta Mean = $18.156\text{ miles}$. Difference = $4\text{ miles}$.
  • SE from StatKey = $0.775$.
  • To find $z$ for $99\%$: Middle = $0.99$, total tails = $0.01$, individual tail = $0.005$. Multiplier = $2.57583$.
  • Calculation: $4 \pm (2.57583 \times 0.775)$.
  • Interval: $2.004$ to $5.996$.
  • Interpretation: We are $99\%$ confident that the difference in mean commute distances between Atlanta and St. Louis is between $2.004\text{ miles}$ and $5.996\text{ miles}$.