Study Notes: Stem-and-Leaf Plots, Normality, and Standardization

Stem-and-Leaf Plots: Overview and Advantages

• A stem-and-leaf plot is a quick viewing mechanism for data that preserves every data element in a compact format.

• Major advantages:- Picture-like visualization of the data.

  • All data elements are included in the plot; nothing is hidden.

  • It behaves similarly to a histogram, providing a sense of distribution while still showing individual values.

• Common use with two-digit data: stems represent the leading digit (the tens place), leaves the trailing digit (the ones place). Example: for data from 20–29, the stem is 2 and leaves are the unit digits.

• Construction example (two-digit data):- Stems: 2 (for 20s), 3 (for 30s), 4 (for 40s), etc.

  • Leaves along each stem show the units: e.g., on the 2 stem you might see 0, 4, 5, 5, 8, 9, etc., representing 20, 24, 25, 25, 28, 29.

• Observing the data: you can spot gaps, clusters, and outliers visually.

• Turning the stem-and-leaf plot on its left side yields a histogram-like view (a “histogram” from the same data).

• With larger ranges, you can adapt the stem to accommodate three-digit data by using the first digit(s) as the stem (see two-digit vs three-digit discussion below).

• Practical workflow: quick stem-and-leaf helps you read the median and approximate mean by reading data off directly; for example, with 12–14 points you can locate the median by counting from top and bottom.

Construction and Interpretation: Two-Digit vs Three-Digit Data

• For data from 0 to 99 (two-digit data): use the tens digit as the stem and the units digit as the leaf.

• Example from transcript (illustrative): data includes values like 20, 24, 25, 25, 28, 29, 46, 52, 61, 60, 65, etc. The stem-leaf shows the tens (2, 4, 5, etc.) and the units as leaves.

• For data that ranges into the hundreds (three-digit data), you can still construct a stem-and-leaf by using the first one or two digits as the stem, depending on the spread of the data. Example: if data range is 140–180, you can use the first two digits (14, 15, 16, 17, 18) as stems with the last digit as leaves, or adjust as needed.

• A multiple-stem (two-group) stem-leaf plot can compare two subgroups (e.g., boys vs girls) side by side: you’ll have two sets of leaves on the same stem or adjacent stems, enabling direct comparison.

• When the data range is narrow (e.g., 140–180), using a two-digit stem for the hundreds/tens portion allows a compact, readable display; if the range were 0–1000, you might avoid stem-leaf altogether and use alternative displays.

Relationship to Histograms

• A stem-and-leaf plot can be rotated to resemble a histogram by turning the plot 90 degrees.

• Advantage: you get the histogram-like distribution while retaining exact data values.

• When comparing two data sets, you can construct a two-group stem-leaf plot and then view histograms for each group by flipping the view for each group.

Central Tendency from Stem-and-Leaf

• With a stem-leaf plot, you can easily identify the median by counting from each end to the middle.

• If there are, for example, 12 data points, the median is between the 6th value from the top and the 7th from the bottom (or the average of the two middle values if the count is even).

• The mean can be approximated or computed exactly by summing all leaves (and accounting for their stems) if the data is small enough to do by hand.

Bell Curve, Empirical Rule, and Standardization Concepts

• Bell (Normal) Curve basics:- About 68% of data lie within one standard deviation of the mean: $P(|X-\mu| \leq \sigma) = 0.68$

  • About 95% lie within two standard deviations: $P(|X-\mu| \leq 2\sigma) = 0.95$

  • About 99.7% lie within three standard deviations: $P(|X-\mu| \leq 3\sigma) = 0.997$

• These are empirical rules for roughly normal data and are often used as quick checks or heuristics.

• Standardization and position vs spread:- The standard deviation is a measure of position relative to the mean as well as spread.

  • Standardizing data allows comparison across different scales (e.g., IQ vs SAT vs ACT).

• Z-score (standard score):- Definition: $z = \frac{X - \mu}{\sigma}$

  • Interpretation: number of standard deviations an observation is from the mean.

• Reverting to the original scale:- If you know the z-score, you can convert back: $X = \mu + z\,\sigma$

• Example scales and their means/sd:- IQ scores: mean $\mu = 100$, standard deviation $\sigma = 16$.

  • SAT scores: mean $\mu = 1500$, standard deviation $\sigma = 300$.

  • ACT scores: mean $\mu = 20$, standard deviation $\sigma = 5$.

• Worked example (IQ example from transcript):- Suppose a group has $\mu = 68$ and $\sigma = 3$, and a student has $X = 72$.

  • Compute z: $z = \frac{72 - 68}{3} = \frac{4}{3} \approx 1.333.$

  • Convert to IQ-scaling: $\text{IQ-score} = 100 + z \cdot 16 = 100 + 1.333 \cdot 16 \approx 121.33.$

  • This illustrates standardization across scales.

• Practical note: standardization is used in many standardized tests (e.g., SAT, ACT, IQ tests) to place scores on common scales.

Normal Distributions and Position on the Z-scale

• Positioning on standardized scales (e.g., IQ with mean 100, SD 16):- A score 2 standard deviations above the mean would be at $100 + 2\times 16 = 132$.

  • Median vs mean: on a perfectly normal distribution they coincide; in real data, skewness affects the mean more than the median.

• Examples of standardization formulas:- Direct z-score: $z = \frac{X - \mu}{\sigma}.$

  • Inverse (finding the original score from a z-score): $X = \mu + z\sigma.$

• Application to standardized tests like SAT/ACT:- SAT: mean 1500, SD 300; a z-score of 1 corresponds to a score of $1500 + 1\times 300 = 1800$ on this scale, etc.

  • ACT: mean 20, SD 5; a z-score of -1 corresponds to $20 + (-1)\times 5 = 15$.

• Practical note: many organizations standardize scores to fixed means and standard deviations to compare across years, populations, or subtests.

• Example: SAT/ACT-style interpretation can be shown on a percentile basis via the standard normal curve, but the exact percentile requires the standard normal CDF $\Phi$.

Normality Assessment Tools and Procedures

• Visual checks:- Histogram shape: bell-like suggests normality but is not conclusive.

  • QQ plot: data points aligning along the reference line indicate normality; deviations indicate non-normality.

• Empirical tests for normality (discussed in transcript):- Ryan–Joiner test (Statdisk example): reports a normality decision with a statement like "Is the data normal? Yes/No" and confidence levels (e.g., 95%, 99%).

  • Shapiro–Wilk test: commonly used online or via software; provides a p-value for normality; easier to interpret in plain language.

• Example workflow with Statdisk (as described):- Enter data column (e.g., column one).

  • Use Normality assessment to obtain a decision (e.g., Fail to reject normality at 95% confidence \rightarrow data is normal with 95% confidence; 99% confidence also provided).

  • If normality is questionable, alternative tests (Ryan–Joiner, Shapiro–Wilk) or graphical checks (QQ plot) are recommended.

• Practical guidance:- The best single method for normality assessment among those discussed is the Ryan–Joiner test in Statdisk (or Shapiro–Wilk online) because they provide explicit normality decisions with confidence levels.

  • Visualization (histogram, QQ plot) is useful but not definitive alone.

• Additional notes on normality assessment:- Normality assessment can also reveal outliers via fences in box plots (see next section).

  • Normality checks are essential when using tests or confidence intervals that assume normality.

Histograms: Regular vs Relative Frequency

• Regular histogram: shows frequency counts on the vertical axis and class intervals on the horizontal axis.

• Relative frequency histogram: shows percentages (relative frequencies) instead of raw counts.- The total area under the histogram becomes 100% (or 1 if expressed as a proportion).

  • When data is displayed as relative frequencies, the height of bars reflects percent, but the shape remains comparable to the regular histogram—the axis labeling changes.

• Why use relative frequency histograms? They align data distributions with probability distributions (probabilities lie between 0 and 1, and areas correspond to probabilities).

• Practical demonstration (as described): switching from frequency to relative frequency changes the axis labels (percentages) but preserves the overall distribution shape and the comparative heights of bars.

• Important test note: the class indicated that the test will cover material up to normality, and that relative frequency histograms would not be included on the test; topics covered include histograms and normality but not this specific histogram variant beyond normality.

Box Plots, Fences, and Outliers

• Box plots display quartiles, medians, and sometimes the mean (denoted by a plus sign in the transcript).

• Outliers in a box plot can be identified using fences:- Lower fence: $Q_1 - 1.5 \times \text{IQR}$

  • Upper fence: $Q_3 + 1.5 \times \text{IQR}$

  • IQR (interquartile range) = $Q<em>3 - Q</em>1$

• Example from transcript:- Given $Q<em>1 = 50, Q</em>3 = 65$, then IQR = $65 - 50 = 15$.

  • Lower fence = $50 - 1.5 \times 15 = 27.5$.

  • Upper fence = $65 + 1.5 \times 15 = 87.5$.

  • Data values outside [27.5, 87.5] are outliers; in the example, a data value of 25 is below the lower fence and would be considered an outlier.

• The box plot also indicates whether data are symmetric or skewed; skewness affects the relation between median and mean:- In a right-skewed distribution, the mean tends to be greater than the median.

  • In a left-skewed distribution, the mean tends to be less than the median.

• The normality assessment can flag outliers via fences and box plots provide a visual of spread and central tendency.

Practical Problem Types and Example Solutions (from the first lecture/test material)

• Normality-identity questions:- Identify which terms are not related to normality. Not related: Interquartile Range (IQR).

  • Related terms include: Q-Q plots, Shapiro–Wilk, Ryan–Joiner, skewness, kurtosis.

• Proportion above a threshold given normal distribution:- If IQs are normally distributed with mean $\mu$ and standard deviation $\sigma$, the proportion above a threshold $x$ is $P(X > x) = 1 - \Phi\left( \frac{x-\mu}{\sigma} \right)$.

  • Example discussion (from transcript): with $\mu = 100$, $\sigma = 15$ vs $\sigma = 16$, find the proportion above 130.

  • For $\sigma = 15$: $z = \frac{130-100}{15} = 2$, so $P(X>130) = 1-\Phi(2) \approx 0.0228 \approx 2.28\%$.

  • For $\sigma = 16$: $z = \frac{130-100}{16} = 1.875$, so $P(X>130) = 1-\Phi(1.875) \approx 0.0303 \approx 3.03\%$.

• Effect of changing the maximum value on the mean and other statistics (outlier/shift effects):- If the largest data value doubles, the range increases and the mean typically increases as well; the IQR and other spread measures would adjust accordingly depending on how the rest of the data behaves.

• Skewness and mean vs. median (interpretation of a stem plot):- A stem plot can reveal right (positive) skew when more data lie toward the left but the tail extends to the right; in such cases, the mean tends to be pulled to the right of the median.

• Box-plot-based outlier checks (continued):- Using fences, no outliers exist if all data lie within [Lower fence, Upper fence].

• Data normalization and normality checks: workflow summary- Start with histograms and QQ plots for a rough view.

  • Use Ryan–Joiner or Shapiro–Wilk for a formal normality test.

  • If normality is questionable, consider nonparametric methods or transform data.

Quick Reference Formulas and Concepts

• Z-score (standard score): $z = \frac{X - \mu}{\sigma}$

• Reconstructing original score from z-score: $X = \mu + z\sigma$

• Normal approximation rules of thumb (empirical rule):- $|X-\mu| \leq \sigma \Rightarrow 68\%\text{ of data}$

  • $|X-\mu| \leq 2\sigma \Rightarrow 95\%\text{ of data}$

  • $|X-\mu| \leq 3\sigma \Rightarrow 99.7\%\text{ of data}$

• Interquartile Range (IQR) and fences:- IQR = $Q<em>3 - Q</em>1$

  • Lower fence = $Q_1 - 1.5 \times \text{IQR}$

  • Upper fence = $Q_3 + 1.5 \times \text{IQR}$

• Coefficient of variation (CV):- $\text{CV} = \frac{\sigma}{\mu}$ (often expressed as a percentage, i.e., multiply by 100% if needed)

• Common standardized-score references:- IQ: $\mu = 100$, $\sigma = 16$

  • SAT: $\mu = 1500$, $\sigma = 300$

  • ACT: $\mu = 20$, $\sigma = 5$

• Proportion above a threshold given normal distribution: $P(X > x) = 1 - \Phi\left( \frac{x-\mu}{\sigma} \right)$

Closing Notes on the Exam Coverage (as described in the transcript)

• The test will cover material from the beginning of these notes up to the point discussed (normality and related topics).

• Relative frequency histograms were introduced but would not be a direct focus on the upcoming test; emphasis remains on normality assessment, histograms, QQ plots, and normality tests.

• Practical tips shared:- You can use calculator tools or software (e.g., Statdisk) to compute descriptive statistics and run normality assessments.

  • Open-book, open-notes format for some problems; focus on understanding concepts and applying formulas rather than memorizing obscure details.

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