Chapter 2: Introduction to Statistics and Data Analysis Study Guide

Overview of Statistical Inference, Samples, Populations, and Probability

  • Historical Context and the Industrial Miracle:

    • Beginning in the 1980s and extending into the 21st century, American industry shifted significant attention toward quality improvement.

    • The Japanese "industrial miracle" of the mid-20th century was largely driven by the widespread implementation of statistical methods and statistical thinking among management personnel to manufacture high-quality products consistently.

  • Use of Scientific Data:

    • Gathering scientific data is a centuries-old practice involved in manufacturing, food product development, computer software engineering, energy systems, and pharmaceuticals.

    • A fundamental distinction exists between mere data collection/summarization and inferential statistics.

    • Inferential statistics provides a structured toolkit designed to make scientific judgments, decisions, and inferences in the presence of natural uncertainty and variation.

  • Sources of Variability in Scientific Data:

    • Variability is inherent in almost all physical and biological systems. If data were always constant and perfectly accurate, statistical analysis would be unnecessary.

    • Batch-to-Batch vs. Within-Batch Variation: In manufacturing (e.g., material density), output varies between production runs (batch-to-batch) as well as within a single batch.

    • Measurement vs. Environmental Variation: Measuring atmospheric sulfur monoxide involves two distinct sources of variation: daily fluctuations at the same locale and measurement error between the observed reading and the true chemical concentration.

    • Subject-to-Subject Variation: In biomedical drug trials (e.g., a hypertension drug achieving an 85%85\% success rate versus an old drug's 80%80\% success rate), patient response varies naturally. Success rates fluctuate across study samples (e.g., a repeated trial might yield 75%75\%), making statistical evaluation of variability critical before approving higher-cost or side-effect-bearing treatments.

  • Populations and Samples:

    • Population: The complete collection of all individual items, observations, or scientific units of a specified type (e.g., all computer boards manufactured under a given process, or all human patients suffering from chronic hypertension).

    • Sample: A subset of observations chosen from a population (e.g., 5050 computer boards selected randomly from an assembly line).

    • Statistical inferences drawn from a sample extend to the overarching population from which the sample was drawn, provided proper sampling methods are used.

  • Types of Scientific Studies:

    • Designed Experiments: The scientist or engineer systematically alters input factors (e.g., temperature, humidity, chemical dosages) according to a planned scheme to evaluate their impact on a response variable.

    • Observational Studies: Data are collected without direct control over input factor levels because factors cannot be preselected or manipulated (e.g., a forest scientist studying wood density across natural tree populations, or an agronomist collecting crop yield data during an unexpected drought).

  • Role of Probability in Inferential Statistics:

    • Probability theory (covered in depth in Chapters 2–6) connects descriptive statistics to inferential methods, quantifying the strength or confidence of statistical conclusions.

    • Deductive Reasoning (Probability): Uses known population characteristics to deduce the behavior or properties of hypothetical samples drawn from that population.

    • Inductive Reasoning (Inferential Statistics): Uses observed sample data alongside probability theory to draw formal conclusions about unknown population parameters.


Fundamental relationship between probability and inferential statistics.
  • Illustrative Examples of Probability in Decision Making:

    • Example 1.1 (Manufacturing Defect Rates):

    • A sample of n=100n = 100 items yields 1010 defective items. The maximum tolerable long-run defective rate for the company is 5%5\% (0.050.05).

    • Assuming a true population defective rate of 5%5\%, the probability (P-valueP\text{-value}) of observing 1010 or more defective items in a random sample of 100100 is calculated as 0.02820.0282

    • Because 0.02820.0282 is small, the engineer concludes that the process defective rate likely exceeds the allowable 5%5\% threshold, signaling a need for process modification.

    • Example 1.2 (Fungus and Seedling Root Growth Study):

    • Conducted at Virginia Polytechnic Institute and State University (Virginia Tech) to evaluate the effect of nitrogen on northern red oak seedlings inoculated with the fungus Pisolithus tinctorus over 140140 days.

    • Data Set (Stem Weights in grams):

      • No Nitrogen (n=10n = 10): 0.320.32, 0.530.53, 0.280.28, 0.370.37, 0.470.47, 0.430.43, 0.360.36, 0.420.42, 0.380.38, 0.430.43

      • Nitrogen (n=10n = 10): 0.260.26, 0.430.43, 0.470.47, 0.490.49, 0.520.52, 0.750.75, 0.790.79, 0.860.86, 0.620.62, 0.460.46


A dot plot of stem weight data.
- Visual inspection via a dot plot reveals that nitrogen-treated observations (∘\circ) generally exceed non-nitrogen observations (×\times).
- If nitrogen had zero effect (both samples coming from the exact same population), the probability of observing data with this degree of separation is approximately 0.030.03. This small P-valueP\text{-value} provides strong evidence that nitrogen increases average stem weight.

Sampling Procedures and Collection of Data

  • Simple Random Sampling:

    • A sampling procedure where every possible sample of size nn has an equal chance of being selected from the population.

    • Sample Size (nn): The total number of elements or observations included in a sample.

    • Biased Sample: Occurs when non-random selection restricts the sample to a sub-population. For instance, selecting 10001000 families exclusively from urban areas for a statewide political survey biases results because rural political preferences differ.

  • Stratified Random Sampling:

    • Used when sampling units are non-homogeneous but naturally divide into homogeneous, non-overlapping groups called strata.

    • Independent random samples are selected within each stratum to ensure all sub-groups are appropriately represented (e.g., dividing a municipal population into ethnic strata before conducting a bond referendum opinion poll).

  • Principles of Experimental Design:

    • Treatments: Specific conditions or factor combinations being compared across populations (e.g., Nitrogen vs. No Nitrogen; Placebo vs. Active Drug; Coated vs. Uncoated materials).

    • Experimental Units: The physical entities or subjects assigned to treatments (e.g., individual tree seedlings, clinical trial patients, metal specimens).

    • Importance of Random Assignment:

    • Assigning experimental units randomly to treatments prevents systematic bias caused by uncontrolled baseline characteristics (e.g., accidentally assigning heavier patients to a placebo group in a hypertension study).

    • Randomization ensures that variability among experimental units is distributed evenly across treatment groups rather than confounding the treatment effects.

  • Example 1.3 (Aluminum Alloy Corrosion Study):

    • Investigates the impact of a chemical corrosion coating and relative humidity on aluminum fatigue life.

    • Response Variable: Corrosion resistance measured in thousands of cycles to failure (higher values indicate lower corrosion).

    • Experimental Design: Four treatment combinations across two factors:

    1. Coating: Uncoated vs. Chemical Corrosion Coating

    2. Humidity: 20%20\% vs. 80%80\% Relative Humidity

    • Eight experimental specimens were prepared, assigning two specimens randomly to each treatment combination.

    • Data Set (Average Cycles to Failure in Thousands):

    • Uncoated at 20%20\% Relative Humidity: 975975

    • Uncoated at 80%80\% Relative Humidity: 350350

    • Chemical Corrosion Coating at 20%20\% Relative Humidity: 17501750

    • Chemical Corrosion Coating at 80%80\% Relative Humidity: 15501550


Corrosion results for Example 1.3.
  • Key Insights: Higher humidity increases corrosion across both groups, but the chemical coating significantly reduces corrosion. Furthermore, the coating mitigates the negative impact of high humidity.

Measures of Location: The Sample Mean and Median

  • Sample Mean:

    • The numerical average of a set of sample observations x1,x2,…,xnx_1, x_2, \dots, x_n.

    • Formula (Definition 1.1):     xˉ=∑i=1nxin=x1+x2+⋯+xnn\bar{x} = \sum_{i=1}^{n} \frac{x_i}{n} = \frac{x_1 + x_2 + \dots + x_n}{n}

    • Physical Interpretation (Centroid): The sample mean serves as the centroid or center of mass of the sample data. In a physical system, it represents the exact fulcrum point where a system of equal weight points balances.


Sample mean as a centroid of the with-nitrogen stem weight.
  • Sample Median:

    • A robust measure of central location arranged to reflect central tendency without distortion from extreme outliers.

    • Formula (Definition 1.2): For ordered observations x1,x2,…,xnx_1, x_2, \dots, x_n:     x~={x(n+1)/2,if n is odd12(xn/2+xn/2+1),if n is even\tilde{x} = \begin{cases} x_{(n+1)/2}, & \text{if } n \text{ is odd} \\ \frac{1}{2}\left(x_{n/2} + x_{n/2+1}\right), & \text{if } n \text{ is even} \end{cases}

  • Sensitivity to Outliers (Mean vs. Median Comparison):

    • For the sample (1.7,2.2,3.9,3.11,14.7)(1.7, 2.2, 3.9, 3.11, 14.7), the single extreme value 14.714.7 pulls the mean to xˉ=5.12\bar{x} = 5.12, whereas the median remains stable at x~=3.9\tilde{x} = 3.9.

    • Stem Weight Summary (Example 1.2 Data):

    • No-Nitrogen: xˉ=0.399 g\bar{x} = 0.399\,\text{g}, x~=0.38+0.422=0.400 g\tilde{x} = \frac{0.38 + 0.42}{2} = 0.400\,\text{g}

    • Nitrogen: xˉ=0.565 g\bar{x} = 0.565\,\text{g}, x~=0.49+0.522=0.505 g\tilde{x} = \frac{0.49 + 0.52}{2} = 0.505\,\text{g}

  • Trimmed Means:

    • A compromise estimator created by eliminating a specified percentage of the highest and lowest sample values prior to computing the arithmetic average.

    • 10%10\% Trimmed Mean Formula: Removes the top 10%10\% and bottom 10%10\% of ordered data.

    • Stem Weight Computations (Example 1.2):

    • No-Nitrogen (10%10\% trimmed mean, removing 0.280.28 and 0.530.53):       xˉtr(10)=0.32+0.37+0.47+0.43+0.36+0.42+0.38+0.438=0.39750 g\bar{x}_{tr(10)} = \frac{0.32 + 0.37 + 0.47 + 0.43 + 0.36 + 0.42 + 0.38 + 0.43}{8} = 0.39750\,\text{g}

    • Nitrogen (10%10\% trimmed mean, removing 0.260.26 and 0.860.86):       xˉtr(10)=0.43+0.47+0.49+0.52+0.75+0.79+0.62+0.468=0.56625 g\bar{x}_{tr(10)} = \frac{0.43 + 0.47 + 0.49 + 0.52 + 0.75 + 0.79 + 0.62 + 0.46}{8} = 0.56625\,\text{g}

    • The median is a limiting extreme case of a trimmed mean where all observations except the central one or two are trimmed.

Measures of Variability

  • Importance of Variability:

    • Controlling and reducing process variability is vital in manufacturing and engineering because product quality and overall profitability depend directly on variance reduction.

    • Measures of location alone are insufficient to summarize a sample; two data sets can share identical sample means yet exhibit vastly different spread, altering conclusions.

  • Sample Range:

    • Defined as the absolute difference between the maximum and minimum observations:     Range=Xmax⁡−Xmin⁡\text{Range} = X_{\max} - X_{\min}

  • Sample Variance and Standard Deviation:

    • Sample Variance Formula (Definition 1.3):     s2=∑i=1n(xi−xˉ)2n−1s^2 = \sum_{i=1}^{n} \frac{(x_i - \bar{x})^2}{n - 1}

    • Sample Standard Deviation Formula:     s=s2s = \sqrt{s^2}

    • Degrees of Freedom (n−1n - 1): Represents the number of independent pieces of information available to compute s2\mathbf{s^2}. Because the sum of deviations about the mean always equals zero:     ∑i=1n(xi−xˉ)=0\sum_{i=1}^{n} (x_i - \bar{x}) = 0     stipulating the first n−1n - 1 deviations fully fixes the final deviation.

  • Step-by-Step Variance Calculation Example:

    • For sample data (5,17,6,4)(5, 17, 6, 4), xˉ=8\bar{x} = 8:

    • Deviations (xi−xˉ)(x_i - \bar{x}): (5−8)=−3(5-8) = -3, (17−8)=9(17-8) = 9, (6−8)=−2(6-8) = -2, (4−8)=−4(4-8) = -4

    • Sum of squared deviations: (−3)2+92+(−2)2+(−4)2=9+81+4+16=110(-3)^2 + 9^2 + (-2)^2 + (-4)^2 = 9 + 81 + 4 + 16 = 110

    • Variance: s2=1104−1=1103=36.67s^2 = \frac{110}{4 - 1} = \frac{110}{3} = 36.67

  • Example 1.4 (pH Meter Bias Test):

    • Sample size n=10n = 10 measurements of a neutral solution (pH=7.0\text{pH} = 7.0):

    • Data: 7.07,7.00,7.10,6.97,7.00,7.03,7.01,7.01,6.98,7.087.07, 7.00, 7.10, 6.97, 7.00, 7.03, 7.01, 7.01, 6.98, 7.08

    • Mean: xˉ=7.0250\bar{x} = 7.0250

    • Variance calculation:       s2=19[(7.07−7.025)2+(7.00−7.025)2+⋯+(7.08−7.025)2]=0.001939s^2 = \frac{1}{9}\left[(7.07 - 7.025)^2 + (7.00 - 7.025)^2 + \dots + (7.08 - 7.025)^2\right] = 0.001939

    • Standard deviation: s=0.001939=0.0440s = \sqrt{0.001939} = 0.0440

  • Measurement Units and Theoretical Context:

    • Variance is expressed in squared linear units (e.g., g2\text{g}^2), whereas standard deviation returns to original linear units (e.g., g\text{g}).

    • For Example 1.2 stem weights, standard deviations are s=0.0728 gs = 0.0728\,\text{g} (No Nitrogen) and s=0.1867 gs = 0.1867\,\text{g} (Nitrogen), confirming that nitrogen increases sample dispersion alongside the mean.

    • Population variance (σ2\sigma^2) and mean (μ\mu) represent true fixed population parameters estimated by sample statistics (s2s^2 and xˉ\bar{x}). Variance is primarily utilized in theoretical inference, while standard deviation is used in practical applications.

Discrete and Continuous Data

  • Continuous Data:

    • Measurements taken along a continuous numerical scale (e.g., chemical yield in grams/pound, tensile strength in megapascals, stem weight in grams).

  • Discrete Data:

    • Count data or categorical classifications representing distinct integers or non-continuous steps (e.g., number of radioactive particles passing a sensor per millisecond, number of oil tankers arriving daily).

  • Binary Data and Sample Proportions:

    • Binary outcomes consist of two mutually exclusive categories (e.g., Success/Failure, Defective/Non-defective, Response/No Response).

    • Categories are coded numerically as 11 (Category 1 / Success) or 00 (Category 2 / Failure).

    • For nn sample items with xx total successes:

    • Sample proportion in Category 1: p^=xn\hat{p} = \frac{x}{n}

    • Sample proportion in Category 2: 1−p^=1−xn1 - \hat{p} = 1 - \frac{x}{n}

    • The sample proportion p^\hat{p} is mathematically identical to the sample mean xˉ\bar{x} of binary 0s and 1s:     xˉ=x1+x2+⋯+xnn=1+1+0+⋯+1n=xn\bar{x} = \frac{x_1 + x_2 + \dots + x_n}{n} = \frac{1 + 1 + 0 + \dots + 1}{n} = \frac{x}{n}

    • Clinical Example: If 2020 out of 5050 patients show improvement under a drug, p^=2050=0.40\hat{p} = \frac{20}{50} = 0.40 (40%40\% success) and 1−p^=0.601 - \hat{p} = 0.60 (60%60\% non-success).

    • Manufacturing Quality Example: A tire manufacturer evaluates 50005000 tires and finds 100100 defective (p^1=1005000=0.02\hat{p}_1 = \frac{100}{5000} = 0.02). After a process change, a second sample of 50005000 yields 9090 defective tires (p^2=905000=0.018\hat{p}_2 = \frac{90}{5000} = 0.018). Inferential statistics determines if the drop from 0.020.02 to 0.0180.018 reflects a genuine population improvement or sampling noise.

Statistical Modeling, Scientific Inspection, and Graphical Diagnostics

  • Statistical Modeling vs. Deterministic Models:

    • Statistical models incorporate random probabilistic variation alongside underlying mathematical assumptions (e.g., assuming sample data follow a Gaussian/Normal distribution).

  • Scatter Plots and Empirical Model Building:

    • Tensile Strength vs. Cotton Percentage Study:

    • Specimens manufactured across four cotton percentages (15%15\%, 20%20\%, 25%25\%, 30%30\%) with five specimens tested per concentration level.

    • Data Set:

      • 15%15\% Cotton: 7,7,9,8,107, 7, 9, 8, 10

      • 20%20\% Cotton: 19,20,21,20,2219, 20, 21, 20, 22

      • 25%25\% Cotton: 21,21,17,19,2021, 21, 17, 19, 20

      • 30%30\% Cotton: 8,7,8,9,108, 7, 8, 9, 10


Scatter plot of tensile strength and cotton percentages.
- A quadratic regression model captures the curvilinear relationship:

      μt,c=β0+β1C+β2C2\mu_{t,c} = \beta_0 + \beta_1 C + \beta_2 C^2       where μt,c\mu_{t,c} represents expected tensile strength given cotton concentration CC.

  • Stem-and-Leaf Plots:

    • Tabular and graphical display splitting observations into leading digits (stems) and trailing digits (leaves).

    • Car Battery Life Study (n=40n = 40 items, guaranteed 3 years):

    • Data Set: 2.22.2, 4.14.1, 3.53.5, 4.54.5, 3.23.2, 3.73.7, 3.03.0, 2.62.6, 3.43.4, 1.61.6, 3.13.1, 3.33.3, 3.83.8, 3.13.1, 4.74.7, 3.73.7, 2.52.5, 4.34.3, 3.43.4, 3.63.6, 2.92.9, 3.33.3, 3.93.9, 3.13.1, 3.33.3, 3.13.1, 3.73.7, 4.44.4, 3.23.2, 4.14.1, 1.91.9, 3.43.4, 4.74.7, 3.83.8, 3.23.2, 2.62.6, 3.93.9, 3.03.0, 4.24.2, 3.53.5

    • Single Stem Plot: Stems = 1,2,3,41, 2, 3, 4

    • Stem 1: Leaves 6,96, 9 (Frequency = 2)

    • Stem 2: Leaves 2,5,6,6,92, 5, 6, 6, 9 (Frequency = 5)

    • Stem 3: Leaves 0,0,1,1,1,1,2,2,2,3,3,3,4,4,4,5,5,6,7,7,7,8,8,9,90, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6, 7, 7, 7, 8, 8, 9, 9 (Frequency = 25)

    • Stem 4: Leaves 1,1,2,3,4,5,7,71, 1, 2, 3, 4, 5, 7, 7 (Frequency = 8)

    • Double-Stem Plot: Splits stems into ⋆\star (leaves 0–4) and ⋅\cdot (leaves 5–9):

    • 1⋅1\cdot: 6,96, 9 (Frequency = 2)

    • 2⋆2\star: 22 (Frequency = 1)

    • 2⋅2\cdot: 5,6,6,95, 6, 6, 9 (Frequency = 4)

    • 3⋆3\star: 0,0,1,1,1,1,2,2,2,3,3,3,4,4,40, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4 (Frequency = 15)

    • 3⋅3\cdot: 5,5,6,7,7,7,8,8,9,95, 5, 6, 7, 7, 7, 8, 8, 9, 9 (Frequency = 10)

    • 4⋆4\star: 1,1,2,3,41, 1, 2, 3, 4 (Frequency = 5)

    • 4⋅4\cdot: 5,7,75, 7, 7 (Frequency = 3)

    • Practical guideline: Select between 55 and 2020 stems based on total sample size.

  • Relative Frequency Distributions and Histograms:

    • Dividing class frequency by sample size nn yields the class relative frequency.

    • Relative Frequency Table for Battery Life Data (n=40n = 40):

    • Interval 1.5−1.91.5 - 1.9 (Midpoint 1.71.7): Frequency =2= 2, Relative Frequency =0.050= 0.050

    • Interval 2.0−2.42.0 - 2.4 (Midpoint 2.22.2): Frequency =1= 1, Relative Frequency =0.025= 0.025

    • Interval 2.5−2.92.5 - 2.9 (Midpoint 2.72.7): Frequency =4= 4, Relative Frequency =0.100= 0.100

    • Interval 3.0−3.43.0 - 3.4 (Midpoint 3.23.2): Frequency =15= 15, Relative Frequency =0.375= 0.375

    • Interval 3.5−3.93.5 - 3.9 (Midpoint 3.73.7): Frequency =10= 10, Relative Frequency =0.250= 0.250

    • Interval 4.0−4.44.0 - 4.4 (Midpoint 4.24.2): Frequency =5= 5, Relative Frequency =0.125= 0.125

    • Interval 4.5−4.94.5 - 4.9 (Midpoint 4.74.7): Frequency =3= 3, Relative Frequency =0.075= 0.075


Relative frequency histogram.


Estimating frequency distribution.
  • Distribution Symmetry and Skewness:

    • Symmetric: Foldable along a central vertical axis into matching halves.

    • Right-Skewed (Positive Skew): Possesses an elongated right tail and concentrated left values.

    • Left-Skewed (Negative Skew): Possesses an elongated left tail and concentrated right values.


Skewness of data.
  • Box-and-Whisker Plots (Box Plots) and Outlier Detection:

    • Encloses the Interquartile Range (IQR=Q3−Q1\mathbf{IQR = Q_3 - Q_1}) within a central box containing the median (Q2Q_2).

    • 1.5×IQR1.5 \times \text{IQR} Rule: Observations falling beyond 1.5×IQR1.5 \times \text{IQR} below Q1Q_1 or above Q3Q_3 are categorized as outliers.

    • Example 1.5 (Cigarette Nicotine Data, n=40n = 40):

    • Data Set: 1.091.09, 1.921.92, 2.312.31, 1.791.79, 2.282.28, 1.741.74, 1.471.47, 1.971.97, 0.850.85, 1.241.24, 1.581.58, 2.032.03, 1.701.70, 2.172.17, 2.552.55, 2.112.11, 1.861.86, 1.901.90, 1.681.68, 1.511.51, 1.641.64, 0.720.72, 1.691.69, 1.851.85, 1.821.82, 1.791.79, 2.462.46, 1.881.88, 2.082.08, 1.671.67, 1.371.37, 1.931.93, 1.401.40, 1.641.64, 2.092.09, 1.751.75, 1.631.63, 2.372.37, 1.751.75, 1.691.69

    • IQR=0.365\text{IQR} = 0.365; Threshold 1.5×IQR=0.54751.5 \times \text{IQR} = 0.5475

    • Outliers identified: 0.720.72 and 0.850.85 in the lower tail; 2.552.55 in the upper tail.


Box-and-whisker plot for Example 1.5.


Stem-and-leaf plot for the nicotine data.
  • Example 1.6 (Paint Can "Ears" Thickness, 3030 samples of size 55):

    • Median =35= 35, Lower Quartile Q1=31Q_1 = 31, Upper Quartile Q3=36Q_3 = 36

    • Asymmetric left block wider than right block; no outliers detected.


Box-and-whisker plot for thickness of paint can “ears.”
  • Quartiles and Percentiles:

    • First Quartile (Q1Q_1 / 25th percentile): Separates the lowest 25%25\% of data.

    • Second Quartile (Q2Q_2 / Median / 50th percentile): Divides the sample into equal halves.

    • Third Quartile (Q3Q_3 / 75th percentile): Separates the top 25%25\% of data.

    • Percentiles: Partition distributions finely (e.g., 95th percentile isolates the top 5%5\%; 1st percentile isolates the bottom 1%1\%).

General Types of Statistical Studies

  • Designed Experiments:

    • Feature active factor level manipulation and random experimental unit allocation.

    • Establish definitive cause-and-effect relationships by neutralizing nuisance variables through randomization.

    • Interaction Effects: Occurs when the response variation across levels of one factor depends on the state of a second factor (demonstrated by non-parallel lines in factor plots, such as Figure 1.3).

  • Observational Studies:

    • Collect monitoring data over time without active factor intervention or randomization (e.g., recording human sodium intake vs. blood cholesterol, or monitoring power plant usage vs. ambient temperature).

    • Limitations: Cannot definitively demonstrate cause-and-effect due to confounding variables (e.g., diet, exercise, plant production volume) and subject to environmental constraints on variable ranges.

  • Retrospective Studies:

    • Utilize historical records compiled over past operational periods.

    • Advantages: Low execution cost.

    • Disadvantages:

    1. Questionable data validity and historical recording accuracy.

    2. Unobserved missing records along time series.

    3. Risk of unrecorded human error.

    4. Uncontrolled historical variable ranges that may not reflect present conditions.

End-of-Chapter Exercises and Practice Datasets

  • Exercise 1.1 (Paint Drying Times in Hours):

    • Data (n=15n = 15): 3.4,2.5,4.8,2.9,3.6,2.8,3.3,5.6,3.7,2.8,4.4,4.0,5.2,3.0,4.83.4, 2.5, 4.8, 2.9, 3.6, 2.8, 3.3, 5.6, 3.7, 2.8, 4.4, 4.0, 5.2, 3.0, 4.8

    • Calculations: xˉ=3.787 hours\bar{x} = 3.787\,\text{hours}, x~=3.6 hours\tilde{x} = 3.6\,\text{hours}, 20%20\% trimmed mean =3.673 hours= 3.673\,\text{hours}.

  • Exercise 1.2 (Cotton Fiber Water Absorbency):

    • Data (n=20n = 20): 18.71,21.41,20.72,21.81,19.29,22.43,20.17,23.71,19.44,20.50,18.92,20.33,23.00,22.85,19.25,21.77,22.11,19.77,18.04,21.1218.71, 21.41, 20.72, 21.81, 19.29, 22.43, 20.17, 23.71, 19.44, 20.50, 18.92, 20.33, 23.00, 22.85, 19.25, 21.77, 22.11, 19.77, 18.04, 21.12

    • Calculations: xˉ=20.7675\bar{x} = 20.7675, x~=20.6100\tilde{x} = 20.6100, 10%10\% trimmed mean =20.7425= 20.7425.

  • Exercise 1.3 (Aircraft Polymer Aging and Tensile Strength in psi):

    • No Aging (n=10n = 10): 227,222,218,217,225,218,216,229,228,221227, 222, 218, 217, 225, 218, 216, 229, 228, 221

    • Aging (n=10n = 10): 219,214,215,211,209,218,203,204,201,205219, 214, 215, 211, 209, 218, 203, 204, 201, 205

    • Comparison: No Aging xˉ=222.1 psi\bar{x} = 222.1\,\text{psi}, x~=221.5 psi\tilde{x} = 221.5\,\text{psi}; Aging xˉ=209.9 psi\bar{x} = 209.9\,\text{psi}, x~=210.0 psi\tilde{x} = 210.0\,\text{psi}. Aging significantly reduces tensile strength.

  • Exercise 1.4 (Steel Rod Spring Flexibility):

    • Company A (n=10n = 10): 9.3,8.8,6.8,8.7,8.5,6.7,8.0,6.5,9.2,7.09.3, 8.8, 6.8, 8.7, 8.5, 6.7, 8.0, 6.5, 9.2, 7.0

    • Company B (n=10n = 10): 11.0,9.8,9.9,10.2,10.1,9.7,11.0,11.1,10.2,9.611.0, 9.8, 9.9, 10.2, 10.1, 9.7, 11.0, 11.1, 10.2, 9.6

    • Results: Company A xˉ=7.95\bar{x} = 7.95, x~=8.25\tilde{x} = 8.25; Company B xˉ=10.26\bar{x} = 10.26, x~=10.15\tilde{x} = 10.15. Company B yields higher flexibility.

  • Exercise 1.5 (Cholesterol Reduction Study):

    • Control (n=10n = 10): 7,3,−4,14,2,5,22,−7,9,57, 3, -4, 14, 2, 5, 22, -7, 9, 5

    • Treatment (n=10n = 10): −6,5,9,4,4,12,37,5,3,3-6, 5, 9, 4, 4, 12, 37, 5, 3, 3

    • Results: Control xˉ=5.6\bar{x} = 5.6, x~=5.0\tilde{x} = 5.0; Treatment xˉ=7.6\bar{x} = 7.6, x~=4.5\tilde{x} = 4.5. The treatment mean is inflated by a single outlier (3737), while medians show similar overall performance.

  • Exercise 1.6 (Silicone Rubber Tensile Strength in MPa vs. Curing Temp):

    • 20∘C20^\circ\text{C} (n=12n = 12): 2.07,2.14,2.22,2.03,2.21,2.03,2.05,2.18,2.09,2.14,2.11,2.022.07, 2.14, 2.22, 2.03, 2.21, 2.03, 2.05, 2.18, 2.09, 2.14, 2.11, 2.02

    • 45∘C45^\circ\text{C} (n=12n = 12): 2.52,2.15,2.49,2.03,2.37,2.05,1.99,2.42,2.08,2.42,2.29,2.012.52, 2.15, 2.49, 2.03, 2.37, 2.05, 1.99, 2.42, 2.08, 2.42, 2.29, 2.01

    • Calculations: 20∘C20^\circ\text{C} xˉ=2.108 MPa\bar{x} = 2.108\,\text{MPa}, s=0.068 MPas = 0.068\,\text{MPa}; 45∘C45^\circ\text{C} xˉ=2.235 MPa\bar{x} = 2.235\,\text{MPa}, s=0.198 MPas = 0.198\,\text{MPa}. Higher temperature increases both mean strength and sample variability.

  • Exercise 1.13 (Electronic Battery Lifetimes in Hours):

    • Sample: 123,116,122,110,175,126,125,111,118,117123, 116, 122, 110, 175, 126, 125, 111, 118, 117

    • xˉ=124.3 hours\bar{x} = 124.3\,\text{hours}, x~=120.0 hours\tilde{x} = 120.0\,\text{hours}. The extreme value 175175 causes the mean to exceed the median.

  • Exercise 1.14 (Tire Inner Diameter in mm):

    • Sample: 572,572,573,568,569,575,565,570572, 572, 573, 568, 569, 575, 565, 570

    • xˉ=571.0 mm\bar{x} = 571.0\,\text{mm}, x~=571.0 mm\tilde{x} = 571.0\,\text{mm}, s2=10.57 mm2s^2 = 10.57\,\text{mm}^2, s=3.25 mms = 3.25\,\text{mm}, Range =10 mm= 10\,\text{mm}. Mean matches target target diameter (570 mm570\,\text{mm}) closely.

  • Exercise 1.15 (Fair Coin Probability Proof):

    • Five coin tosses yielding HHHHH has probability (12)5=0.03125\left(\frac{1}{2}\right)^5 = 0.03125

    • Evaluating P=0.03125P = 0.03125 under a two-tailed test (P=0.0625P = 0.0625) shows moderate, but not overwhelming, evidence of bias.

  • Exercise 1.16 (Sum of Deviations Proof):

    • Show that ∑i=1n(xi−xˉ)=0\sum_{i=1}^{n} (x_i - \bar{x}) = 0:     ∑i=1n(xi−xˉ)=∑i=1nxi−∑i=1nxˉ=nxˉ−nxˉ=0\sum_{i=1}^{n} (x_i - \bar{x}) = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} \bar{x} = n\bar{x} - n\bar{x} = 0

  • Exercise 1.17 (Smoking Impact on Sleep Onset in Minutes):

    • Smokers (n=12n = 12): 69.3,56.0,22.1,47.6,53.2,48.1,52.7,34.4,60.2,43.8,23.2,13.869.3, 56.0, 22.1, 47.6, 53.2, 48.1, 52.7, 34.4, 60.2, 43.8, 23.2, 13.8

    • Nonsmokers (n=15n = 15): 28.6,25.1,26.4,34.9,29.8,28.4,38.5,30.2,30.6,31.8,41.6,21.1,36.0,37.9,13.928.6, 25.1, 26.4, 34.9, 29.8, 28.4, 38.5, 30.2, 30.6, 31.8, 41.6, 21.1, 36.0, 37.9, 13.9

    • Results: Smokers xˉ=43.70 min\bar{x} = 43.70\,\text{min}, s=16.93 mins = 16.93\,\text{min}; Nonsmokers xˉ=30.32 min\bar{x} = 30.32\,\text{min}, s=7.13 mins = 7.13\,\text{min}. Smoking increases sleep latency and variation.

  • Exercise 1.21 (Power Failure Durations in Minutes):

    • Data (n=45n = 45 observations from 1313 to 158158 minutes).

    • xˉ=74.02 min\bar{x} = 74.02\,\text{min}, x~=78.00 min\tilde{x} = 78.00\,\text{min}, s=37.16 mins = 37.16\,\text{min}.

  • Exercise 1.23 (Automobile Hydrocarbon Emissions in ppm):

    • 1980 Models (n=20n = 20): xˉ=378.4 ppm\bar{x} = 378.4\,\text{ppm}

    • 1990 Models (n=20n = 20): xˉ=160.9 ppm\bar{x} = 160.9\,\text{ppm}

    • Shows significant reduction in average emissions and variance across decade models.

  • Exercise 1.27 (Bearing Wear yy vs. Load xx in lbs):

    • Loads tested: 700 lb700\,\text{lb}, 1000 lb1000\,\text{lb}, 1300 lb1300\,\text{lb}

    • Means: yˉ1=210\bar{y}_1 = 210, yˉ2=325\bar{y}_2 = 325, yˉ3=375\bar{y}_3 = 375

    • Individual variability shows substantial overlap across load categories, emphasizing the importance of evaluating standard deviations alongside means.

  • Exercises 1.28 & 1.31 (Injection Molding Shrinkage in cm×104\text{cm} \times 10^4):

    • Low Mold Temperature:

    • Low Velocity Mean xˉ=72.69\bar{x} = 72.69, High Velocity Mean xˉ=71.61\bar{x} = 71.61

    • High Mold Temperature:

    • Low Velocity Mean xˉ=76.10\bar{x} = 76.10, High Velocity Mean xˉ=93.22\bar{x} = 93.22

    • Conclusion: A strong interaction exists between injection velocity and mold temperature: increasing injection velocity slightly reduces shrinkage at low temperatures, but substantially increases shrinkage at high temperatures.