Unit 1 Study Guide: Exploring One-Variable Data

Core Concepts: Unit 1 - Exploring One-Variable Data

Data Types and Variable Classifications

Variables are characteristics that can take on different values for different individuals or objects in a data set. Understanding the distinction between variable types is essential for appropriate statistical analysis.

Categorical (Qualitative) Variables

Categorical variables place individuals or observations into distinct groups or categories based on qualitative attributes. Values do not represent measurable quantities, even if numbers are assigned as labels.

• Examples:
  • Movie Genre (Action, Comedy, Drama)
  • Name of a Movie
  • Payment Method (Credit Card, Cash, Cryptocurrency)

Quantitative (Numerical) Variables

Quantitative variables take on numerical values for which arithmetic operations (such as calculating means or standard deviations) make logical sense. They represent measured or counted quantities.

• Discrete Quantitative Variables: Arise from counting processes and take on isolated numerical values (typically integers).
  • Examples: Number of times a movie is watched ($0, 1, 2, 3, \dots$), number of students in a classroom.
• Continuous Quantitative Variables: Arise from measurement processes and can take on any real value within a continuous range or interval.
  • Examples: Time spent watching a movie ($112.5\,\text{minutes}$), age of a viewer ($16.4\,\text{years}$).

Multiple Choice Practice: Solutions & Detailed Concepts

Question 1: Categorical vs. Quantitative Variable Classification

Netflix gathers data on user activity when streaming a movie. Consider the following five variables:

1. I. Genre of the movie being watched: Categorical (groups movies into descriptive genres).
2. II. Time spent watching the movie: Quantitative (a continuous numerical duration measurement).
3. III. Age of the account holder: Quantitative (a continuous or discrete numerical age measurement).
4. IV. Name of the movie: Categorical (identifying text label/title).
5. V. Number of times the account holder watches the movie: Quantitative (a discrete numerical count of viewings).

Correct Answer Analysis

• I and IV are categorical.
• II, III, and V are quantitative.

Correct Choice: (C) I and IV are categorical and the rest are quantitative

Question 2: Analyzing Skewness in Daycare Age Distributions

Distribution Shape Analysis

• The histogram displays the ages of children at a local daycare.
• The peak (mode) occurs in the bin corresponding to ages $1$ to $2$ years.
• The frequencies decrease as age increases, stretching out a long tail to the right up to age $7$.
• Consequently, the distribution is skewed to the right (positively skewed).

Mathematical Relationships in Right-Skewed Distributions

• Mean vs. Median: In a right-skewed distribution, high values in the right tail pull the mean in the direction of the skew. Therefore, the mean age is strictly greater than the median age ($\text{Mean} > \text{Median}$).
• Mode vs. Mean: The mode represents the peak frequency ($1$\text{--}$2\,\text{years}$), whereas the mean is pulled rightward by older ages ($4$\text{--}$7\,\text{years}$). Thus, the mode is less than the mean ($\text{Mode} < \text{Mean}$).

Evaluating Statements

• Statement I: The mean age is greater than the median age. (TRUE)
• Statement II: The median age is greater than the mean age. (FALSE)
• Statement III: The mode age is less than the mean age. (TRUE)

Correct Choice: (C) I and III only

Question 3: Effect of Removing an Observation on the Sample Mean

Given Data

• Initial sample size: $n_1 = 10$
• Initial mean age: $\bar{x}_1 = 16\,\text{years old}$
• Age of student leaving the room: $x_{\text{removed}} = 7\,\text{years old}$

Step-by-Step Calculation

1. Find the total sum of the ages of the initial 10 students:

$\text{Sum}_1 = n_1 \times \bar{x}_1 = 10 \times 16 = 160\,\text{years}$

2. Calculate the new total sum after the 7-year-old leaves:

$\text{Sum}_2 = 160 - 7 = 153\,\text{years}$

3. Calculate the new mean for the remaining 9 students:

$n_2 = 9$

$\bar{x}_2 = \frac{\text{Sum}_2}{n_2} = \frac{153}{9} = 17\,\text{years old}$

Correct Choice: (A) 17 years old

Question 4: Comparing Standard Deviation Across Discrete Data Sets

Standard deviation ($S_x$) measures the average distance of data values from their mean. A larger dispersion from the center yields a larger standard deviation.

Evaluation of Options

• (A) $2, 2, 2, 2, 2$:

  • All values are identical. Dispersion = $0$.
  • $S_x = 0$

• (B) $1, 1, 1, 10, 10$:

  • Mean: $\bar{x} = \frac{1+1+1+10+10}{5} = \frac{23}{5} = 4.6$
  • Deviations squared sum: $3 \times (1 - 4.6)^2 + 2 \times (10 - 4.6)^2 = 3(-3.6)^2 + 2(5.4)^2 = 3(12.96) + 2(29.16) = 38.88 + 58.32 = 97.2$
  • Sample Variance: $S_x^2 = \frac{97.2}{4} = 24.3$
  • $S_x = \sqrt{24.3} \approx 4.93$

• (C) $5, 6, 7, 8, 9$:

  • Mean: $\bar{x} = 7$
  • Deviations squared sum: $(-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 = 4 + 1 + 0 + 1 + 4 = 10$
  • Sample Variance: $S_x^2 = \frac{10}{4} = 2.5$
  • $S_x = \sqrt{2.5} \approx 1.58$

• (D) $9, 9, 9, 10, 10$:

  • Values are clustered tightly around $\bar{x} = 9.4$. Standard deviation is less than $0.6$

Correct Choice: (B) 1, 1, 1, 10, 10

Question 5: Comparing Graphical Distributions via Back-to-Back Stemplots

Key: $16 \mid 3 = 163\,\text{minutes}$

Extracted Data Values

• Action Movies ($n = 10$): $109, 113, 115, 132, 134, 143, 150, 155, 157, 163$
• Comedy Movies ($n = 10$): $84, 87, 89, 95, 97, 100, 108, 115, 124, 134$

Descriptive Statistics Calculations

Action Movies

• Minimum: $109$
• Maximum: $163$
• Range: $\text{Max} - \text{Min} = 163 - 109 = 54$
• Median ($M$): Average of 5th and 6th values: $\frac{134 + 143}{2} = 138.5$
• First Quartile ($Q_1$): Median of lower 5 values ($109, 113, 115, 132, 134$) = $115$
• Third Quartile ($Q_3$): Median of upper 5 values ($143, 150, 155, 157, 163$) = $155$
• Interquartile Range ($IQR$): $Q_3 - Q_1 = 155 - 115 = 40$

Comedy Movies

• Minimum: $84$
• Maximum: $134$
• Range: $\text{Max} - \text{Min} = 134 - 84 = 50$
• Median ($M$): Average of 5th and 6th values: $\frac{97 + 100}{2} = 98.5$
• First Quartile ($Q_1$): Median of lower 5 values ($84, 87, 89, 95, 97$) = $89$
• Third Quartile ($Q_3$): Median of upper 5 values ($100, 108, 115, 124, 134$) = $115$
• Interquartile Range ($IQR$): $Q_3 - Q_1 = 115 - 89 = 26$

Evaluation of Options

• (A) The third quartile of Comedy ($Q_{3,\text{Comedy}} = 115$) equals the first quartile of Action ($Q_{1,\text{Action}} = 115$). (TRUE)
• (B) The IQR of Action ($40$) is larger than the IQR of Comedy ($26$). (TRUE)
• (C) The range of Action ($54$) is larger than the range of Comedy ($50$). (TRUE)
• (D) All the statements above are true. (TRUE)

Correct Choice: (D) All the statements above are true

Question 6: Converting Bar Graphs to Pie Charts

Frequency and Relative Frequency Distribution

Sample Size: $N = 200\,\text{adults}$

• 401(k): Frequency = $60$$\text{Relative Frequency} = \frac{60}{200} = 0.30 = 30\%$

• Individual Stocks: Frequency = $40$$\text{Relative Frequency} = \frac{40}{200} = 0.20 = 20\%$

• Mutual Funds: Frequency = $50$$\text{Relative Frequency} = \frac{50}{200} = 0.25 = 25\%$

• Real Estate: Frequency = $30$$\text{Relative Frequency} = \frac{30}{200} = 0.15 = 15\%$

• Cryptocurrency: Frequency = $20$$\text{Relative Frequency} = \frac{20}{200} = 0.10 = 10\%$

Pie Chart Verification

Pie charts require relative frequencies (percentages) adding up to $100\%$. Examining Pie Chart (C):

• 401(k) = $30\%$
• Individual Stocks = $20\%$
• Mutual Funds = $25\%$
• Real Estate = $15\%$
• Cryptocurrency = $10\%$

Correct Choice: (C)

Question 7: Summary Statistics and Outlier Identification Rules

Given summary statistics for $n = 20$ golf scores:

Evaluation of Statements

• Statement (A): 25% of scores were below 70.
  • By definition, $Q_1 = 70$ marks the cutoff for the lower $25\%$ of data.
• Statement (B): This distribution is approximately symmetric.
  • The mean ($75.1$), median ($75$), and mode ($75$) are virtually equal, indicating clear symmetry.
• Statement (C): According to the 1.5 x IQR rule, there are no outliers.
  • $IQR = Q_3 - Q_1 = 78.5 - 70 = 8.5$
  • $1.5 \times IQR = 1.5 \times 8.5 = 12.75$
  • $\text{Lower Boundary} = Q_1 - 12.75 = 70 - 12.75 = 57.25$
  • $\text{Upper Boundary} = Q_3 + 12.75 = 78.5 + 12.75 = 91.25$
  • Minimum score ($65 > 57.25$) and Maximum score ($89 < 91.25$) both lie strictly inside the boundaries. Thus, no outliers are present.

Correct Choice: (D) All of the above statements are true.

Question 8: Interpreting Histograms and Sample Medians

Histogram Frequency Table

Sample Size: $n = 33\,\text{homes}$. Bin widths represent thousands of dollars ($\$1000$\text{s}):

• Bins and Frequencies:
  • $[0, 500): 18$
  • $[500, 1000): 8$
  • $[1000, 1500): 2$
  • $[1500, 2000): 2$
  • $[2000, 2500): 2$
  • $[2500, 3000): 0$
  • $[3000, 3500): 0$
  • $[3500, 4000): 1$

Finding Median Position

• In an ordered list of $33$ values, the median is the $\frac{33+1}{2} = 17\text{th}$ value.
• Because the first interval ($[0, 500)$) contains $18$ observations, the $17\text{th}$ value falls completely within this first bin.
• Therefore, the median home price is guaranteed to be less than or equal to $500\,\text{thousand dollars}$ ($\le \$500,000$).

Correct Choice: (C) The median price is not greater than $500,000

Question 9: Conceptual Definition of Standard Deviation

Given standard deviation $S_x = \$788,417$:

• Definition: Standard deviation measures the characteristic or typical distance between individual data values and the sample mean.

Correct Choice: (A) The approximate mean distance between the listed home price and the mean home price.

Question 10: Transforming Stemplot Data into Binned Histograms

Stemplot Data Conversion

• Interval $[20, 40):$ Stems 2 & 3 ($28, 32, 35, 38$) $= 4\,\text{observations}$
• Interval $[40, 60):$ Stems 4 & 5 ($41, 43, 45, 47, 49, 52, 54, 56, 58$) $= 9\,\text{observations}$
• Interval $[60, 80):$ Stems 6 & 7 ($60, 62, 65, 67, 69, 72, 74, 76, 79$) $= 9\,\text{observations}$
• Interval $[80, 100):$ Stems 8 & 9 ($82, 85, 88, 92, 95$) $= 5\,\text{observations}$
• Interval $[100, 120):$ Stems 10 & 11 ($105, 110, 118$) $= 3\,\text{observations}$

Histogram (A) exactly matches these bar heights ($4, 9, 9, 5, 3$).

Correct Choice: (A)

Free Response Practice: Solutions & Analysis

Question 11: Fuel Efficiency Comparison (Small Cars vs. Sporty Cars)

Data Sets

• Small Cars ($n = 14$): $33, 33, 33, 34, 34, 34, 34, 35, 36, 37, 37, 37, 45, 50$
• Sporty Cars ($n = 27$): $29, 29, 29, 29, 29, 29, 29, 29, 30, 30, 30, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31, 32, 32, 32, 32, 33$

Part a) Comparative Graph Construction

To graph and compare the distributions, calculate the five-number summary for both vehicle types:

Back-to-Back Stemplot Comparison

Key: $3 \mid 4 = 34\,\text{mpg}$

\begin{array}{r|c|l}\n\text{Small Cars} & \text{Stem} & \text{Sporty Cars} \\\n\hline\n& 2 & 9\ 9\ 9\ 9\ 9\ 9\ 9\ 9 \\\n7\ 7\ 7\ 6\ 5\ 4\ 4\ 4\ 4\ 3\ 3\ 3 & 3 & 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 1\ 1\ 1\ 1\ 1\ 2\ 2\ 2\ 2\ 3 \\\n5\ 0 & 4 & \\\n& 5 &\n\end{array}

Part b) Comprehensive Comparison of Distributions (SOCS)

• Shape: The distribution of fuel efficiencies for small cars is strongly skewed to the right due to extreme high values ($45$ and $50\,\text{mpg}$). In contrast, the distribution for sporty cars is roughly unimodal and symmetric, centered around $30\,\text{mpg}$.
• Outliers: Small cars have two high outliers at $45\,\text{mpg}$ and $50\,\text{mpg}$. Sporty cars contain no outliers.
• Center: Small cars exhibit higher median fuel efficiency ($34.5\,\text{mpg}$) and mean ($\bar{x} \approx 37.29\,\text{mpg}$) than sporty cars, which have a median of $30\,\text{mpg}$ and mean ($\bar{x} = 30.33\,\text{mpg}$).
• Spread: Fuel efficiency in small cars is significantly more variable than in sporty cars across all measures:
  • Range: Small cars range $17\,\text{mpg}$ ($50 - 33$) vs. Sporty cars range $4\,\text{mpg}$ ($33 - 29$).
  • IQR: Small cars $IQR = 3\,\text{mpg}$ ($37 - 34$) vs. Sporty cars $IQR = 2\,\text{mpg}$ ($31 - 29$).
  • Standard Deviation: Small cars $S_x \approx 5.14\,\text{mpg}$ vs. Sporty cars $S_x \approx 1.21\,\text{mpg}$.

Part c) Outlier Identification via 1.5 x IQR Rule for Small Cars

1. Determine $Q_1$, $Q_3$, and $IQR$:

   • $Q_1 = 34\,\text{mpg}$
   • $Q_3 = 37\,\text{mpg}$
   • $IQR = Q_3 - Q_1 = 37 - 34 = 3\,\text{mpg}$

2. Calculate Outlier Fences:

   • $1.5 \times IQR = 1.5 \times 3 = 4.5\,\text{mpg}$
   • $\text{Lower Fence} = Q_1 - 4.5 = 34 - 4.5 = 29.5\,\text{mpg}$
   • $\text{Upper Fence} = Q_3 + 4.5 = 37 + 4.5 = 41.5\,\text{mpg}$

3. Identify Outliers:

   • Data values less than $29.5\,\text{mpg}$ or greater than $41.5\,\text{mpg}$ are classified as outliers.
   • Minimum value is $33\,\text{mpg} \ge 29.5\,\text{mpg}$ (no lower outliers).
   • Values $45\,\text{mpg}$ and $50\,\text{mpg}$ exceed $41.5\,\text{mpg}$.

Conclusion: There are two high outliers in the small cars data set: 45 mpg and 50 mpg.

Question 12: Normal Distributions and Comparative Standardization ($z$-Scores)

Given summary table for Normally Distributed Exam Scores:

Standardized score formula:

$z = \frac{x - \mu}{\sigma}$

Part a) Comparative Performance Analysis

• Nicole's Performance (Morning Class, score = $90$):

$z_{\text{Nicole}} = \frac{90 - 82.6}{4.65} = \frac{7.4}{4.65} \approx 1.5914$

Nicole scored approximately $1.59$ standard deviations above her class mean.

• Steven's Performance (Afternoon Class, score = $87$):

$z_{\text{Steven}} = \frac{87 - 75.8}{7.48} = \frac{11.2}{7.48} \approx 1.4973$

Steven scored approximately $1.50$ standard deviations above his class mean.

Conclusion & Justification

Nicole's score is more impressive. Because both distributions are normally distributed, Nicole's higher $z$-score ($1.59 > 1.50$) indicates that her score lies relatively farther above her class average than Steven's score does relative to his class average.

Part b) Finding an Equivalent Comparative Score

1. Calculate $z$-score for a student scoring 75 in the Morning Class:

$z_{\text{morning}} = \frac{75 - 82.6}{4.65} = \frac{-7.6}{4.65} \approx -1.6344$

2. Equate $z$-scores and solve for equivalent score ($x$) in the Afternoon Class:

$z_{\text{afternoon}} = \frac{x - 75.8}{7.48} = -1.6344$

$x - 75.8 = -1.6344 \times 7.48$

$x - 75.8 = -12.2253$

$x = 75.8 - 12.2253 = 63.5747$

Conclusion

A student in the afternoon class would need a score of approximately 63.57 (or 63.6) to achieve the same comparative relative grade.