AP Statistics Unit 1 Complete Study Guide

Categorical Data Analysis and Two-Way Tables

Two-way table showing mode of transportation by grade level

Variable Identification and Classification:
- Mode of Transportation: Categorical variable with three categories (Bus, Car, Walk/Bike).
- Grade Level: Categorical variable with two categories (9th grade, 12th grade).
Joint Percentage Calculation:
- To find the percentage of students who are in 9th grade and ride the bus, divide the count of 9th-grade bus riders by the overall sample total:
- $\text{Joint Percentage} = \frac{42}{150} \times 100\% = 28\%$
Conditional Distributions of Transportation Mode by Grade Level:
- 9th Grade Group ( $n = 100$ ):
  - $\text{Bus} = \frac{42}{100} = 0.42 = 42\%$
  - $\text{Car} = \frac{36}{100} = 0.36 = 36\%$
  - $\text{Walk/Bike} = \frac{22}{100} = 0.22 = 22\%$
- 12th Grade Group ( $n = 50$ ):
  - $\text{Bus} = \frac{14}{50} = 0.28 = 28\%$
  - $\text{Car} = \frac{26}{50} = 0.52 = 52\%$
  - $\text{Walk/Bike} = \frac{10}{50} = 0.20 = 20\%$
Association Between Grade Level and Mode of Transportation:
- There is an association between grade level and mode of transportation because the conditional distribution of transportation mode varies across grade levels.
- A higher proportion of 9th graders ride the bus ( $42\%$ ) compared to 12th graders ( $28\%$ ).
- A substantially higher proportion of 12th graders travel by car ( $52\%$ ) compared to 9th graders ( $36\%$ ).
- The proportions of students walking or biking are very similar between 9th grade ( $22\%$ ) and 12th grade ( $20\%$ ).
Mosaic Plot Geometry:
- In a mosaic plot, the width of each bar represents the relative sample size of the respective category of the explanatory variable.
- Since there are $100$ 9th graders ( $\frac{100}{150} = \frac{2}{3}$ of the sample) and $50$ 12th graders ( $\frac{50}{150} = \frac{1}{3}$ of the sample), the rectangle for 9th grade will be twice as wide as the rectangle for 12th grade (a $2:1$ width ratio).

Comparing Distributions of Quantitative Data

Boxplots comparing distance walked in October and November

Summary statistics table for October and November walks

Variable Classification:
- Distance Walked (in miles): Quantitative (Continuous) variable.
- Month: Categorical variable with two categories (October, November).
Comparative Analysis of Distributions (October vs. November):
- Center: The median distance walked in November ( $2.11\,\text{miles}$ ) is higher than the median distance walked in October ( $1.57\,\text{miles}$ ). Similarly, the mean distance in November ( $2.277\,\text{miles}$ ) exceeds the mean distance in October ( $2.025\,\text{miles}$ ).
- Spread: The overall range of distance walked is larger in November ( $\text{Range} = 3.02 - 0.53 = 2.49\,\text{miles}$ ) than in October ( $\text{Range} = 3.25 - 1.23 = 2.02\,\text{miles}$ ). The standard deviation is slightly higher in November ( $0.741\,\text{miles}$ ) than in October ( $0.707\,\text{miles}$ ). However, the interquartile range (IQR) is larger in October ( $IQR = 3.01 - 1.5 = 1.51\,\text{miles}$ ) than in November ( $IQR = 3 - 2 = 1\,\text{mile}$ ).
- Shape: The October distribution is strongly skewed right, as indicated by the median ( $1.57\,\text{miles}$ ) being pulled close to $Q1$ ( $1.5\,\text{miles}$ ) with a long right tail extending to $3.25\,\text{miles}$ . The November distribution is skewed left due to a low minimum value ( $0.53\,\text{miles}$ ), with the middle $50\%$ of distances tightly concentrated between $2\,\text{miles}$ and $3\,\text{miles}$ .
- Outliers: November contains a potential low outlier at $0.53\,\text{miles}$ , whereas October displays no clear outliers.
Combined Data Set Calculations:
- Combined Mean:
  - $\sum x_{\text{Oct}} = 21 \times 2.025 = 42.525\,\text{miles}$
  - $\sum x_{\text{Nov}} = 19 \times 2.277 = 43.263\,\text{miles}$
  - $\text{Total Distance} = 42.525 + 43.263 = 85.788\,\text{miles}$
  - $\text{Total Sessions} = 21 + 19 = 40$
  - $\bar{x}_{\text{combined}} = \frac{85.788}{40} = 2.1447\,\text{miles}$
- Combined Range:
  - $\text{Overall Maximum} = \max(3.25, 3.02) = 3.25\,\text{miles}$
  - $\text{Overall Minimum} = \min(1.23, 0.53) = 0.53\,\text{miles}$
  - $\text{Range}_{\text{combined}} = 3.25 - 0.53 = 2.72\,\text{miles}$

Outlier Rules, Standardized Scores, and Linear Transformations

Verification of Outliers in October ( $1.5 \times IQR$ Rule):
- $Q1 = 1.5\,\text{miles}$
- $Q3 = 3.01\,\text{miles}$
- $IQR = Q3 - Q1 = 3.01 - 1.5 = 1.51\,\text{miles}$
- $1.5 \times IQR = 1.5 \times 1.51 = 2.265\,\text{miles}$
- $\text{Lower Boundary} = Q1 - 1.5 \times IQR = 1.5 - 2.265 = -0.765\,\text{miles}$
- $\text{Upper Boundary} = Q3 + 1.5 \times IQR = 3.01 + 2.265 = 5.275\,\text{miles}$
- Since all observed values in October fall within the range $[-0.765, 5.275]$ (Minimum $= 1.23$ , Maximum $= 3.25$ ), there are no outliers in October.
Identification of Outliers in November ( $2 \times SD$ Rule):
- $\bar{x} = 2.277\,\text{miles}$
- $SD = 0.741\,\text{miles}$
- $2 \times SD = 2 \times 0.741 = 1.482\,\text{miles}$
- $\text{Lower Boundary} = \bar{x} - 2 \times SD = 2.277 - 1.482 = 0.795\,\text{miles}$
- $\text{Upper Boundary} = \bar{x} + 2 \times SD = 2.277 + 1.482 = 3.759\,\text{miles}$
- Because the minimum value of $0.53\,\text{miles}$ is less than $0.795\,\text{miles}$ , $0.53\,\text{miles}$ is identified as an outlier under the $2 \times SD$ rule.
Standardized Score Calculation and Interpretation:
- For a walk of $x = 2.25\,\text{miles}$ on November 21:
- $z = \frac{x - \bar{x}}{SD}$
- $z = \frac{2.25 - 2.277}{0.741} = \frac{-0.027}{0.741} \approx -0.0364$
- Interpretation: The walk distance of $2.25\,\text{miles}$ is approximately $0.0364$ standard deviations below the mean walk distance for the month of November.
Impact of Adding an Additional Data Value:
- An omitted walk of $0.75\,\text{miles}$ is added to October's data set ( $\bar{x} = 2.025$ , $SD = 0.707$ ):
- Effect on Mean: The mean will decrease because the added distance of $0.75\,\text{miles}$ is smaller than the current mean of $2.025\,\text{miles}$ .
- Effect on Standard Deviation: The standard deviation will increase because $0.75\,\text{miles}$ is $1.275\,\text{miles}$ away from the mean, which is significantly further than the current standard deviation of $0.707\,\text{miles}$ , thereby increasing overall variability around the mean.
Linear Transformations of Summary Statistics:
- Transformation formula: $\text{Activity Score} = 10 + 8 \times \text{distance}$
- Median Activity Score:
  - Measures of position are affected by both addition and multiplication.
  - $\text{Median}_{\text{score}} = 10 + 8 \times \text{Median}_{\text{distance}}$
  - $\text{Median}_{\text{score}} = 10 + 8 \times 2.11 = 10 + 16.88 = 26.88$
- Interquartile Range (IQR) of Activity Scores:
  - Measures of spread are affected only by multiplication/division, not by adding constant base values.
  - $IQR_{\text{distance}} = Q3 - Q1 = 3 - 2 = 1\,\text{mile}$
  - $IQR_{\text{score}} = 8 \times IQR_{\text{distance}}$
  - $IQR_{\text{score}} = 8 \times 1 = 8$