Data Diagnostics Worksheet
DATA DIAGNOSTICS WORKSHEET
Precalculus • Analyzing Univariate Data, Spread & Outliers
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Date:
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Sample Mean (\(\bar{x}\)) \(\bar{x} = \frac{1}{n}\sum x_i\)
Sample Std. Dev. (\(s\)) \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}\)
Spread & IQR \(\text{IQR} = Q_3 - Q_1\)
1.5 × IQR Fences \(Q_1 - 1.5(\text{IQR}) \text{ to } Q_3 + 1.5(\text{IQR})\)
Case Study 1: Drone Battery Lifespan Testing Sample Size: \(n = 10\)
An engineering lab records the continuous flight time (in hours) of 10 identical lithium-polymer drone prototypes until complete discharge under controlled conditions:
12, 14, 15, 15, 16, 17, 18, 19, 21, 33
1 Five-Number Summary Computation
Determine each parameter for the ordered sample above. Show any mid-point calculations.
Minimum
First Quartile (\(Q_1\))
Median (\(\tilde{x}\))
Third Quartile (\(Q_3\))
Maximum
2 Interquartile Range & Outlier Boundary Test
a) Calculate \(\text{IQR} = Q_3 - Q_1\):
b) Calculate \(1.5 \times \text{IQR}\):
c) Lower Fence \(= Q_1 - 1.5(\text{IQR})\):
d) Upper Fence \(= Q_3 + 1.5(\text{IQR})\):
e) Conclusion: Are there statistical outliers? Justify by comparing point(s) to the fences:
3 Statistical Sensitivity Analysis
Calculate the sample mean \(\bar{x}\) of the 10 batteries. Compare it to the median \(\tilde{x}\). Which statistic better represents typical flight time, and why?
Sample Mean \(\bar{x}\):
Comparison (\(\bar{x}\) vs \(\tilde{x}\)) & Skewness Direction:
Resistance to Extreme Values Explanation:
Precalculus • Section 15.4 Data Analysis Page 1 of 2
GRAPHICAL PROFILES & VARIANCE MECHANICS
Modified Boxplots • Sum of Squares • Linear Transformations
Data Diagnostics Worksheet
4 Construct a Modified Box-and-Whisker Plot
Dataset from Page 1
Plot the 5-number summary on the scaled axis. Mark outliers with an asterisk (*) and terminate the right whisker at the highest data value that is not an outlier.
Plot whiskers, box, median, and outlier marker here:
10
12
14
16
18
20
22
24
26
28
30
32
34
36
What value marks the end of the upper whisker? Explain why:
5 Sample Variance & Standard Deviation Calculation
Sample: 5, 8, 9, 12, 16
Complete the deviation table for this 5-point dataset to determine the sample variance \(s^2\) and standard deviation \(s\).
| Data Value (\(x_i\)) | Mean (\(\bar{x}\)) | Deviation \((x_i - \bar{x})\) | Squared Dev. \((x_i - \bar{x})^2\) |
|---|
| 5 | | | |
| 8 | | | |
| 9 | | | |
| 12 | | | |
| 16 | | | |
| \(\sum x_i = \) | \(\sum(x_i - \bar{x}) = 0\) | \(\sum(x_i - \bar{x})^2 = \) | |
Sample Variance: \(s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}\)
Sample Standard Deviation: \(s = \sqrt{s^2}\)
6 Analytical Extension: Linear Transformations of Data
A dataset has sample mean \(\bar{x} = 24\) and standard deviation \(s = 4.5\). Each value is transformed into \(y_i = 3x_i - 10\).
a) Find new mean \(\bar{y}\) (show calculation):
b) Find new std. dev. \(s_y\) (show calculation):
c) Explain why shifting by a constant changes the mean but has zero effect on the standard deviation:
Precalculus • Section 15.4 Data Analysis Page 2 of 2
Data Diagnostics Answer Key
DATA DIAGNOSTICS • ANSWER KEY
Teacher Reference
Precalculus • Section 15.4 Data Analysis Solutions & Rubrics
Complete Solution Guide
Grading • 25 Total Points
Common Misconception 1
Students often extend the right whisker to the fence (25) or to the outlier (33) rather than the largest non-outlier data point (21).
Common Misconception 2
In linear transformations \(y = 3x - 10\), students often mistakenly subtract 10 from standard deviation. Constant shifts have 0 effect on spread.
Degrees of Freedom Notice
Emphasize division by \(n - 1 = 4\) for sample variance \(s^2\), not \(n = 5\) (which yields biased population variance \(\sigma^2\)).
Dataset: Drone Battery Hours (\(n = 10\)) Ordered: 12, 14, 15, 15, 16, 17, 18, 19, 21, 33
1 Five-Number Summary Computation
[5 Points • 1 pt each]
Lower half: \(\{12, 14, 15, 15, 16\}\) • Upper half: \(\{17, 18, 19, 21, 33\}\)
Minimum
12
First Quartile (\(Q_1\))
15
Median (\(\tilde{x}\))
16.5
\(\frac{16+17}{2}\)
Third Quartile (\(Q_3\))
19
Maximum
33
2 Interquartile Range & Outlier Boundary Test
[5 Points • 1 pt each]
a) \(\text{IQR} = Q_3 - Q_1\): \(19 - 15 = 4\text{ hours}\)
b) \(1.5 \times \text{IQR}\): \(1.5 \times 4 = 6\)
c) Lower Fence \(= Q_1 - 1.5(\text{IQR})\): \(15 - 6 = 9\text{ hours}\)
d) Upper Fence \(= Q_3 + 1.5(\text{IQR})\): \(19 + 6 = 25\text{ hours}\)
e) Conclusion & Formal Inequality Justification:
Yes, the observation \(x = 33\) is a high outlier because \(33 > 25\) (it strictly exceeds the upper fence). There are no low outliers because all remaining points are greater than the lower fence (\(\text{Min} = 12 \ge 9\)). Outlier threshold: \(x \notin [9, 25]\).
3 Statistical Sensitivity Analysis
[5 Points total]
Sample Mean \(\bar{x}\): \(\bar{x} = \frac{180}{10} = 18.0\text{ hrs}\)
Center Comparison & Skewness: \(\bar{x} (18.0) > \tilde{x} (16.5) \implies\) Right-Skewed (Positively Skewed).
Resistance Explanation & Real-World Recommendation:
The median (16.5 hrs) is the superior measure of center. The mean is a non-resistant statistic heavily inflated by the single outlier at 33 hours. In reality, 70% of prototype units lasted \(\le 17\) hours. Reporting a mean of 18 hours would mislead drone customers.
