Which Professor? Suppose Professor Alpha and Professor Omega each teach Introductory Biology. You need to decide which professor to take the class from and have just completed your Introductory Statistics course. Records obtained from past students indicate that students in Professor Alpha’s class have a mean score of 80% with a standard deviation of 5%, while past students in Professor Omega’s class have a mean score of 80% with a standard deviation of 10%. Decide which instructor to take for Introductory Biology using a statistical argument.
Table of contents
- 1. Intro to Stats and Collecting Data1h 14m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically2h 5m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables3h 6m
- 6. Normal Distribution and Continuous Random Variables2h 11m
- 7. Sampling Distributions & Confidence Intervals: Mean3h 23m
- Sampling Distribution of the Sample Mean and Central Limit Theorem19m
- Distribution of Sample Mean - Excel23m
- Introduction to Confidence Intervals15m
- Confidence Intervals for Population Mean1h 18m
- Determining the Minimum Sample Size Required12m
- Finding Probabilities and T Critical Values - Excel28m
- Confidence Intervals for Population Means - Excel25m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 25m
- 9. Hypothesis Testing for One Sample3h 29m
- 10. Hypothesis Testing for Two Samples4h 50m
- Two Proportions1h 13m
- Two Proportions Hypothesis Test - Excel28m
- Two Means - Unknown, Unequal Variance1h 3m
- Two Means - Unknown Variances Hypothesis Test - Excel12m
- Two Means - Unknown, Equal Variance15m
- Two Means - Unknown, Equal Variances Hypothesis Test - Excel9m
- Two Means - Known Variance12m
- Two Means - Sigma Known Hypothesis Test - Excel21m
- Two Means - Matched Pairs (Dependent Samples)42m
- Matched Pairs Hypothesis Test - Excel12m
- 11. Correlation1h 24m
- 12. Regression1h 50m
- 13. Chi-Square Tests & Goodness of Fit2h 21m
- 14. ANOVA1h 57m
3. Describing Data Numerically
Standard Deviation
Problem 3.4.12
Textbook Question
Triathlon Roberto finishes a triathlon (750-meter swim, 5-kilometer run, and 20-kilometer bicycle) in 63.2 minutes. Among all men in the race, the mean finishing time was 69.4 minutes with a standard deviation of 8.9 minutes. Zandra finishes the same triathlon in 79.3 minutes. Among all women in the race, the mean finishing time was 84.7 minutes with a standard deviation of 7.4 minutes. Who did better in relation to their gender?
Verified step by step guidance1
Identify the relevant statistics for each athlete's gender group: for Roberto (men), the mean finishing time \(\mu_m = 69.4\) minutes and standard deviation \(\sigma_m = 8.9\) minutes; for Zandra (women), the mean finishing time \(\mu_w = 84.7\) minutes and standard deviation \(\sigma_w = 7.4\) minutes.
Calculate the z-score for Roberto's finishing time using the formula:
\[Z = \frac{X - \mu}{\sigma}\]
where \(X\) is Roberto's time (63.2 minutes), \(\mu\) is the men's mean, and \(\sigma\) is the men's standard deviation.
Calculate the z-score for Zandra's finishing time using the same formula, but with the women's statistics:
\[Z = \frac{X - \mu}{\sigma}\]
where \(X\) is Zandra's time (79.3 minutes), \(\mu\) is the women's mean, and \(\sigma\) is the women's standard deviation.
Interpret the z-scores: since finishing time is a measure where lower is better, a more negative z-score indicates a better performance relative to the group. Compare Roberto's and Zandra's z-scores to see who performed better relative to their gender group.
Conclude who did better by identifying whose z-score is lower (more negative), indicating a better relative performance in the triathlon.
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
3mPlay a video:
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Z-Score
A z-score measures how many standard deviations a data point is from the mean of its distribution. It standardizes different data points, allowing comparison across different groups or scales by showing relative performance.
Recommended video:
Guided course
Z-Scores From Given Probability - TI-84 (CE) Calculator
Mean and Standard Deviation
The mean is the average value of a dataset, while the standard deviation quantifies the amount of variation or dispersion from the mean. Together, they describe the central tendency and spread of finishing times in the race.
Recommended video:
Guided course
Calculating Standard Deviation
Comparing Performance Across Groups
To compare individuals from different groups with distinct averages and variability, we use standardized scores like z-scores. This approach accounts for group differences, enabling fair comparison of relative performance.
Recommended video:
Performing Hypothesis Tests: Proportions Example 1
Watch next
Master Calculating Standard Deviation with a bite sized video explanation from Patrick
Start learningRelated Videos
Related Practice
Textbook Question
8
views
