Table of contents
- 1. Intro to Stats and Collecting Data55m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically1h 45m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables2h 33m
- 6. Normal Distribution and Continuous Random Variables1h 38m
- 7. Sampling Distributions & Confidence Intervals: Mean1h 3m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- 9. Hypothesis Testing for One Sample1h 1m
- 10. Hypothesis Testing for Two Samples2h 8m
- 11. Correlation48m
- 12. Regression1h 4m
- 13. Chi-Square Tests & Goodness of Fit1h 20m
- 14. ANOVA1h 0m
6. Normal Distribution and Continuous Random Variables
Standard Normal Distribution
Problem 6.6.16b
Textbook Question
Eye Color Based on a study by Dr. P. Sorita at Indiana University, assume that 12% of us have green eyes. In a study of 650 people, it is found that 86 of them have green eyes.
b. Is 86 people with green eyes significantly high?

1
Step 1: Define the problem in terms of probability. We are tasked with determining if 86 people with green eyes out of 650 is significantly high, given that the population proportion of green eyes is 12% (p = 0.12). This involves using the concept of a sampling distribution for proportions.
Step 2: Calculate the mean of the sampling distribution for the proportion of green eyes. The mean is given by \( \mu_p = p \), where \( p = 0.12 \).
Step 3: Calculate the standard deviation of the sampling distribution for the proportion. The formula is \( \sigma_p = \sqrt{\frac{p(1-p)}{n}} \), where \( p = 0.12 \), \( 1-p = 0.88 \), and \( n = 650 \).
Step 4: Convert the observed number of green-eyed people (86) into a sample proportion \( \hat{p} \). The formula is \( \hat{p} = \frac{x}{n} \), where \( x = 86 \) and \( n = 650 \).
Step 5: Compute the z-score to determine how many standard deviations \( \hat{p} \) is from \( \mu_p \). The formula is \( z = \frac{\hat{p} - \mu_p}{\sigma_p} \). Compare the z-score to a critical value (e.g., \( z = 1.96 \) for a 5% significance level) to decide if 86 is significantly high.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hypothesis Testing
Hypothesis testing is a statistical method used to determine if there is enough evidence to reject a null hypothesis. In this context, the null hypothesis might state that the proportion of people with green eyes in the sample is equal to the expected proportion of 12%. By comparing the observed number of individuals with green eyes to what is expected, we can assess whether the difference is statistically significant.
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Step 1: Write Hypotheses
Significance Level
The significance level, often denoted as alpha (α), is the threshold for determining whether a result is statistically significant. Commonly set at 0.05, it represents a 5% risk of concluding that a difference exists when there is none. In this scenario, if the proportion of green-eyed individuals in the sample significantly exceeds the expected 12%, we would reject the null hypothesis at the chosen significance level.
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Step 4: State Conclusion Example 4
Proportion and Sample Size
Proportion refers to the fraction of a population that exhibits a certain characteristic, such as having green eyes. In this case, we compare the observed proportion of green-eyed individuals (86 out of 650) to the expected proportion of 12%. The sample size is crucial because larger samples tend to provide more reliable estimates of population parameters, affecting the power of the hypothesis test.
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Sampling Distribution of Sample Proportion
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