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
7. Sampling Distributions & Confidence Intervals: Mean
Introduction to Confidence Intervals
Problem 5.4.34
Textbook Question
Which Is More Likely? Assume that the fertility rates in Exercise 32 are normally distributed. Are you more likely to randomly select a state with a fertility rate of less than 65 or to randomly select a sample of 15 states in which the mean of the state fertility rates is less than 65? Explain.

1
Step 1: Identify the key components of the problem. The problem involves comparing probabilities under a normal distribution. Specifically, we are comparing the likelihood of a single state's fertility rate being less than 65 versus the likelihood of the mean fertility rate of a sample of 15 states being less than 65.
Step 2: Recall the properties of the normal distribution. For a single state's fertility rate, the distribution is given as N(μ, σ²), where μ is the population mean and σ is the population standard deviation. For the sample mean of 15 states, the sampling distribution of the mean is also normal, but with a mean μ and a standard error of σ/√n, where n is the sample size.
Step 3: Standardize the values to calculate probabilities. For a single state, calculate the z-score using the formula z = (X - μ) / σ, where X = 65. For the sample mean, calculate the z-score using the formula z = (X̄ - μ) / (σ / √n), where X̄ = 65 and n = 15.
Step 4: Compare the two z-scores. The z-score for the sample mean will generally be larger in magnitude (further from 0) than the z-score for a single state because the standard error (σ/√n) is smaller than the population standard deviation (σ). This means the probability of the sample mean being less than 65 is smaller than the probability of a single state's fertility rate being less than 65.
Step 5: Conclude the comparison. Since the z-score for the sample mean is larger in magnitude, the area under the normal curve to the left of 65 (i.e., the probability) is smaller for the sample mean. Therefore, it is more likely to randomly select a single state with a fertility rate less than 65 than to randomly select a sample of 15 states with a mean fertility rate less than 65.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Normal Distribution
Normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In this context, understanding that fertility rates are normally distributed allows us to apply statistical methods to determine probabilities related to these rates.
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Sampling Distribution
The sampling distribution is the probability distribution of a statistic (like the mean) obtained from a large number of samples drawn from a specific population. When selecting a sample of 15 states, the mean fertility rate will have its own distribution, which can be analyzed to determine the likelihood of it being less than a certain value, such as 65.
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Central Limit Theorem
The Central Limit Theorem states that the distribution of the sample mean will approach a normal distribution as the sample size increases, regardless of the shape of the population distribution. This theorem is crucial for understanding why the mean of a sample of 15 states can be analyzed using normal distribution properties, even if the original data is not perfectly normal.
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