Chapter 10 Part 1: Point and Interval Estimation
Welcome back to math class! Today, we're starting Chapter 10, which covers the fascinating world of estimation. We'll explore how to estimate population parameters using sample data, focusing on point estimation and interval estimation. Remember, understanding these concepts is crucial for making informed decisions based on data.
10.1 Point Estimation of the Population Mean
A point estimator is a single value that estimates a population parameter. Here are some common point estimators:
- Sample Mean ($\bar{x}$): Estimates the population mean ($\mu$)
- Sample Proportion ($\hat{p}$): Estimates the population proportion ($p$)
- Sample Standard Deviation ($s$): Estimates the population standard deviation ($\sigma$)
10.2 Interval Estimation of the Population Mean
An interval estimate provides a range of values within which the population parameter is likely to fall. This range is called a confidence interval.
When the population standard deviation ($\sigma$) is known and the sample is drawn from a normal population, or if the sample size ($n$) is greater than 30, the confidence interval for the population mean is calculated as follows:
$$ \bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} $$Where:
- $\bar{x}$ is the sample mean
- $z_{\alpha/2}$ is the critical value from the standard normal distribution corresponding to the desired confidence level (e.g., for a 95% confidence interval, $z_{\alpha/2} = 1.96$)
- $\sigma$ is the population standard deviation
- $n$ is the sample size
Let's work through some examples!
Example 1: Wall Clock Lifespan
A production manager wants to test new wall clocks. The designer claims they have a mean life of 16 years with a variance of 25. If the claim is true, and we take a sample of 46 wall clocks, what is the probability that the mean clock life will be greater than 17.4 years? (Round to four decimal places.)
Here's how we solve it:
First, calculate the Z-score:
$$ Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} $$ $$ Z = \frac{17.4 - 16}{\frac{5}{\sqrt{46}}} = \frac{1.4}{\frac{5}{6.782}} = \frac{1.4}{0.737} \approx 1.90 $$Using a Z-table, we find that the area to the left of Z = 1.90 is 0.9713. Therefore, the probability that the mean clock life would be *greater* than 17.4 years is:
$$ 1 - 0.9713 = 0.0287 $$Example 2: Airborne Viruses
A scientist claims that 7% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 685 viruses would be greater than 8%? (Round to four decimal places.)
We use a similar approach, but with proportions:
$$ Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} $$ $$ Z = \frac{0.08 - 0.07}{\sqrt{\frac{0.07(0.93)}{685}}} = \frac{0.01}{\sqrt{\frac{0.0651}{685}}} = \frac{0.01}{\sqrt{0.000095}} = \frac{0.01}{0.00975} \approx 1.03 $$The area to the left of Z = 1.03 is approximately 0.8485. Thus, the probability of the proportion being *greater* than 8% is:
$$ 1 - 0.8485 = 0.1515 $$Example 3: Voter Preferences
Suppose a sample of 500 is used to estimate the fraction of voters favoring a candidate. If the true population proportion is 0.4, what is the probability that the error of estimation will be less than 0.05?
We're looking for the probability that the sample proportion falls between 0.35 and 0.45.
$$ Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} $$ For $\hat{p} = 0.45$: $$ Z = \frac{0.45 - 0.40}{\sqrt{\frac{0.4(0.6)}{500}}} = \frac{0.05}{\sqrt{0.00048}} \approx 2.28 $$ For $\hat{p} = 0.35$: $$ Z = \frac{0.35 - 0.40}{\sqrt{\frac{0.4(0.6)}{500}}} = \frac{-0.05}{\sqrt{0.00048}} \approx -2.28 $$The area between -2.28 and 2.28 is approximately $0.9884 - (1 - 0.9884) = 0.9774 $. Therefore, the probability is about 97.74%.
Keep practicing, and you'll master these concepts in no time! Good luck with your studies!