Estimation and Confidence Intervals
- Understand the difference between point estimation and interval estimation.
- Learn how confidence intervals provide a range of likely values.
- Compute confidence intervals for means and proportions.
- Visualize confidence intervals with probability distributions.
Point Estimation vs. Interval Estimation
Point estimation provides a single value as an estimate of an unknown population parameter.
- Example: Using the sample mean \( \bar{x} \) to estimate the population mean \( \mu \).
Interval estimation gives a range of values where the parameter is likely to be found, with a confidence level.
Confidence Intervals
A confidence interval (CI) is an interval estimate of a population parameter. It has the form:
\[ \text{Estimate} \pm (\text{Margin of Error}) \]The general formula for a confidence interval for the population mean \( \mu \) is:
\[ \bar{x} \pm z^* \frac{\sigma}{\sqrt{n}} \]where:
- \( \bar{x} \) = sample mean
- \( \sigma \) = population standard deviation (or \( s \) if unknown)
- \( n \) = sample size
- \( z^* \) = critical value from the standard normal table
Common Confidence Levels and z-values:
| Confidence Level | z-value |
|---|---|
| 90% | 1.645 |
| 95% | 1.960 |
| 99% | 2.576 |
Derivation
From the Central Limit Theorem, the sampling distribution of \( \bar{x} \) follows a normal distribution:
\[ \bar{X} \sim N\left( \mu, \frac{\sigma}{\sqrt{n}} \right) \]Rearrange the probability statement for a normal distribution:
\[ P\left( -z^* \leq \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \leq z^* \right) = C \]Multiplying through by \( \sigma/\sqrt{n} \):
\[ P\left( \bar{x} - z^* \frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{x} + z^* \frac{\sigma}{\sqrt{n}} \right) = C \]This shows the confidence interval formula.
Visualization of Confidence Intervals
Examples
Example 1: A sample of 50 students has an average test score of 78, with a standard deviation of 10. Compute the 95% confidence interval for the population mean.
Using:
\[ \bar{x} = 78, \quad s = 10, \quad n = 50, \quad z^* = 1.96 \] \[ \text{Margin of Error} = 1.96 \times \frac{10}{\sqrt{50}} = 2.77 \] \[ CI = (78 - 2.77, 78 + 2.77) = (75.23, 80.77) \]Interpretation: We are 95% confident that the true mean score is between 75.23 and 80.77.
Exercises
- Question 1: A sample of 40 people has a mean height of 170 cm with a standard deviation of 8 cm. Find the 95% confidence interval for the population mean.
- Question 2: A survey finds that 60% of 500 respondents support a policy. Compute a 99% confidence interval for the proportion.
- Question 3: A sample of size 36 has a mean weight of 65 kg. Assume \( \sigma = 5 \). Compute a 90% confidence interval.
- Question 4: A quality control study finds that 30 defective items are found in a sample of 200. Compute a 95% confidence interval for the defect rate.
- Question 5: A researcher estimates the mean reaction time of a drug to be 2.5 seconds, with a sample of 100 trials and a standard deviation of 0.4. Find a 95% confidence interval.
- Answer 1: \( CI = (167.5, 172.5) \).
- Answer 2: \( CI = (0.553, 0.647) \).
- Answer 3: \( CI = (63.63, 66.37) \).
- Answer 4: \( CI = (0.107, 0.193) \).
- Answer 5: \( CI = (2.42, 2.58) \).