Math Problem Statement

Discuss the Formula for Confidence Interval with an example

Solution

A Confidence Interval (CI) is a range of values used to estimate the true value of a population parameter, such as the mean or proportion, based on sample data. The formula for a confidence interval depends on the parameter being estimated (e.g., population mean, population proportion) and the type of data (e.g., normal distribution or binomial).

General Formula for Confidence Interval

For a population mean with a known population standard deviation (σ\sigma), the confidence interval is given by:

CI=xˉ±Zα/2×σnCI = \bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}

Where:

  • xˉ\bar{x} = sample mean
  • Zα/2Z_{\alpha/2} = Z-value corresponding to the desired confidence level (e.g., for 95% confidence, Zα/21.96Z_{\alpha/2} \approx 1.96)
  • σ\sigma = population standard deviation
  • nn = sample size

For a population mean with an unknown population standard deviation, the confidence interval is:

CI=xˉ±tα/2×snCI = \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}

Where:

  • tα/2t_{\alpha/2} = t-value from the t-distribution, based on the desired confidence level and degrees of freedom (df=n1df = n - 1)
  • ss = sample standard deviation
  • The rest of the variables are the same as above.

Example of Confidence Interval for a Population Mean

Let’s say you are conducting a study on the average height of a population of adult women. You take a sample of 50 women, and you find the following:

  • Sample mean (xˉ\bar{x}) = 160 cm
  • Sample standard deviation (ss) = 8 cm
  • Sample size (nn) = 50
  • You want a 95% confidence interval.
  1. Find the critical value: For a 95% confidence level, the tt-value for 49 degrees of freedom (since n1=501=49n-1 = 50-1 = 49) can be found from a t-distribution table. For 95% confidence, the tα/2t_{\alpha/2} value is approximately 2.0096.

  2. Calculate the margin of error: ME=tα/2×sn=2.0096×850=2.0096×1.13142.275ME = t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 2.0096 \times \frac{8}{\sqrt{50}} = 2.0096 \times 1.1314 \approx 2.275

  3. Calculate the confidence interval: CI=xˉ±ME=160±2.275CI = \bar{x} \pm ME = 160 \pm 2.275 This gives us: CI=(157.725,162.275)CI = (157.725, 162.275)

Thus, the 95% confidence interval for the population mean height is approximately (157.73 cm, 162.28 cm). This means we are 95% confident that the true mean height of adult women in the population lies within this range.

Interpretation:

This confidence interval implies that if we were to repeatedly sample from the population and compute a confidence interval each time, about 95% of those intervals would contain the true mean height.


Would you like a more detailed breakdown of any of the steps or concepts used?

Here are 5 related questions:

  1. How does the sample size impact the width of the confidence interval?
  2. How do you choose between using a Z-distribution or a t-distribution?
  3. What does it mean when we say a confidence interval has a "95% confidence"?
  4. How can you calculate the confidence interval for a population proportion?
  5. What happens if the confidence interval includes zero?

Tip: The larger your sample size, the narrower the confidence interval, because more data leads to a more precise estimate of the population parameter.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Interval
Sample Mean
Standard Deviation
T-Distribution
Z-Distribution

Formulas

CI = x̄ ± Z_{α/2} × (σ / √n)
CI = x̄ ± t_{α/2} × (s / √n)

Theorems

Central Limit Theorem
Confidence Interval Theorem

Suitable Grade Level

Grades 10-12