Interval Estimation for Population Mean (Sigma Known)

TL;DR

When you know the population standard deviation ($\sigma$), you can estimate the population mean ($\mu$) using an interval, not just a single point. This interval gives a range where we're confident the true mean lies. The width of this interval depends on your desired confidence level and the margin of error.

1. The Mental Model

Imagine trying to guess someone's exact height. It's hard! But saying their height is "between 5'8" and 5'10"" gives you a confident range. Interval estimation does the same for a population mean, providing a range based on your sample data.

2. The Core Material

Interval estimation helps you construct a range, called an interval estimate, within which you believe the true population mean ($\mu$) lies. Instead of just a single number (a point estimate), you get a lower and upper bound. This is useful because it reflects the uncertainty inherent in using a sample to understand an entire population.

What is an Interval Estimate?

An interval estimate is a range of values used to estimate a population parameter. For example, you might say, "We are 95% confident that the mean rent per month is between $720 and $780."

The general form of an interval estimate for a population mean is:

Sample Mean ± Margin of Error

In this topic, we're focusing on the case where the population standard deviation ($\sigma$) is known. This is important because it tells us which formula and distribution to use.

How to Construct the Interval When $\sigma$ is Known

To create an interval estimate for $\mu$ when $\sigma$ is known, you need to calculate the margin of error. The formula for the interval estimate is:

$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

Where:
* $\bar{x}$ is your sample mean.
* $z_{\alpha/2}$ is the z-score corresponding to your desired confidence level ($1-\alpha$). This value determines how many standard errors you go out from the mean.
* $\sigma$ is the known population standard deviation.
* $n$ is the sample size.

The term $z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ is your margin of error.

Understanding Confidence Levels and $z_{\alpha/2}$

Confidence levels tell you the probability that your interval will contain the true population mean. Common confidence levels are 90%, 95%, and 99%.

$\alpha$ (alpha) is the probability that the interval does not contain the population mean. So, for a 95% confidence level, $\alpha = 0.05$. We divide $\alpha$ by 2 ($\alpha/2$) because the z-values represent the area in the tails of the standard normal distribution.

Here are the $z_{\alpha/2}$ values for common confidence levels:

(Note: Your source material provided "0. 576" for 90% confidence level. The correct standard $z_{\alpha/2}$ for 90% confidence (meaning 0.05 in each tail) is 1.645 when looking up 0.9500 in the Z-table. We'll use the correct value for examples.)

What "Confidence" Means

When we say "we are 95% confident," it means that if we were to take many, many samples and construct an interval estimate for each, about 95% of those intervals would contain the true population mean ($\mu$). It doesn't mean there's a 95% chance the true mean is within this particular interval. It's about the reliability of the method.

graph TD
    A["Does population σ (sigma) (Standard Deviation) be known?"] --> B{Known Case};
    B --> C["Use Z-distribution"];
    C --> D["Formula: x̄ ± Z_α/2 * (σ/√n)"];
    D --> E["Calculate Margin of Error"];
    E --> F["Construct Interval: x̄ - Margin of Error to x̄ + Margin of Error"];
    A --> G{Unknown Case};
    G --> H["Use T-distribution"];
    H --> I["(Covered in a separate topic)"];

3. Worked Example

Let's use the Discount Sounds example provided in your source.

Discount Sounds wants to estimate the mean annual sales for its 260 retail outlets. A sample of stores gives a mean annual sales of $\bar{x} = \$41,100$. Assume the population standard deviation ($\sigma$) for sales is known to be $\$4,000$, and the sample size ($n$) is 40.

Let's construct a 95% confidence interval for the population mean annual sales.

1. Identify knowns:

   • Sample mean ($\bar{x}$) = $\$41,100$
   • Population standard deviation ($\sigma$) = $\$4,000$
   • Sample size ($n$) = 40
   • Confidence level = 95%

2. Find $z_{\alpha/2}$ for 95% confidence:

   • From the table above, for 95% confidence, $z_{\alpha/2} = 1.96$.

3. Calculate the standard error of the mean ($\frac{\sigma}{\sqrt{n}}$):

   • Standard Error = $4000 / \sqrt{40} = 4000 / 6.3245 \approx 632.46$

4. Calculate the margin of error ($z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$):

   • Margin of Error = $1.96 \times 632.46 \approx 1239.62$

   (Note: Your source briefly shows 1,470 for 95% confidence, likely from an example where the sample size or sigma was different. We're directly calculating here with the provided $\sigma$ and n=40. Let's recalculate with n that would yield 1470 margin of Error with the $4000 standard deviation ($1470 / 1.96 = 750$ so $\sigma/\sqrt{n} = 750 => \sqrt{n} = 4000/750 = 5.333 => n = 28.44$. If $n=28$, margin of error is $1.96 * (4000/\sqrt{28}) = 1.96 * 755.93 = 1481.6$. We will stick to the $n=40$ given in the example in your source text for the sample calculation.)

5. Construct the interval estimate:

   • Lower limit = $\bar{x}$ - Margin of Error = $\$41,100 - \$1239.62 = \$39,860.38$
   • Upper limit = $\bar{x}$ + Margin of Error = $\$41,100 + \$1239.62 = \$42,339.62$

So, we are 95% confident that the mean annual sales for Discount Sounds outlets are between $\$39,860.38$ and $\$42,339.62$.

4. Key Takeaways

• An interval estimate provides a range rather than a single value for the population mean.
• This method specifically applies when the population standard deviation ($\sigma$) is known.
• The confidence level determines the $z_{\alpha/2}$ value, influencing the width of your interval.
• A higher confidence level (e.g., 99% vs. 95%) results in a wider interval because you need to be "more sure."
• The margin of error quantifies the precision of your estimate.
• The formula relies on the sample mean, known population standard deviation, sample size, and the appropriate z-score.
• $z_{\alpha/2}$ marks the boundary for the central area of the standard normal distribution corresponding to your confidence level.

Common mistakes to avoid:
- Confusing $\sigma$ (population standard deviation) with $s$ (sample standard deviation). This topic only uses $\sigma$.
- Using the wrong $z_{\alpha/2}$ value for your chosen confidence level.
- Misinterpreting confidence: It's about the method's reliability, not the probability that this specific interval contains the mean.
- Forgetting to divide $\alpha$ by 2 when finding $z_{\alpha/2}$.

5. Now Try It

You've taken a sample of 30 customer wait times at a call center, finding a sample mean of 6.2 minutes. From historical data, the population standard deviation ($\sigma$) of wait times is known to be 2 minutes. Provide a 90% confidence interval for the true population mean wait time. What values did you get for the margin of error, and the lower and upper bounds of the interval?