Understanding Confidence Intervals for Normal Mean

This question asks us to calculate a 95% confidence interval for the unknown mean ($\mu$) of a normal distribution. We are given specific information about the distribution and a sample drawn from it.

• The population distribution is normal.
• The population variance ($\sigma^2$) is known and equals 1.
• The sample size ($n$) is 100.
• The sample mean ($\bar{x}$) is 5.
• The desired confidence level is 95%.

Since the population variance ($\sigma^2$) is known and the sample size is large ($n \ge 30$) (or the population is normal, which is given here), we use the z-distribution to construct the confidence interval for the population mean ($\mu$).

Formula for Confidence Interval (Known Variance)

The formula for a $(1 - \alpha) \times 100\%$ confidence interval for the population mean ($\mu$) when the population variance ($\sigma^2$) is known is:

Where:

• \(\bar{x}\) is the sample mean.
• \(\sigma\) is the population standard deviation (the square root of the variance).
• \(n\) is the sample size.
• \(z_{\alpha/2}\) is the z-score corresponding to the desired confidence level. It is the value such that the area to its right under the standard normal curve is \(\alpha/2\).

Step-by-Step Calculation

Let's plug in the given values into the formula.

1. Identify the given values:
   • Sample mean, \(\bar{x} = 5\).
   • Population variance, \(\sigma^2 = 1\).
   • Population standard deviation, \(\sigma = \sqrt{1} = 1\).
   • Sample size, \(n = 100\).
   • Confidence level = 95%, which means \(1 - \alpha = 0.95\).
2. Calculate the significance level, \(\alpha\):
   • \(\alpha = 1 - 0.95 = 0.05\).
3. Find \(\alpha/2\):
   • \(\alpha/2 = 0.05 / 2 = 0.025\).
4. Find the critical z-score, \(z_{\alpha/2} = z_{0.025}\):
   • For a 95% confidence interval, we need the z-score that leaves 0.025 area in the upper tail (and 0.025 in the lower tail), total \(\alpha = 0.05\) in both tails. This corresponds to a cumulative area of \(1 - \alpha/2 = 1 - 0.025 = 0.975\) to the left of \(z_{\alpha/2}\).
   • Looking up 0.975 in the standard normal (z) table, we find that the corresponding z-score is 1.96. So, \(z_{0.025} = 1.96\).
5. Calculate the standard error of the mean:
   • Standard Error \( = \frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt{100}} = \frac{1}{10} = 0.1 \).
6. Calculate the margin of error:
   • Margin of Error \( = z_{\alpha/2} \times \text{Standard Error} = 1.96 \times 0.1 = 0.196 \).
7. Construct the confidence interval:
   • Lower bound \( = \bar{x} - \text{Margin of Error} = 5 - 0.196 = 4.804 \).
   • Upper bound \( = \bar{x} + \text{Margin of Error} = 5 + 0.196 = 5.196 \).

The 95% confidence interval for \(\mu\) is \((4.804, 5.196)\).

Confidence Interval Result

Based on the sample data and the known variance, we are 95% confident that the true population mean (\(\mu\)) lies between 4.804 and 5.196.

Comparing with Options

Let's compare our calculated interval \((4.804, 5.196)\) with the given options:

Our calculated interval matches Option 2.

Revision Table: Key Concepts

Additional Information on Confidence Intervals

Constructing confidence intervals is a fundamental aspect of inferential statistics. It allows us to estimate population parameters based on sample data and quantify the uncertainty of our estimate.

• The width of the confidence interval is influenced by the confidence level, the standard deviation (or estimate), and the sample size.
• A higher confidence level (e.g., 99% vs. 95%) leads to a wider interval, as we need to be more certain that the interval contains the true parameter.
• A larger sample size ($n$) leads to a narrower interval, as the sample mean becomes a more precise estimate of the population mean.
• If the population variance ($\sigma^2$) is unknown and the sample size is small ($n < 30$), the t-distribution is used instead of the z-distribution.
• The interpretation of a 95% confidence interval is that if we were to take many random samples and compute a confidence interval for each, about 95% of those intervals would contain the true population mean.