To compute a 95% confidence interval (CI) for the population mean given that the population is normally distributed with a known standard deviation $\sigma = 5$, a sample mean $\bar{X} = 20$, and a sample size $N = 25$, we'll follow these steps:

Step-by-Step Process:

1. Determine the Z-score for a 95% confidence level:

   • For a 95% confidence interval, the corresponding Z-score (critical value) is found using standard normal distribution tables. The Z-value for 95% confidence is approximately $Z = 1.96$.

2. Calculate the standard error of the mean (SE): $SE = \frac{\sigma}{\sqrt{N}} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1$

3. Compute the margin of error (ME): $ME = Z \times SE = 1.96 \times 1 = 1.96$

4. Calculate the confidence interval bounds: The formula for the confidence interval is: $\left( \bar{X} - ME, \bar{X} + ME \right)$ Substituting the values: $\left( 20 - 1.96, 20 + 1.96 \right) = \left( 18.04, 21.96 \right)$

Conclusion:

The 95% confidence interval for the population mean is $(18.04, 21.96)$. This means we are 95% confident that the true population mean lies within this interval.

Interpretation:

If we were to take many samples of size 25 and compute a 95% confidence interval for each sample, approximately 95% of those intervals would contain the true population mean.

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Would you like further details on the steps or any clarifications?

Related Questions:

1. What is the relationship between confidence intervals and sample size?
2. How does increasing the confidence level affect the confidence interval width?
3. Why do we use Z-scores instead of T-scores in this case?
4. What are the implications of having a larger standard deviation in confidence interval estimation?
5. How would the confidence interval change if we had a larger sample size?

Tip:

Always check whether the population standard deviation is known or unknown before deciding to use Z-scores or T-scores when constructing confidence intervals.