To calculate the 95% confidence interval for the population mean, we use the following formula:

$\text{Confidence Interval} = \bar{x} \pm Z \cdot \frac{\sigma}{\sqrt{n}}$

where:

• $\bar{x}$ is the sample mean (75),
• $Z$ is the Z-score corresponding to the desired confidence level (1.96 for 95%),
• $\sigma$ is the standard deviation of the sample (10),
• $n$ is the sample size (100).

Step-by-Step Calculation

1. Calculate the standard error (SE): $\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{100}} = \frac{10}{10} = 1$

2. Determine the margin of error (MOE): $\text{MOE} = Z \cdot \text{SE} = 1.96 \cdot 1 = 1.96$

3. Construct the confidence interval: $\text{Confidence Interval} = \bar{x} \pm \text{MOE} = 75 \pm 1.96$

   This results in: $(75 - 1.96, 75 + 1.96) = (73.04, 76.96)$

Answer

The 95% confidence interval for the population mean is: $(73.04, 76.96)$

Would you like more details on any part of this calculation, or do you have any questions?

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1. How does the sample size affect the confidence interval?
2. What if the confidence level was changed to 99%? How would that affect the interval?
3. Why do we use the Z-score in this problem instead of the t-score?
4. What would happen if the standard deviation was larger?
5. How do confidence intervals change with smaller sample sizes?

Tip: Remember, increasing the sample size decreases the standard error, resulting in a narrower confidence interval, assuming other factors remain constant.