To construct a 95% confidence interval for the population mean $\mu$, we will use the formula for the confidence interval:

$CI = \bar{X} \pm Z_{\alpha/2} \cdot \left( \frac{s}{\sqrt{N}} \right)$

Where:

• $\bar{X}$ is the sample mean,
• $s$ is the sample standard deviation,
• $N$ is the sample size,
• $Z_{\alpha/2}$ is the critical value for a 95% confidence level, which is approximately 1.96 for large samples (normal distribution).

Now, let's go through each sample outcome from the image you provided:

(a)

• $\bar{X} = 5.2$,
• $s = 0.7$,
• $N = 157$.

$CI = 5.2 \pm 1.96 \cdot \left( \frac{0.7}{\sqrt{157}} \right)$

(b)

• $\bar{X} = 100$,
• $s = 9$,
• $N = 620$.

$CI = 100 \pm 1.96 \cdot \left( \frac{9}{\sqrt{620}} \right)$

(c)

• $\bar{X} = 20$,
• $s = 3$,
• $N = 220$.

$CI = 20 \pm 1.96 \cdot \left( \frac{3}{\sqrt{220}} \right)$

(d)

• $\bar{X} = 1020$,
• $s = 50$,
• $N = 329$.

$CI = 1020 \pm 1.96 \cdot \left( \frac{50}{\sqrt{329}} \right)$

(e)

• $\bar{X} = 7.3$,
• $s = 1.2$,
• $N = 105$.

$CI = 7.3 \pm 1.96 \cdot \left( \frac{1.2}{\sqrt{105}} \right)$

(f)

• $\bar{X} = 33$,
• $s = 6$,
• $N = 220$.

$CI = 33 \pm 1.96 \cdot \left( \frac{6}{\sqrt{220}} \right)$

Would you like me to calculate these intervals for you in detail?

Here are 5 questions related to this topic:

1. How is the sample size related to the width of the confidence interval?
2. What is the difference between using a Z-distribution and a t-distribution for confidence intervals?
3. What effect does increasing the sample standard deviation have on the confidence interval?
4. How would the confidence interval change if you wanted a 99% confidence level instead of 95%?
5. Why do we assume normality for large sample sizes when constructing confidence intervals?

Tip: A larger sample size reduces the margin of error, leading to a narrower confidence interval.