Let's solve each part step by step, using the provided sample data.

Given Data:

• Sample mean, $\bar{y} = 3.39$
• Sample standard deviation, $s = 0.29$
• Sample size, $n = 30$ for parts (a) and (b), and $n = 60$ for part (c).

Part 1: 95% Confidence Interval for the Mean

To calculate a confidence interval (CI), we use the following formula: $CI = \bar{y} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$ Where:

• $\bar{y}$ is the sample mean,
• $t_{\alpha/2}$ is the critical value from the $t$-distribution,
• $s$ is the sample standard deviation,
• $n$ is the sample size.

For a 95% confidence level and $n = 30$, the degrees of freedom $df = n - 1 = 29$.

From the $t$-distribution table, $t_{0.025, 29} \approx 2.045$.

Now calculate the margin of error: $ME = 2.045 \times \frac{0.29}{\sqrt{30}} \approx 0.108$

So, the 95% confidence interval is: $CI = 3.39 \pm 0.108 = (3.282, 3.498)$

Part 2: 90% Confidence Interval for the Mean

For a 90% confidence level, we use a different critical value. From the $t$-distribution table, $t_{0.05, 29} \approx 1.699$.

Now calculate the margin of error: $ME = 1.699 \times \frac{0.29}{\sqrt{30}} \approx 0.090$

So, the 90% confidence interval is: $CI = 3.39 \pm 0.090 = (3.300, 3.480)$

Part 3: 95% Confidence Interval for $n = 60$

If we increase the sample size to 60, the formula remains the same, but now $n = 60$. The degrees of freedom $df = 59$.

From the $t$-distribution table, $t_{0.025, 59} \approx 2.000$.

Now calculate the margin of error: $ME = 2.000 \times \frac{0.29}{\sqrt{60}} \approx 0.075$

So, the new 95% confidence interval is: $CI = 3.39 \pm 0.075 = (3.315, 3.465)$

Summary:

• a) 95% CI for $n = 30$: $(3.282, 3.498)$
• b) 90% CI for $n = 30$: $(3.300, 3.480)$
• c) 95% CI for $n = 60$: $(3.315, 3.465)$

Do you want more details or have any questions about these calculations?

Here are 5 related questions to deepen your understanding:

1. How does increasing the sample size affect the confidence interval?
2. What is the relationship between confidence level and the width of the confidence interval?
3. Why do we use the $t$-distribution instead of the normal distribution here?
4. What happens to the margin of error if the sample standard deviation is larger?
5. How would the confidence interval change if the sample mean were lower?

Tip: Larger sample sizes tend to decrease the margin of error, leading to narrower confidence intervals, meaning more precise estimates.