Let's go step by step to solve this problem for each confidence level.

Step 1: Calculate the Sample Proportion ($\hat{p}$)

The sample proportion is calculated by:

$\hat{p} = \frac{\text{Number of people who said "It is not important"}}{\text{Total number of surveyed people}}$

Here, we know:

• 150 people said "It is not important."
• The total sample size is 600.

So, the sample proportion is:

$\hat{p} = \frac{150}{600} = 0.25$

Thus, for both parts (a) and (b), the sample proportion is 25%.

Step 2: Margin of Error Formula

The margin of error (MOE) is calculated using the formula:

$MOE = z^* \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Where:

• $z^*$ is the critical value corresponding to the given confidence level.
• $\hat{p}$ is the sample proportion (0.25).
• $n$ is the sample size (600).

Now, we need to calculate the MOE for each confidence level.

(a) 85% Level of Confidence

For an 85% confidence level, the critical value $z^*$ is approximately 1.44 (you can find this value from a standard normal distribution table).

Substitute the values into the MOE formula:

= 1.44 \times \sqrt{\frac{0.25 \times 0.75}{600}} = 1.44 \times \sqrt{\frac{0.1875}{600}} = 1.44 \times \sqrt{0.0003125} = 1.44 \times 0.01768 = 0.0255$$ Thus, the margin of error is **0.0255**, or approximately **2.6%**. ### (b) 98% Level of Confidence For a 98% confidence level, the critical value $$z^*$$ is approximately **2.33**. Substitute the values into the MOE formula: $$MOE = 2.33 \times \sqrt{\frac{0.25 \times (1 - 0.25)}{600}} = 2.33 \times \sqrt{\frac{0.25 \times 0.75}{600}} = 2.33 \times \sqrt{\frac{0.1875}{600}} = 2.33 \times \sqrt{0.0003125} = 2.33 \times 0.01768 = 0.0412$$ Thus, the margin of error is **0.0412**, or approximately **4.1%**. ### Final Answers (a) For an 85% confidence level: - Sample proportion = **25%** - MOE = **2.6%** (b) For a 98% confidence level: - Sample proportion = **25%** - MOE = **4.1%** --- Do you need any more details or explanations? Here are 5 follow-up questions to explore this topic further: 1. How does increasing the confidence level affect the margin of error? 2. What is the significance of the sample proportion in statistical analysis? 3. How would the margin of error change if the sample size were larger? 4. How is the critical value $$z^*$$ determined for various confidence levels? 5. What are some potential sources of bias in surveys like the one described? **Tip**: Confidence level and sample size are key factors affecting the margin of error in surveys. Larger sample sizes and lower confidence levels yield smaller margins of error.