https://file.aiquickdraw.com/mathbot/file/1727187975995x6btues8.jpg

To solve the given problem, we need to calculate the confidence intervals for a population proportion using the given data.

### Given Data:
- Sample proportion (\( \hat{p} \)) = 0.47 (47%)
- Sample size (\( n \)) = 450
- Confidence levels: 95% and 99%

#### Formula for Confidence Interval:
\[
\text{Confidence Interval} = \hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
Where:
- \( \hat{p} \) is the sample proportion (0.47),
- \( n \) is the sample size (450),
- \( z \) is the z-value corresponding to the desired confidence level (1.96 for 95% and 2.576 for 99%).

### Step-by-step:
1. **Find the margin of error:**
   \[
   \text{Margin of error} = z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
   \]

2. **95% Confidence Interval (z = 1.96):**
   \[
   \text{Margin of error} = 1.96 \cdot \sqrt{\frac{0.47 \times 0.53}{450}}
   \]

3. **99% Confidence Interval (z = 2.576):**
   \[
   \text{Margin of error} = 2.576 \cdot \sqrt{\frac{0.47 \times 0.53}{450}}
   \]

Let's calculate these values.### Results:
1. **95% Confidence Interval**: (0.42, 0.52)
2. **99% Confidence Interval**: (0.41, 0.53)

### Explanation for part (c):
- As the confidence level increases from 95% to 99%, the margin of error increases. This makes the interval wider, which means we are more confident that the true population proportion lies within the interval, but at the cost of a larger range.

Let me know if you want further details or have any questions!

#### Relative questions:
1. What is the effect of increasing the sample size on the confidence interval?
2. How is the margin of error affected by a higher confidence level?
3. Can the sample proportion ever be outside the confidence interval?
4. Why does the z-value change for different confidence levels?
5. What would happen if the sample proportion were 50% instead of 47%?

#### Tip:
Larger sample sizes decrease the margin of error, leading to more precise confidence intervals.

The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. Provide a 95% confidence interval and a 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses, and explain the effect of increasing the confidence level on the margin of error.

Learn how to calculate the 95% and 99% confidence intervals for the population proportion of college students who work to pay for tuition and living expenses. This problem involves sample proportion, z-values for confidence levels, and margin of error, and is suitable for high school or college-level statistics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation