To calculate the confidence intervals for the population proportion based on the sample proportion, we will use the formula for a confidence interval (CI) for a population proportion:

$CI = \hat{p} \pm z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

• $\hat{p}$ is the sample proportion (in this case, $\hat{p} = 0.49$),
• $z$ is the critical value (which changes depending on the confidence level),
• $n$ is the sample size ($n = 125$).

Step 1: 75% Confidence Interval

For the 75% confidence interval, the critical value $z = 1.150$.

The formula becomes:

$CI_{75} = 0.49 \pm 1.150 \cdot \sqrt{\frac{0.49 \cdot (1 - 0.49)}{125}}$

Let's compute this.

Step 2: 90% Confidence Interval

For the 90% confidence interval, the critical value $z = 1.645$.

The formula becomes:

$CI_{90} = 0.49 \pm 1.645 \cdot \sqrt{\frac{0.49 \cdot (1 - 0.49)}{125}}$

Let me calculate both of these confidence intervals.The confidence intervals for the population proportion based on the sample are as follows:

1. 75% Confidence Interval:

   • Lower bound: 0.439
   • Upper bound: 0.541

2. 90% Confidence Interval:

   • Lower bound: 0.416
   • Upper bound: 0.564

These intervals provide a range of values within which the true population proportion is likely to lie with the given levels of confidence.

Would you like to see these intervals plotted on a graph or need further details?

Here are 5 related questions to consider:

1. How do confidence intervals change with larger sample sizes?
2. What is the effect of a higher confidence level on the width of the interval?
3. How is the critical value for a confidence interval determined?
4. What are the assumptions behind using a normal approximation for confidence intervals of proportions?
5. How does the standard error of the sample proportion change with different values of $\hat{p}$?

Tip: Higher confidence levels result in wider intervals to ensure greater certainty that the population parameter lies within the range.