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MATH1127: Introduction to Statistics (60022) Lesson 8.3 A Population Proportion
Unit 3 Chapter 8: Lesson 8.3 Assignment
Score: $71.2 / 100 \quad 11 / 14$ answered
Question 2
Out of 500 people sampled, 240 had children. Based on this, construct a $99 \%$ confidence interval for the true population proportion of people with children.
Preliminary:
a. Is it safe to assume that $n \leq 5 \%$ of all people with children?
Yes
No
b. Verify $n \hat{p}(1-\widehat{p}) \geq 10$. Round your answer to one decimal place.
\[
n \hat{p}(1-\hat{p})=
\]
Confidence Interval: What is the $99 \%$ confidence interval to estimate the population proportion? Round your answer to three decimal places.
\[
< p<
\]
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Step 1 ：First, we need to calculate the sample proportion $\hat{p}$, which is the number of successes (people with children) divided by the total number of trials (total people sampled). In this case, $\hat{p} = \frac{240}{500} = 0.48$.

Step 2 ：Next, we need to check if it is safe to assume that $n \leq 5\%$ of all people with children. Since we don't know the total population size, we can't directly check this condition. However, if the population size is at least 20 times the sample size (which is a common rule of thumb), then this condition would be satisfied. Since $20 \times 500 = 10000$, as long as the total population size is at least 10000, it is safe to assume that $n \leq 5\%$ of all people with children.

Step 3 ：Now, we need to verify that $n \hat{p}(1-\hat{p}) \geq 10$. Substituting $n = 500$ and $\hat{p} = 0.48$ into the inequality, we get $500 \times 0.48 \times (1 - 0.48) = 124.8$, which is greater than 10. So, this condition is satisfied.

Step 4 ：Finally, we can construct the $99\%$ confidence interval for the population proportion. The formula for the confidence interval is $\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $z$ is the z-score corresponding to the desired level of confidence. For a $99\%$ confidence level, the z-score is approximately 2.576.

Step 5 ：Substituting $\hat{p} = 0.48$, $n = 500$, and $z = 2.576$ into the formula, we get $0.48 \pm 2.576 \sqrt{\frac{0.48(1-0.48)}{500}}$.

Step 6 ：Calculating the expression inside the square root gives $\sqrt{\frac{0.48(1-0.48)}{500}} \approx 0.022$.

Step 7 ：Multiplying this by the z-score gives $2.576 \times 0.022 \approx 0.057$.

Step 8 ：So, the $99\%$ confidence interval for the population proportion is $0.48 \pm 0.057$, or equivalently, $(0.423, 0.537)$ when rounded to three decimal places.