MATH1127: Introduction to Statistics (60022)
Lesson 8.3 A Population Proportion
Unit 3 Chapter 8: Lesson 8.3 Assignment
If $n=120$ and $\widehat{p}=0.45$, construct a $90 \%$ confidence interval about the population proportion. Round your answers to three decimal places.
Preliminary:
a. Is it safe to assume that $n \leq 0.05$ of all subjects in the population?
No
Yes
b. Verify $n \hat{p}(1-\hat{p}) \geq 10$. Round your answer to one decimal place.
\[
n \widehat{p}(1-\widehat{p})=
\]
Confidence Interval: What is the $90 \%$ confidence interval to estimate the population proportion? Round your answer to three decimal places.
$
Solution
Step 1 ：The problem is asking for a 90% confidence interval for the population proportion. The formula for a confidence interval for a population proportion is given by \(\widehat{p} \pm z \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\), where \(\widehat{p}\) is the sample proportion, \(n\) is the sample size, and \(z\) is the z-score corresponding to the desired level of confidence. In this case, \(n=120\) and \(\widehat{p}=0.45\). The z-score for a 90% confidence interval is approximately 1.645.
Step 2 ：Before we calculate the confidence interval, we need to answer the preliminary questions. The first question is asking if the sample size is less than or equal to 5% of the population size. We don't have information about the population size, so we can't answer this question.
Step 3 ：The second preliminary question is asking to verify if \(n \hat{p}(1-\hat{p}) \geq 10\). This is a condition for using the normal approximation to the binomial distribution. If this condition is met, we can use the formula for the confidence interval mentioned above.
Step 4 ：Let's calculate \(n \hat{p}(1-\hat{p})\). The value of \(n \hat{p}(1-\hat{p})\) is approximately 29.7, which is greater than 10. So, it is safe to use the normal approximation to the binomial distribution.
Step 5 ：Now, let's calculate the confidence interval. The 90% confidence interval for the population proportion is approximately (0.375, 0.525).
Step 6 ：Final Answer: The 90% confidence interval for the population proportion is approximately \(\boxed{(0.375, 0.525)}\).
From Solvely APP
Solution
Step 1 ：The problem is asking for a 90% confidence interval for the population proportion. The formula for a confidence interval for a population proportion is given by \(\widehat{p} \pm z \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\), where \(\widehat{p}\) is the sample proportion, \(n\) is the sample size, and \(z\) is the z-score corresponding to the desired level of confidence. In this case, \(n=120\) and \(\widehat{p}=0.45\). The z-score for a 90% confidence interval is approximately 1.645.
Step 2 ：Before we calculate the confidence interval, we need to answer the preliminary questions. The first question is asking if the sample size is less than or equal to 5% of the population size. We don't have information about the population size, so we can't answer this question.
Step 3 ：The second preliminary question is asking to verify if \(n \hat{p}(1-\hat{p}) \geq 10\). This is a condition for using the normal approximation to the binomial distribution. If this condition is met, we can use the formula for the confidence interval mentioned above.
Step 4 ：Let's calculate \(n \hat{p}(1-\hat{p})\). The value of \(n \hat{p}(1-\hat{p})\) is approximately 29.7, which is greater than 10. So, it is safe to use the normal approximation to the binomial distribution.
Step 5 ：Now, let's calculate the confidence interval. The 90% confidence interval for the population proportion is approximately (0.375, 0.525).
Step 6 ：Final Answer: The 90% confidence interval for the population proportion is approximately \(\boxed{(0.375, 0.525)}\).
