Assistance with Homework ### Constructing and Interpreting Confidence Intervals**Scenario:**Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 36 students, she finds 3 who eat cauliflower. Your task is to obtain and interpret a 90% confidence interval for the proportion of students who eat cauliflower on Jane's campus using Agresti and Coull’s method.**Instructions:**Construct and interpret the 90% confidence interval. Select the correct choice below and fill in the answer boxes within your choice. (Round to three decimal places as needed.)- **Option A:** - There is a 90% chance that the proportion of students who eat cauliflower in Jane’s sample is between \[ \_\_ \] and \[ \_\_ \]. - **Option B:** - One is 90% confident that the proportion of students who eat cauliflower on Jane’s campus is between \[ \_\_ \] and \[ \_\_ \]. - **Option C:** - The proportion of students who eat cauliflower on Jane’s campus is between \[ \_\_ \] and \[ \_\_ \] 90% of the time. - **Option D:** - There is a 90% chance that the proportion of students who eat cauliflower on Jane’s campus is between \[ \_\_ \] and \[ \_\_ \].### Agresti and Coull’s Method for Constructing Confidence IntervalsTo address issues such as the distribution of $\hat{p}$ not following a normal distribution, A. Agresti and B. Coull proposed a modified approach to constructing confidence intervals for a proportion. A $(1-\alpha) \times 100\%$ confidence interval for $p$ is given by the bounds shown below, where $p = \frac{x + 2}{n + 4}$ and $x$ is the number of successes in $n$ trials.- **Lower bound:** \[ p - z_{\frac{\alpha}{2}} \sqrt{\frac{p(1-p)}{n+4}} \]- **Upper bound:** \[ p + z_{\frac{\alpha}{2}} \sqrt{\frac{p(1-p)}{n+4}} \]The method provides a more accurate interval,

Given : n = 36 x = 3 Agresti and Couli's method for constructing confidence intervals : p^ = x+2n+4 = 3+236+4 = 540 = 0.125 CI = p^ ± Zα/2*p^(1-p^)n+4 At 90% CI, Zα/2 = 1.645 90% CI = 0.125 ± 1.645*0.125(1-0.125)36+4 = 0.125 ± 1.645*0.05229 = 0.125 ± 0.086 = (0.039, 0.211) There is a 90% chance that the proportion of students who eat cauliflower in Jane's sample is between 0.039 and 0.211