Fatimah wants to estimate the proportion of students at her school who ate breakfast every day last week. She takes an SRS of 80 of the 1000 students at her school and finds that 20 of those sampled ate breakfast every day last week. Based on this sample, which of the following is a \( 95 \% \) confidence interva for the proportion of all students at the school who ate breakfast every day last week? Choose 1 answer: (A) \( 20 \pm 0.080 \) (B) \( 20 \pm 0.095 \) (C) \( 0.25 \pm 0.080 \) (D) \( 0.25 \pm 0.095 \)

To estimate the proportion of students who ate breakfast every day last week, we first calculate the sample proportion. Fatimah found that 20 out of 80 students ate breakfast, giving us a sample proportion \( p̂ = 20/80 = 0.25 \). Now we can calculate the confidence interval for the population proportion. The formula for a confidence interval for proportions is given by \( p̂ \pm z \cdot \sqrt{\frac{p̂(1 - p̂)}{n}} \), where \( z \) is the z-score corresponding to the desired confidence level (1.96 for 95%). This leads us to calculate the standard error and thus the interval. Upon evaluating, we find that the correct answer is: **(D) \( 0.25 \pm 0.095 \)** – The interval suggests that we’re fairly certain the true proportion of breakfast eaters lies between approximately 0.155 and 0.345. Now you’re equipped with a solid understanding of estimating proportions!