Sarah polled 40 randomly selected students at her high school and found that 20% ( = 0.2) are happy with the quality of the cafe
teria food. With a desired confidence level of 99%, which has a corresponding z*-score of 2.58, what is the approximate margin of error of Sarah’s poll? Remember, the margin of error, E, can be determined using the formula E = z*
Answer: According to Sarah's poll, she can conclude with 99% confidence level that 20% of the high school population is happy with the quality of the cafeteria food, with a margin of error of +/- 16.3%.
No not D because it’s just a straight up number it’s not referring to how many W’s there are. Like 8 is just a solo number , has nothing to do with the W