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a) What is the margin of error (ME) for this confidence interval? ME = 2.984 (Type an integer or decimal rounded to three decimal places

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a) What is the margin of error (ME) for this confidence interval? ME = 2.984 (Type an integer or decimal rounded to three decimal places as needed.) b) If a 95% confidence interval is created, would the margin of error be larger or smaller? Explain. The margin of error would be smaller because the critical value decreases as the confidence level decreases, and the margin of error decreases as the critical value decreases. c) Explain what the calculated interval means in this context. O A. With 99% confidence, a student in a traditional math class is expected to score between 3.384 points lower and 2.584 points higher on the standardized test than a student in a nontraditional math class. C B. With 99% confidence, students in nontraditional math classes will score, on average, between 3.384 points lower and 2.584 points higher on the standardized test than students in traditional math classes. O C. With 99% confidence, a student in a nontraditional math class is expected to score between 3.384 points lower and 2.584 points higher on the standardized test than a student in a traditional math class. O D. With 99% confidence, students in traditional math classes will score, on average, between 3.384 points lower and 2.584 points higher on the standardized test than students in nontraditional math classes. d) Does this result suggest that students who learn mathematics with nontraditional methods will have significantly higher mean scores on standardized tests than those in traditional classes? Explain. No. Because the interval contains zero, there is insufficient evidence to conclude that students who learn mathematics with nontraditional methods will have higher mean scores on standardized tests than those in traditional classes.The scores on a standardized test for 328 students in nontraditional math classes were compared to the scores of 271 students in traditional math classes. Computer software was used to create a confidence interval for the difference in mean scores. Complete parts a through d. Conf level: 99% Variable: (NonTrad) - u(Trad) Interval: ( - 3.384, 2.584)

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