Afield test for a new exam was given to randomly selected seniors. the exams were graded, and the sample mean and sample standard deviation were calculated. based on the results, the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 6% of 80%.
is the confidence interval at 90%, 95%, or 99%? what is the margin of error? calculate the confidence interval and explain what it means in terms of the situation.

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ince the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 6% of 80%, that means there is a 90% confidence interval. 9 / 10 = .90 = 90% 90% = 1.645 1.645 * .9 = 1.4805 Margin of Error - 1.4805
and The margin of error is the distance from x-bar, the center of the confidence interval ,to the end of the interval. This distance is 6% of 80% which is given by: 80×6100=4.8 Therefore the margin of error is 4.8% and the confidence interval is: ({80 - 4.8}%, {80 + 4.8}%)

Chope you get it right gl

Then why is the question on here if you know the answer