Just need help with the last question please, which is the one about the confidence interval and the terms of the situation?? ASAP and ill mark a brainliest

A field 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 5% of 75%.

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.

Question

06-07-2018
Mathematics

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Just need help with the last question please, which is the one about the confidence interval and the terms of the situation?? ASAP and ill mark a brainliest

A field 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 5% of 75%.

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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markguevara03 markguevara03 · Answer 1 · 2018-07-06T22:53:29+02:00

well, since 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 - 1.4805and 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}%)