Does anyone understand this. Please help!!

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

27-06-2018
Mathematics

contestada

Does anyone understand this. Please help!!

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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11beehshahbaz 11beehshahbaz · Answer 1 · 2018-06-27T21:06:51+02:00

The Claim of the exam creator is:

Nine times out of ten, seniors will have an average score within 5% of 75%.

We can rewrite this statement as:

Nine times out of ten or 90% of the times, seniors will have an average score within 5% of 75% i.e from 70% to 80%.

Thus we can say, the exam creator is 90% confident that the seniors will have an average score between 70% to 80%.

From here we can conclude that the Confidence Level is 90%.

Margin of error means by how much points the population parameters will vary from the sample mean/sample proportion. We are provided this information in the question:

Seniors will have an average score within 5% of 75%.

Thus the margin of Error will be 5% and the sample proportion is 75%.

So the Confidence Interval is 75-5 to 75+5
i.e
70% to 80%

The exam creator is 90% confident that the seniors will have an average scores between 70% to 80%