Question

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 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.

Answer 1

So the fisrt thing we need to calculate is that 9 times out of 10 is 90% which will be the confidence interval. 
Margin of error is 6% = .06 
Confidence interval is 80 +/- .06 = .74 to .86. or 74% to 86$ 
In 9 times out of 10 the confidence interval for similar samples of seniors will contain the true mean of the population of senior scores. 

Answer 2

Answer:

Margin of error: 6%; Confidence Interval Percentage: 90%; Confidence Interval: 74% and 86%

Step-by-step explanation:

Margin of error: 6%  --- This is already given in the question.  

Confidence interval is at 90%. 9/10=.90

The exam creator is saying that they are 90% confident that the seniors test scores will be between 74% and 86%.

0.8 - 0.06 = 0.74

0.8 + 0.06 = 0.86

The confidence interval is (0.74, 0.86)