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 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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i)

The confidence interval is at 90%

ii)

The margin of error is 5%

iii)

The 90% confidence interval is (70%, 80%)

iv)

We are 90% confident that the average score of seniors in the field test is between 70% and 80%.

Step-by-step explanation:

i)

The confidence interval is at 90%

We are informed that the exam creator claims that on this particular exam, nine times out of ten, seniors will have an average score within 5% of 75%. This implies that we are 9 times out of 10 confident that seniors will have an average score within 5% of 75%.

The level of confidence is thus;

(9/10)*100 = 90%

ii)

The margin of error is 5% or equivalently 0.05

We are informed that the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 5% of 75%.

Since the average score is within 5%, the margin of error is 5%

iii)

A confidence interval is calculated using the formula;

point estimate ± margin of error

Our point estimate is 75%

Our margin of error is 5%

The 90% confidence interval is thus;

75% ± 5% = (70%, 80%)

iv)

The 90% confidence interval is interpreted as;

We are 90% confident that the average score of seniors in the field test is between 70% and 80%.

This is a confidence interval for the mean, the level of confidence is 90% and our confidence interval is (70%, 80%).

ok then what happened.

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step-by-step explanation: