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 4% of 70%.

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

The margin of error is 2.08. The confidence interval is 67.92% – 72.08%. This represents the range of the scores that 90% of the students that took the exam scored.

Explanation:

The standard deviation is 4%, and the sample size is 10.

The confidence interval is 90%, since 9 times out of 10, the students had an average score within 4% of 70%.

To calculate the margin of error, we know that the confidence interval is 90% and the z-score for this interval is 1.645 so, we use the formula:

`1.645*((x)/(√(n) ) )`

where x = standard deviation (which is 4% in this case)

n = the sample size (10 in this case)

`1.645 * ((4)/(√(10) ) )= 2.08`

The margin of error is 2.08. The confidence interval is 67.92% – 72.08%. This represents the range of the scores that 90% of the students that took the exam scored.

Answer 2

All our answers lie in the above statement.

Confidence Level:
The creator claims that 9 out 10 students will have the average score in the said range. Or in other words we can say that the creator is 90% confident about the result of the field test. So the confidence level is 90%.

Margin of Error:
The average score lies within 4% of 70%. This means the margin of error is 4% i.e. the average scores can deviate from 70% by 4% .

Confidence Interval:
Lower Limit = 70% - 4% = 66%
Upper Limit = 70% + 4% = 74%

Interpretation:
The exam creator is 90% confident that the average scores of seniors will be between 66% and 74%.