Suppose that a standardized biology exam has a mean score of 80% correct, with a standard deviation of 3. The school administration believes that the new class of freshmen have a different mean exam score. A sample of 65 students from the freshmen class is used and a mean score of 76% correct is obtained. Calculate a 99% confidence interval. List the lower bound (rounded to 2 decimal places) and the upper bound (rounded to 2 decimal places).

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

A sample of 65 students from the freshmen class is used and a mean score of 76% correct is obtained.

This means that

99% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

0.6236*100 = 62.36%

0.8964*100 = 89.64%

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

at least two

step-by-step explanation: