The principal randomly selected six students to take an aptitude test. their scores were: 81.4 78.6 71.3 77.8 76.9 77.7 determine a 90% confidence interval for the mean score for all students.

The confidence interval is .

We first find the mean.  Add together all of the data points and divide by 6, the number of data points; the mean is 77.28.

Next we find the standard deviation.  Find the difference between each data point and the mean; square it; find the sum; divide by the number of data points; take the square root.  The standard deviation is 3.32.

To find the margin of error, we calculate the z-score associated with this level of confidence.  100-90 = 10% = 0.1; 0.1/2 = 0.05; 1-0.05 = 0.95.  Using a z-table (http://www.z-table.com) we see that this is between two scores, 1.64 and 1.65; we will use 1.645.

The margin of error is given by
z * (σ/√n) = 1.645*(3.32/√6) = 2.23.

Thus the confidence interval is 77.28 +/ 2.23.

Wow. that is a really cool question.