A. a researcher claims that 20% of students spend more than 10 hours a week on homework asignments and studying. a random sample of 53 students were surveyed and 18 of them said they spend more than 10 hours a week on homework assignments and studying. construct a 95% confidence interval to test the researcher's clain ie, is the value 20% within the confidence interval? you must show the value of e, p, and za/2 rounded to 4 significant digits.

The confidence interval is
, or 
(0.212, 0.4672); no, his claim is not accurate, as 20% is not in the interval.

To construct the interval we first find p:
18/53 = 0.3396

To calculate :
Convert 95% to a decimal:  95/100 = 0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this value is associated with a z-score of 1.96.

We next calculate E:


Our confidence interval is then given by

The reporter forgot to multiply the decimal by 100, making it 25%