Question

To obtain an estimate of the proportion of "full time" university students who have a part time job in excess of 30 hours per week, the student union decides to interview a random sample of full time students. They want the length of their 90% confidence interval to be no greater than 0.2 with standard deviation is known to be 3 hours. What size of the sample, n should be taken? Round up your answer to the nearest whole number.

Answer 1

Answer:

2,436 students

Step-by-step explanation:

At a 90% confidence level, the z-score is 1.645 and the confidence interval is given by:

$x\pm z\frac{s}{\sqrt n}$

Where s is the standard deviation, and  n is the sample size.

If they want the length of their confidence interval to be no greater than 0.2, it must be no further than 0.1 from the mean 'X':

$0.1>1.645\frac{3}{\sqrt n}\\\sqrt n>1.645*30\\n>2,435.42$

Rounding up to the next whole number, the sample size should be 2,436 students.