Homework SourseA company started a new recreation program for its employees
A company started a new recreation program for its employees in the hope that a little recreation would improve an employee\'s performance at work. To determine whether the high cost of the program is justified, the president of the company wishes to estimate the proportion of the employees who participate in the recreational activities. In a random sample of 200 employees, 60 were found to regularly participate in the recreation program. Find the 95% confidence interval for the true proportion of employees who participate in the new recreation program
Solution
(0.2365 , 0.3635)
| sample size, n=200 sample proportion, p = 60 / 200 = 0.3 we use normal approximation, for this we check that both np and n(1-p) > 5. Since n*p = 60 > 5 and n*(1-p) = 140 > 5, we can take binomial random variable as normally distributed, with mean = p = 0.3 and std deviation = root( p * (1-p) /n ) = 0.032404 |
| For constructing Confidence interval, Margin of Error (ME) = z x SD = 0.0635 95% confidence interval is given by: Sample Mean +/- (Margin of Error) 0.3 +/- 0.0635 = (0.2365 , 0.3635) |
