A company started a new recreation program for its employees in the hope that a

Question

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

Explanation / Answer

(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)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.