In: Statistics and Probability
A sample of 22 offshore oil-workers took part in a simulated
escape exercise. The sample yielded an average escape time of 388.7
min. and standard deviation of 22.4 min. The 95% confidence
interval for the true average of escape time is:
(379.34 , 398.06)
(378.768 , 398.632)
(380.482 , 396.918)
Solution :
Given that,
= 388.7
s =22.4
n =22
degrees of freedom = df = n - 1 = 22- 1 =21
a ) At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2= 0.05 / 2 = 0.025
t /2,df = t0.025,21 =2.080 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.080* ( 22.4/ 22)
= 9.9335
The 95% confidence interval estimate of the population mean is,
- E < < + E
388.7 -9.9335 < < 388.7+ 9.9335
378.768< < 398.632
( 378.768, 398.632 )