A sample of 22 offshore oil-workers took part in a simulated escape exercise. The sample yielded...

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)

Solutions

Expert Solution

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 / $\sqrt$ n)

= 2.080* ( 22.4/ $\sqrt$ 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 )

orchestra answered 11 months ago

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