A random sample of 12 light bulbs has a mean life of 1421 hours with a standard deviation of 68 hours. Construct a 95% confidence interval for the mean life, μ, of all light bulbs of this type. Assume the population has a normal distribution.
Group of answer choices
(1378.2, 1463.8)
(1383.7, 1458.3)
(1382.5, 1459.5)
(1377.8, 1464.2)
(1381.7, 1460.2)
Solution :
Given that,
= 1421
s =68
n =12
Degrees of freedom = df = n - 1 =12 - 1 = 11
a ) At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2= 0.05 / 2 = 0.025
t /2,df = t0.025,11 =2.201 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.201 * ( 68/ 12)
= 43.2
The 95% confidence interval is,
- E < < + E
1421 - 43.2 < < 1421+ 43.2
(1377.8, 1464.2)
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