Question

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)

Answer #1

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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