Suppose that the standard deviation of the tube life for a particular brand of TV picture tube is known to be \sigma=400, but that the mean operating life is not known. Determine (i) the 95 percent and (ii) the 90 percent confidence intervals for estimating the population mean.

ediculeN

ediculeN

Answered question

2021-08-03

Suppose that the standard deviation of the tube life for a particular brand of TV picture tube is known to be σ=400, but that the mean operating life is not known. Overall, the operating life of the tubes is assumed to be approximately normally distributed.
For a sample of n=15, the mean operating life is X=8900 hr. Determine (i) the 95 percent and (ii) the 90 percent confidence intervals for estimating the population mean.

Answer & Explanation

Tolnaio

Tolnaio

Beginner2021-08-10Added 1 answers

Step 1
Confidence interval is obtained as
x± tn1.α{2}×sn
If population standard deviation is known.
Confidence interval is obtained as
x± z{α}{2}×σn
if the population standard deviation is known.
i) Here x=8900
σ=400
x=0.05(5%)
n=15
So, CI=(x± z{α}{2}×σn)
=8900± z0.0052×40015
=8900±1.96×40015
=8900±78415
=8900±7843.87
=8900±202.60
CI=(8900202.60, 8900+202.60)
=(8697.4, 9102.6)
i.c. Confidence interval for population mean is (8697.4, 9102.6)
Step 2
ii) n=15
x=8900
σ=400
α=0.10
CI=x± z{α}{2}×σn
=8900± z0.102×40015
=8900±1.645×40015
=8900±65815
=8900±6583.87
=8900±658170.02
Step 3
The confidence interval is obtained as
8900±170.02
I.e.
CI=(8900170.02, 8900+170.02)
CI=(8729.98, 9070.02)
Hence the confidence interval is
(8729.98, 9070.02)
Here the value of z{0.05}{2}=1.96 is obtained from z table
Also
The value of z{0.10}{2}=1.645<

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