If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result

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Question:

The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038 hours.

If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result?

Given Information:

standard deviation = σ = 30 hours

confidence level = 99%

Margin of error = 6 hours

Required Information:

sample size = n = ?

sample size = n ≈ 165

Step-by-step explanation:

We know that margin of error is given by

Margin of error = z*(σ/√n)

Where z is the corresponding confidence level score, σ is the standard deviation and n is the sample size

√n = z*σ/Margin of error

squaring both sides

n = (z*σ/Margin of error)²

For 99% confidence level the z-score is 2.576

n = (2.576*30/6)²

n = 164.73

since number of bulbs cannot be in fraction so rounding off yields

n ≈ 165

Therefore, a sample size of 165 bulbs is needed to ensure a margin of error not greater than 6 hours.

b

step-by-step explanation:

the word superfluous means unnecessary