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Question

A simple random sample of kitten toasters is to be taken to determine the mean operational lifetime in hours. Assume that the lifetimes are normally distributed with population standard deviation (30) hours.

Find the sample size needed so that the numbers have confidence interval for the mean lifetime will have a margin of error seven

Answer 1

If we're given:

σ = population standard deviation = 30

E = margin of error = 7

confidence level = 95% or .95

(95% is our usual assumption, but it can also commonly be 90% or 99%. Since we don't have one, we'll use this for now)

The equation we must use to find n, the sample size, needed for the mean lifetime is:

n = ( z_α/2 x α) ²

( E )²

where α = 1 - confidence level = 1 - 0.95 = 0.05;

thus α/2 = 0.05/2 = 0.025. Find the critical value on the z-score table (which will be 1.96)

then, z_α/2 = 1.96

Now, by plugging everything in:

n = ( 1.96 x 30) ²

( 7 )²

n = (58.8)²

49

n = 3457.44

49

n = 70.56

Thus, the sample size must be approximately 71.

Answer 2

If the population standard deviation is σ, the standard deviation of the sample mean is σ/√n, where n is the number in the sample. in this case, we want to solve 30/√n = 7.

Solving the for n gives n = (30 / 7)^2 ≈ 18.37. Sample size always rounds up, so this would be 19. However, this is just the sample size needed to that ONE standard deviation of the mean is under 7.

Normally we'd be looking for a sample size that gives ±7 for the bounds of a 95% confidence interval. That requires an interval of ±1.96σ_x-av . That means we would solve 30/√n = 7/1.96. This gives us n = 70.56, so we would need a sample of 71.