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A random sample of 12 graduates of a certain secretarial school typed an average of 79.3 words per minute with a standard deviation of 7.8 words per minute. assuming a normal distribution for the number of words typed per minute, find a 95% confidence interval for the average number of words typed by all graduates of this school.

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merlynthewhizz
We will use the following  formula to work out the confidence interval

Upper limit = μ + z* (σ/√n)
Lower limit = μ - z* (σ/√n)

We have
μ = 79.3
σ = 7.8
n = 12
z* is the z-score for 95% confidence level = 1.96

Substitute these into the formula, we have

Upper limit = 79.3 + 1.96 (7.8/√12) = 83.7
Lower limit = 79.3 - 1.96 )7.8/√12) = 74.9

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