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Known mean population and an unknown variance

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A normal population has a known mean 50 and unknown variance.

a) A random sample of n=16 is selected from this population and the sample results are x-bar = 52 and s= 8. How unusual are these results? Explain.

b) A random sample of n=30 is selected from this population and the sample results are x-bar = 52 and s=8. How unusual are these results. Explain.

c) A random sample of n=100 is selected from this population and the sample results are x-bar = 52 and s=8. How unusual are these results? Explain

d) Compare your answers to parts (a)-(c) and explain why they are the same or differ

https://brainmass.com/statistics/probability/known-mean-population-unknown-variance-352311

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Solution:

(a) t = (x-bar - μ)/(s/√n) = (51.5 - 50)/(8/√16) = 0.75 and t = (52.5 - 50)/(8/√16) = 1.25
P(x-bar = 52) = P(51.5 < x-bar < 52.5) = P(0.75 < t < 1.25) = 0.1172
Probability that a sample has x-bar = 52 and s = 8 is 0.1172. This result is not unusual since 0.1172 is not a low probability.

(b) t = (x-bar - μ)/(s/√n) = (51.5 - ...

Solution Summary

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