Question 137046This question is from textbook A Brief Introduction to Probability and Statistics
: A random sample of n=64 observations has a mean x(bar)=29.1 and a standard deviation s=3.9.
a) Give the point estimate of the population mean mu and find the margin of error for your estimate.
b) Find a 90% confidence interval for mu. what does "90% confident" mean?
c) Find a 90% lower confidence bound for the population mean mu. Why is this bound different from the lower confidence limit in part b?
d) How many observations do you need to estimate mu to within .5, with probability equal to .95?
This question is from textbook A Brief Introduction to Probability and Statistics
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A random sample of n=64 observations has a mean x(bar)=29.1 and a standard deviation s=3.9.
a) Give the point estimate of the population mean mu and find the margin of error for your estimate.
point estimate: 29.1
margin of error: s/sqrt(n) = 3.9/sqrt(64) = 0.4875
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b) Find a 90% confidence interval for mu. what does "90% confident" mean?
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Comment: I am using z=1.645 for 90% CI
29.1-1.45*0.4875 < mu < 20.1+0.4875
90% confident means just that; It does not mean there is a 90% probability.
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c) Find a 90% lower confidence bound for the population mean mu. Why is this bound different from the lower confidence limit in part b?
???
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d) How many observations do you need to estimate mu to within .5, with probability equal to .95?
n = [z*s/E]^2
n = [1.645*3.9/0.5]^2 = 164.634; rounding up, n = 165
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Cheers,
Stan H.
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