A tire manufacturer wishes to test a sample of its best-selling tire for average lifetime mileage. When the manufacturing process is operating properly, the standard deviation of the mileage is known to be sd=2,500. An SRS of 100 tires is selected and is found to have mean mileage x-bar= 59,500. You wish to construct a 95% confidence interval for mu.
A. Identify mu, the parameter about which we want to draw conclusions.
B. Verify that the necessary conditions exist for construction of a confidence interval.
I just want to make sure i'm right, is A the mean 59,500? and i'm confused on B.

Question

A tire manufacturer wishes to test a sample of its best-selling tire for average lifetime mileage. When the manufacturing process is operating properly, the standard deviation of the mileage is known to be sd=2,500. An SRS of 100 tires is selected and is found to have mean mileage x-bar= 59,500. You wish to construct a 95% confidence interval for mu. 
A. Identify mu, the parameter about which we want to draw conclusions.
B. Verify that the necessary conditions exist for construction of a confidence interval.
I just want to make sure i'm right, is A the mean 59,500? and i'm confused on B.

anonymous · Answer

For B I'm pretty sure you have to use the equation x-bar plus or minus the margin of error, but is that how you verify it? Do I just explain what the numbers represent or should I plug the numbers in?

mathmale · Answer

Generally, n=the number of SRS, must be 30 or more for us to calculate the desired confidence interval.
Please use the Greek letter for 'mu' to represent the population mean and x-bar (x with a bar over it) to represent the sample mean.  You should write "x-bar = mu = 59500.  Yes, the sample mean and the population mean are the same.  

However, the sample std. dev. and the population std. dev. differ.  Remember how to find the sample std. dev. from the population std. dev?

anonymous · Answer

you find the sd by doing sd/Sqrt. 100 to find the sample std. dev. right?

anonymous · Answer

for B would you verify by showing that equation or is there something else I need to do to verify the necessary conditions exist for construction of a confidence interval. @mathmale

mathmale · Answer

"you find the sd by doing sd/Sqrt. 100 to find the sample std. dev. right?"   Yes.  Good.

I'd prefer you do some Internet lookups to verify what I said in regard to Part B.
Generally, n>30 is good enough to use methods here that pertain to normal distributions.

Seems to me that a search for "sample normal distributions" would produce not only "n>30" but also any other condition or conditions that might apply.

anonymous · Answer

yeah I just searched it and the conditions I found were 1. needs to be a simple random sample 2. needs to be normally distributed. Thank you! @mathmale