Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (for 95%, use .025) and its Z score.
95% = mean ± Z SEm
SEm = SD/√n
For other intervals, adjust Z score accordingly.
Go to a local library; collect a sample of books consisting of the ages of book (based on copyright dates). Plan and describe the sampling procedure, then use the results to construct a confidence interval estimate of the mean age of all books in the library. Please, post and share your results with the class.
Based on copyright dates, my sample of books consisted of ages 1992, 1999, 2004, 2013, 2002 and 2014.
The sample mean is: 2004
The sample standard deviation: is 8.4
n=6
2 answers
How would you solve this with the same information. Please I need help Im clueless!!
Use the mean and the standard deviation obtained from the last module and test the claim that the mean age of all books in the library is greater than 2005. Share your results with the class
Use the mean and the standard deviation obtained from the last module and test the claim that the mean age of all books in the library is greater than 2005. Share your results with the class