a sociologist is found that in a random sample of 50 retired men, the average number if jobs they had during their lifetimes was 7.2. the population standard deviation is 2.1.
find the 95% confidence interval of the mean number of jobs;
+36916 Studying a z-score table , you will find z = -1.96 = .025
and z = +1.96 = .9750 .9750 - .025 = .95 = 95%
Standard deviation 2.1 ( 1.96 standard deviations above and below the mean (7.2) will encompass 95% )
7.2 - 2.1 ( - 1.96) = 3.084 jobs
7.2 + 2.1 (+1.96) = 11.316 jobs so I THINK the 95% confidence interval is 3.084 to 11.316 jobs (not positive of this answer)
