1. Interpret a 95% confidence interval of (3.355, 3.445) for the population mean. 2. A nursing school wants to estimate the true mean annual income of its alumni. It randomly samples 200 of its alumni. The mean annual income was $52,500 with a standard deviation of $1,800. Find a 95% confidence interval for the true mean annual income of the nursing school alumni. Write a statement about the confidence level and the interval you find. Solution Given: 1. 95% confidence interval of (3.355, 3.445) for the population mean probability that population mean lies in the interval (3.355,3.445) is 0.95 or you are 95% confident that unknown population parameter (here it is mean) lies in the range given 2.A nursing school wants to estimate the true mean annual income of its alumni. It randomly samples 200 of its alumni mean annual income = $52,500 standard deviation = $1,800. Confidence interval for mean, given sample from population with unknown mean & variance we can calulate t value t= n (xbar - u)/s sample size =200 , therefore d.f of t value =199 CI = ( xbar - t(0.975,199)* s/n , xbar + t(0.975,199)* s/n ) = ( 52500 - 1.975 * 1800/200 , 52500 - 1.975 * 1800/200 ) = ( 52248.62 , 52751.38 ).