Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Advanced Math Topics 9.2/9.3 Estimating the Population Mean From a Large Sample Remember this from last chapter? How would you find the limits of the sample means that lie in the middle 95% of the normal distribution? .475 .475 xz y μx Lower Limit x 1.96 Upper Limit y x 1.96 If you did not know the mean of a population (μ), but you knew a sample mean ( x ), it would help to have a formula for μ. Rearrange the formulas above to get… x 1.96 y x 1.96 y y 95% Confidence Interval for μ If x is known for a large sample (y > 30), and σ is the sample standard deviation, then there is a 95% chance that the population mean (μ), lies within the interval… x 1.96 y and x 1.96 μ y You may want to know a different interval than 95%... Lower Boundary 90% Confidence Interval 99% Confidence Interval x 1.645 x 2.58 y y Upper Boundary x 1.645 x 2.58 y y A sample survey of 81 movie theaters showed that the average length of the main feature film was 90 minutes with a standard deviation of 20 minutes. Find a… a) 90% confidence interval for the mean of the population. x 1.645 x 1.645 y 20 90 1.645 81 y 20 90 1.645 81 The 90% confidence interval for μ is 86.34 to 93.66 minutes. b) 95% confidence interval for the mean of the population. x 1.96 y 20 90 1.96 81 x 1.96 90 1.96 y 20 81 The 95% confidence interval for μ is 85.64 to 94.36 minutes. From the HW P. 445 1) A sample of 49 banks found that the average charge for a money order was $2.75 for amounts up to $1000. The standard deviation was $0.25. Find a 90% confidence interval for the average charge for a money order. The 90% confidence interval is $2.69 to $2.81. HW P. 445 #1-8