1. Lecture 14 Dustin Lueker

2.  This interval will contain μ with a 100(1-α)% confidence ◦ If we are estimating µ, then why it is unreasonable for us to know σ?  Thus we replace σ by s (sample standard deviation)  This formula is used for a large sample size (n≥30)  If we have a sample size less than 30 a different distribution is used, the t-distribution, we will get to this later 2STA 291 Summer 2010 Lecture 14

3.  Incorrect statement ◦ With 95% probability, the population mean will fall in the interval from 3.5 to 5.2  To avoid the misleading word “probability” we say ◦ We are 95% confident that the true population mean will fall between 3.5 and 5.2 3STA 291 Summer 2010 Lecture 14

4.  Changing our confidence level will change our confidence interval ◦ Increasing our confidence level will increase the length of the confidence interval  A confidence level of 100% would require a confidence interval of infinite length  Not informative  There is a tradeoff between length and accuracy ◦ Ideally we would like a short interval with high accuracy (high confidence level) 4STA 291 Summer 2010 Lecture 14

5.  Start with the confidence interval formula assuming that the population standard deviation is known  Mathematically we need to solve the above equation for n 5STA 291 Summer 2010 Lecture 14

6.  The width of a confidence interval ◦ as the confidence level increases ◦ as the error probability increases ◦ as the standard error increases ◦ as the sample size n increases  Why? 6STA 291 Summer 2010 Lecture 14

7.  To account for the extra variability of using a sample size of less than 30 the student’s t- distribution is used instead of the normal distribution 7STA 291 Summer 2010 Lecture 14

8.  t-distributions are bell-shaped and symmetric around zero  The smaller the degrees of freedom the more spread out the distribution is  t-distribution look much like normal distributions  In face, the limit of the t-distribution is a normal distribution as n gets larger STA 291 Summer 2010 Lecture 148

9.  Need to know α and degrees of freedom (df) ◦ df = n-1  α=.05, n=23 ◦ t α/2 =  α=.01, n=17 ◦ t α/2 =  α=.1, n=20 ◦ t α/2 = 9STA 291 Summer 2010 Lecture 14

10.  A sample of 12 individuals yields a mean of 5.4 and a variance of 16. Estimate the population mean with 98% confidence. 10STA 291 Summer 2010 Lecture 14

11.  The sample proportion is an unbiased and efficient estimator of the population proportion ◦ The proportion is a special case of the mean 11STA 291 Summer 2010 Lecture 14

12.  ABC/Washington Post poll (December 2006) ◦ Sample size of 1005 ◦ Question  Do you approve or disapprove of the way George W. Bush is handling his job as president?  362 people approved  Construct a 95% confidence interval for p  What is the margin of error? 12STA 291 Summer 2010 Lecture 14

13.  As with a confidence interval for the sample mean a desired sample size for a given margin of error (ME) and confidence level can be computed for a confidence interval about the sample proportion ◦ This formula requires guessing before taking the sample, or taking the safe but conservative approach of letting =.5  Why is this the worst case scenario? (conservative approach) 13STA 291 Summer 2010 Lecture 14

14.  If we wanted B=2%, using the sample proportion from the Washington Post poll, recall that the sample proportion was.36 ◦ n=2212.7, so we need a sample of 2213  What do we get if we use the conservative approach? 14STA 291 Summer 2010 Lecture 14

15. 15  To calculate the confidence interval, we use the Central Limit Theorem (np and nq ≥ 5) ◦ What if this isn’t satisfied?  Instead of the typical estimator, we will use  Then the formula for confidence interval becomes STA 291 Summer 2010 Lecture 14

16.  Suppose a student in an advertising class is studying the impact of ads placed during the Super Bowl, and wants to know what the proportion of students on campus watched it. She takes a random sample of 25 students and finds that all 25 watched the Super Bowl. ◦ Find a 95% confidence interval using first method learned for p ◦ Find a 95% confidence interval using the new method if np, nq condition fails STA 291 Summer 2010 Lecture 1416