Today’s lesson

1. Today’s lesson • Confidence intervals for the expected value of a random variable. • Determining the sample size needed to have a specified probability of a Type II error and probability of a Type I error.

2. Confidence Interval for the Mean of a Normally Distributed Random Variable • ASS-U-ME that Y is normally distributed with unknown mean μ and known standard deviation (say 100). • Confidence interval for the mean is the set of “reasonable values” based on the data observed.

3. Example Problem 1 • The random variable Y is normally distributed with unknown mean and standard deviation 100. A random sample of 25 observations is taken from Y and has mean value 515. What is the 99 percent confidence interval for the mean of Y?

4. Solution to Problem 1 • Find the standard error of the mean: • standard deviation of Y/square root of sample size=100/square root of 25=20 • Find the factor for the size of the confidence interval: • use 1.960 for 95 percent CI • use 2.576 for 99 percent CI

5. Solution to Problem 1 • Multiply standard error and factor: • 2.576*20=51.52 • Add and subtract this product to the mean: • Left end point is mean-51.52=463.48 • Right end point is mean+51.52=566.52 • In real life, round off CI to “non-obsessive” numbers.

6. Statement of Solution • The 99 percent confidence interval for the expected value of Y (mean of Y) is the interval between 463.48 and 566.52. • The difference between the left and right end points of the CI is a measure of how much sampling variability is present in the experimental results.

7. Example Problem 2 • Test the following two situations and select the answer that describes your conclusions: • I. Test H0: E(Y)=500, α=0.01 against H1: E(Y) not equal to 500. • II. Test H0: E(Y)=600, α=0.01 against H1: E(Y) not equal to 600.

8. Example Problem 2 Options • A) Reject null hypothesis I and reject null hypothesis II. • B) Reject null hypothesis I and accept null hypothesis II. • C) Accept null hypothesis I and reject null hypothesis II. • D) Accept null hypothesis I and accept null hypothesis II.

9. Solution to Problem 2 • Use the confidence interval calculated in problem 1, the interval between 463.48 and 566.52. • Null mean in I is 500, which is in 99 percent CI; hence accept in I. • Null mean in II is 600, which is not in 99 percent CI; hence reject in II. • Answer is C.

10. Hints and Reminders • READ YOUR COMPUTER OUTPUT. • Use the numbers that the computer calculates. • MAKE SURE THAT THE PARAMETER IN THE QUESTION AND THE PARAMETER YOU ARE CALCULATING ARE THE SAME!

11. Example Question 3 • The random variable Y is normally distributed with unknown mean and unknown standard deviation. A random sample of 4 observations is taken from Y The mean is 515, and the unbiased estimate of the variance is 8100. What is the 99 percent confidence interval for the mean of Y?

12. Solution to Problem 3 • Recognize that this problem requires the use of Student’s t (the standard deviation is not known). • Find the estimated standard error of the mean: • square root of the unbiased estimate of the variance/square root of sample size=90/square root of 4=45

13. Solution to Problem 3 Continued • Find the degrees of freedom for the estimate of the unknown standard deviation: • size of sample-1=4-1=3. • Find the factor for the size of the confidence interval: • stretch 1.960 for 95 percent CI; for 3 df, 3.182 • stretch 2.576 for 99 percent CI; for 3 df, 5.841

14. Solution to Problem 3 Continued • Multiply standard error and factor: • 5.841*45=262.8 • Add and subtract this product to the mean: • Left end point is mean-262.8=252.2 • Right end point is mean+262.8=777.8 • In real life, round off CI to “non-obsessive” numbers.

15. How to Use Student t Confidence Interval for Mean • Exactly the same as the use of the normal confidence interval for the mean.

16. Example Problem 4 • Test the following two situations and select the answer that describes your conclusions: • I. Test H0: E(Y)=500, α=0.01 against H1: E(Y) not equal to 500. • II. Test H0: E(Y)=600, α=0.01 against H1: E(Y) not equal to 600.

17. Example Problem 4 Options • A) Reject null hypothesis I and reject null hypothesis II. • B) Reject null hypothesis I and accept null hypothesis II. • C) Accept null hypothesis I and reject null hypothesis II. • D) Accept null hypothesis I and accept null hypothesis II.

18. Solution to Example Problem 4 • Use the confidence interval calculated in problem 3, the interval between 252.2 and 777.8. • Null mean in I is 500, which is in 99 percent CI; hence accept in I. • Null mean in II is 600, which is also in 99 percent CI; hence reject in II. • Answer is D.

19. Determining Sample Size • Design in a statistical study is crucial for success. • Key issue is how large does the sample size have to be. • There is a key formula for determining the sample size.

20. One-sample test sample size parameters • ASS-U-ME sampling for Y, a normally distributed random variable. • Null hypothesis values • E0, expected value of Y under the null • σ0, standard deviation of Y (a SINGLE value drawn from Y) under the null • α, the level of significance • |zα|, the percentile from the standard normal corresponding to α.

21. One-sample test sample size parameters (continued) • Alternative hypothesis values • E1, expected value of Y under the alternative • σ1, standard deviation of Y (a SINGLE value drawn from Y) under the alternative • β, the probability of a Type II error. • |zβ|, the percentile from the standard normal corresponding to β.

22. Sample Size Formula • Use a sample size n that is as large or larger than:

23. Example Problem Scenario • A research team will test the null hypothesis that E(Y)=1000 at the 0.05 level of significance against the alternative that E(Y)<1000. When the null hypothesis is true, Y has a normal distribution with standard deviation 600.

24. Standard Warm-up Problem • What is the standard deviation of the mean of 900 observations under the null hypothesis? • Solution: • The standard error is the standard deviation of Y under H0 divided by the square root of the sample size. • 600/square root of 900=600/30=20.

25. A Sometime Warm-up Problem • What is the critical value of the null hypothesis in the scenario when using the average of 900 observations as the test statistic. • Solution • E0 sign |zα|*standard error of test statistic • left sided test, hence use - • 1000 - 1.645*20=967.1

26. Example Problem 5 • What is the probability of a Type II error for an alternative in which Y is normally distributed, E(Y)=950, and its standard deviation is 600 using the average of a random sample of 900 as the test statistic of the null hypothesis in the scenario?

27. Solution to Example Problem 5 • The probability of a Type II error β is equal to Pr1{Accept H0}= Pr1{Test statistic>critical value}= Pr1{Sample mean>967.1}. • Under alternative, sample mean is • normal • with mean 950 • with standard error 20

28. Solution to Example Problem 5 Continued • The problem now becomes: what is the probability that a normally distributed random variable with mean 950 and standard deviation 20 is larger than 967.1? • Find standard units value of 967.1. • (967.1-950)/20=0.855 • What is Pr{Z>0.855}? • This is about 0.197.

29. Example Problem 6 • What is the smallest sample size so that the probability of a Type II error is 0.05 when the (alternative) distribution of Y is normally, E(Y)=950, and its standard deviation is 600. The test statistic is the average of a random sample of n as the test statistic of the null hypothesis in the scenario.

30. Solution • Use formula: • For null, • E0=1000, σ0=600, α=0.05, |zα|=1.645 • For alternative, • E1=950, σ1=600, β=0.05, |zβ|=1.645

31. Solution Continued • Plug and chug: • square root of sample size required is 39.48 • Required sample size is the square of 39.48,which is 1558.6. • Round up to 1559. • This is optimistic. How do you account for nonresponse?

32. Review of Today’s lecture • One sample procedures (today confidence intervals). • Determining Sample Size • When you solve problem, think about the meaning of the answer.