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 H 0 : E(Y)=500, α=0.01 against H 1 : E(Y) not equal to 500. II. Test H 0 : E(Y)=600, α=0.01 against H 1 : 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 H 0 : E(Y)=500, α=0.01 against H 1 : E(Y) not equal to 500. II. Test H 0 : E(Y)=600, α=0.01 against H 1 : 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 –E 0, expected value of Y under the null –σ 0, standard deviation of Y (a SINGLE value drawn from Y) under the null –α, the level of significance –|z α |, the percentile from the standard normal corresponding to α.

21. One-sample test sample size parameters (continued) Alternative hypothesis values –E 1, expected value of Y under the alternative –σ 1, standard deviation of Y (a SINGLE value drawn from Y) under the alternative –β, the probability of a Type II error. –|z β |, 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 H 0 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 –E 0 sign |z α |*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 Pr 1 {Accept H 0 }= Pr 1 {Test statistic>critical value}= Pr 1 {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, –E 0 =1000, σ 0 =600, α=0.05, |z α |=1.645 For alternative, –E 1 =950, σ 1 =600, β=0.05, |z β |=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.