1. 6.3 One- and Two- Sample Inferences for Means

2. If σ is unknown Estimate σ by sample standard deviation s The estimated standard error of the mean will be Using the estimated standard error we have a confidence interval of The multiplier needs to be bigger than Z (e.g., 1.96 for 95% CI). The confidence interval needs to be wider to take into account the added uncertainty in using s to estimate s. The correct multipliers were figured out by a Guinness Brewery worker.

3. What is the correct multiplier? “t” 100(1- a )% confidence interval when s is unknown 95% CI =100(1- 0.05 )% confidence interval when s is unknown

4. Properties of t distribution The value of t depends on how much information we have about s. The amount of information we have about s depends on the sample size. The information is “degrees of freedom” and for a sample from one normal population this will be: df=n-1.

5. t curve and z curve Both the standard normal curve N(0,1) (the z distribution), and all t(v) distributions are density curves, symmetric about a mean of 0, but t distributions have more probability in the tails. As the sample size increases, this decreases and the t distribution more closely approximates the z distribution. By n = 1000 they are virtually indistinguishable from one another.

6. Quantiles of t distribution t table is given in the book: Table B.4 It depends on the degrees of freedom as well df probability t 5 0.90 1.476 10 0.95 1.812 20 0.99 2.528 25 0.975 2.060 ∞ 0.9751.96

7. Confidence interval for the mean when s is unknown

8. Example Noise level, n=12 74.0 78.6 76.8 75.5 73.8 75.6 77.3 75.8 73.9 70.2 81.0 73.9 1.Point estimate for the average noise level of vacuum cleaners; 2.95% Confidence interval

9. Solution n=12, Critical value with df=11 95% CI:

13. On the other hand t 0.95,9 =1.833, t 0.05,9 =-1.83 If we have a test statistic value that is either too small ( 1.83), then we have strong evidence against H 0. t=1.74 which is not too small or too big (compared to the cutoff values above/ ”critical values”), then we cannot reject the null.

14. Another method: Rejection Regions Alternative Hypotheses  >  0  <  0   0 Rejection Regions z>z  ----------------- t>t  z<-z  --------------- t<-t  z>z  /2 or z<-z  /2 ------------------ t>t  /2 or t<-t  /2

15. Rejection Region method and p-value method For Ha:  <  0, if z test statistic is less than -1.645, then the p-value is less than 0.05. Comparing the p-value to 0.05 is the same as comparing the z value to -1.645. For t tests we can also find some critical values corresponding to level of  that we can compare to our test statistic. Test statistic in the rejection region is the same as p-value is less than .

16. Paired Data 1 2 3 4 5 6 T=top water zinc concentration (mg/L) B=bottom water zinc (mg/L) 123456 Top0.4150.2380.3900.4100.6050.609 Bottom0.4300.2660.5670.5310.7070.716 1982 study of trace metals in South Indian River. 6 random locations

17. 6.3.2 Paired Mean Difference

23. Assumptions: The population of differences follows a normal distribution. A normal plot of differences, d’s, should be fairly straight. Note: We don’t need B or T to be normal.

24. Tell when to reject H 0 : μ = 120 using a t-test. Answers would be of the form reject H 0 when t < -1.746 or maybe reject H 0 when |t| > 1.746 or maybe reject H 0 when t > 1.746 (a) H A : μ < 120, α = 0.05, n=20 (b) H A : μ > 120, α = 0.10, n=18 (c) H A : μ ≠ 120, α = 0.01, n=9 Rejection region exercise

25. Answers would be of the form 0.10 < p < 0.05 p < 0.001 p > 0.8 After finding the p-value in each case, tell whether to reject or not reject H 0 at the α = 0.05 level. (a) H A : μ > 120, n=7, t = -2.58 (b) H A : μ < 120, n=7, t = -2.58 (c) H A : μ ≠ 120, n=7, t = -2.58 Find p-value exercise

26. 6.3.3 Large Sample Comparisons of Two Means

29. 6.3.4 Small Sample Comparisons of Two Means