1. June 23, 2008Stat 111 - Lecture 13 - One Mean1 Inference for a Population Mean Confidence Intervals and Tests with unknown variance and Two- sample Tests Statistics 111 - Lecture 13

2. June 23, 2008Stat 111 - Lecture 13 - One Mean2 Administrative Notes Homework 4 due Wednesday Homework 5 assigned tomorrow The final is ridiculously close (next Thursday)

3. June 23, 2008Stat 111 - Lecture 13 - One Mean3 Outline Review: Confidence Intervals and Hypothesis Tests which assume known variance Population variance unknown: t-distribution Confidence intervals and Tests using the t-distribution Small sample situation Two-Sample datasets: comparing two means Testing the difference between two samples when variances are known Moore, McCabe and Craig: 7.1-7.2

4. June 23, 2008Stat 111 - Lecture 13 - One Mean4 Chapter 5: Sampling Distribution of Distribution of values taken by sample mean in all possible samples of size n from the same population Standard deviation of sampling distribution: Central Limit Theorem: Sample mean has a Normal distribution These results all assume that the sample size is large and that the population variance is known

5. June 23, 2008Stat 111 - Lecture 13 - One Mean5 Chapter 6: Confidence Intervals We used sampling distribution results to create two different tools for inference Confidence Intervals: Use sample mean as the center of an interval of likely values for pop. mean  Width of interval is a multiple Z * of standard deviation of sample mean Z * calculated from N(0,1) table for specific confidence level (eg. 95% confidence means Z * =1.96) We assume large sample size to use N(0,1) distribution, and we assume that  is known (usually just use sample SD s)

6. June 23, 2008Stat 111 - Lecture 13 - One Mean6 Chapter 6: Hypothesis Testing Compare sample mean to a hypothesized population mean  0 Test statistic is also a multiple of standard deviation of the sample mean p-value calculated from N(0,1) table and compared to  -level in order to reject or accept null hypothesis Eg. p-value < 0.05 means we reject null hypothesis We again assume large sample size to use N(0,1) distribution, and we assume that  is known

7. June 23, 2008Stat 111 - Lecture 13 - One Mean7 Unknown Population Variance What if we don’t want to assume that population SD  is known? If  is unknown, we can’t use our formula for the standard deviation of the sample mean: Instead, we use the standard error of the sample mean: Standard error involves sample SD s as estimate of 

8. June 23, 2008Stat 111 - Lecture 13 - One Mean8 t distribution If we have small sample size n and we need to use the standard error formula because the population SD  is unknown, then: The sample mean does not have a normal distribution! Instead, the sample mean has a T distribution with n - 1 degrees of freedom What the heck does that mean?!?

9. June 23, 2008Stat 111 - Lecture 13 - One Mean9 t distribution t distribution looks like a normal distribution, but has “thicker” tails. The tail thickness is controlled by the degrees of freedom The smaller the degrees of freedom, the thicker the tails of the t distribution If the degrees of freedom is large (if we have a large sample size), then the t distribution is pretty much identical to the normal distribution Normal distribution t with df = 5 t with df = 1

10. June 23, 2008Stat 111 - Lecture 13 - One Mean10 Known vs. Unknown Variance Before: Known population SD  Sample mean is centered at  and has standard deviation: Sample mean has Normal distribution Now: Unknown population SD  Sample mean is centered at  and has standard error: Sample mean has t distribution with n-1 degrees of freedom

11. June 23, 2008Stat 111 - Lecture 13 - One Mean11 New Confidence Intervals If the population SD is unknown, we need a new formula for our confidence interval Standard error used instead of standard deviation t distribution used instead of normal distribution If we have a sample of size n from a population with unknown , then our 100·C % confidence interval for the unknown population mean  is: The critical value is calculated using a table for the t distribution (back of textbook)

12. June 23, 2008Stat 111 - Lecture 13 - One Mean12 Tables for the t distribution If we want a 100·C% confidence interval, we need to find the value so that we have a probability of C between -t * and t * in a t distribution with n-1 degrees of freedom Example: 95% confidence interval when n = 14 means that we need a tail probability of 0.025, so t * =2.16 = 0.95 = 0.025 t*t*-t* df = 13

13. June 23, 2008Stat 111 - Lecture 13 - One Mean13 Example: NYC blackout baby boom Births/day from August 1966: Before: we assumed that  was known, and used the normal distribution for a 95% confidence interval: Now: let  be unknown, and used the t distribution with n-1 = 13 degrees of freedom to calculate our a 95% confidence interval: Interval is now wider because we are now less certain about our population SD 

14. June 23, 2008Stat 111 - Lecture 13 - One Mean14 Another Example: Calcium in the Diet Daily calcium intake from 18 people below poverty line (RDA is 850 mg/day) Before: used known  = 188 from previous study, used normal distribution for 95% confidence interval: Now: let  be unknown, and use the t distribution with n-1 = 17 degrees of freedom to calculate our a 95% confidence interval: Again, Wider interval because we have an unknown 

15. June 23, 2008Stat 111 - Lecture 13 - One Mean15 New Hypothesis Tests If the population SD is unknown, we need to modify our test statistics and p-value calculations as well Standard error used in test statistic instead of standard deviation t distribution used to calculate the p-value instead of standard normal distribution

16. June 23, 2008Stat 111 - Lecture 13 - One Mean16 Example: Calcium in Diet Daily calcium intake from 18 people below poverty line Test our data against the null hypothesis that  0 = 850 mg (recommended daily allowance) Before: we assumed known  = 188 and calculated test statistic T= -2.32 Now:  is actually unknown, and we use test statistic with standard error instead of standard deviation:

17. June 23, 2008Stat 111 - Lecture 13 - One Mean17 Example: Calcium in Diet Before: used normal distribution to get p-value = 0.02 Now:  is actually unknown, and we use t distribution with n-1 = 17 degrees of freedom to get p-value ≈ 0.04 With unknown , we have a p-value that is closer to the usual threshold of  = 0.05 than before Normal distribution T= -2.32 prob = 0.01 T= 2.32 t 17 distribution T= -2.26 prob ≈ 0.02 T= 2.26

18. June 23, 2008Stat 111 - Lecture 13 - One Mean18 Review Known population SD  Use standard deviation of sample mean: Use standard normal distribution Unknown population SD  Use standard deviation of sample mean: Use t distribution with n-1 d.f.

19. June 23, 2008Stat 111 - Lecture 13 - Means19 Small Samples We have used the standard error and t distribution to correct our assumption of known population SD  However, even t distribution intervals/tests not as accurate if data is skewed or has influential outliers Rough guidelines from your textbook: Large samples (n> 40): t distribution can be used even for strongly skewed data or with outliers Intermediate samples (n > 15): t distribution can be used except for strongly skewed data or presence of outliers Small samples (n < 15): t distribution can only be used if data does not have skewness or outliers What can we do for small samples of skewed data?

20. June 23, 2008Stat 111 - Lecture 13 - Means20 Techniques for Small Samples One option: use log transformation on data Taking logarithm of data can often make it look more normal Another option: non-parametric tests like the sign test Not required for this course, but mentioned in text book if you’re interested

21. June 23, 2008Stat 111 - Lecture 13 - Means21 Comparing Two Samples Up to now, we have looked at inference for one sample of continuous data Our next focus in this course is comparing the data from two different samples For now, we will assume that these two different samples are independent of each other and come from two distinct populations Population 1:  1,  1 Sample 1:, s 1 Population 2:  2,  2 Sample 2:, s 2

22. June 23, 2008Stat 111 - Lecture 13 - Means22 Blackout Baby Boom Revisited Nine months (Monday, August 8th) after Nov 1965 blackout, NY Times claimed an increased birth rate Already looked at single two-week sample: found no significant difference from usual rate (430 births/day) What if we instead look at difference between weekends and weekdays? SunMonTueWedThuFriSat 452470431448467377 344449440457471463405 377453499461442444415 356470519443449418394 399451468432 WeekdaysWeekends

23. June 23, 2008Stat 111 - Lecture 13 - Means23 Two-Sample Z test We want to test the null hypothesis that the two populations have different means H 0 :  1 =  2 or equivalently,  1 -  2 = 0 Two-sided alternative hypothesis:  1 -  2  0 If we assume our population SDs  1 and  2 are known, we can calculate a two-sample Z statistic: We can then calculate a p-value from this Z statistic using the standard normal distribution Next class, we will look at tests that do not assume known  1 and  2

24. June 23, 2008Stat 111 - Lecture 13 - Means24 Two-Sample Z test for Blackout Data To use Z test, we need to assume that our pop. SDs are known:  1 = s 1 = 21.7 and  2 = s 2 = 24.5 We can then calculate a two-sided p-value for Z=7.5 using the standard normal distribution From normal table, P(Z > 7.5) is less than 0.0002, so our p- value = 2  P(Z > 7.5) is less than 0.0004 We reject the null hypothesis at  -level of 0.05 and conclude there is a significant difference between birth rates on weekends and weekdays Next class: get rid of assumption of known  1 and  2

25. June 23, 2008Stat 111 - Lecture 13 - Means25 Next Class – Lecture 14 More on Comparing Means between Two Samples Moore, McCabe and Craig: 7.1-7.2