1. Two-Sample Inference Procedures with Means

2. Remember: We will be interested in the difference of means, so we will use this to find standard error.

3. Suppose we have a population of adult men with a mean height of 71 inches and standard deviation of 2.6 inches. We also have a population of adult women with a mean height of 65 inches and standard deviation of 2.3 inches. Assume heights are normally distributed. Describe the distribution of the difference in heights between males and females (male- female). Normal distribution with  x-y =6 inches &  x-y =3.471 inches

4. 7165 Female Male 6 Difference = male - female  = 3.471

5. a)What is the probability that the height of a randomly selected man is at most 5 inches taller than the height of a randomly selected woman? b) What is the 70th percentile for the difference (male-female) in heights of a randomly selected man & woman? P((x M -x F ) < 5) = normalcdf(-∞,5,6,3.471) =.3866 (x M -x F ) = invNorm(.7,6,3.471) = 7.82

6. a) What is the probability that the mean height of 30 men is at most 5 inches taller than the mean height of 30 women? b) What is the 70th percentile for the difference (male-female) in mean heights of 30 men and 30 women? 6.332 inches P((x m – x w )< 5) =.0573

7. Two-Sample Procedures with means two treatments two populationsThe goal of these inference procedures is to compare the responses to two treatments or to compare the characteristics of two populations. We have INDEPENDENT samples from each treatment or population When we compare, what are we interested in?

8. Assumptions: two SRS’s two randomly assignedHave two SRS’s from the populations or two randomly assigned treatment groups Samples are independent Both distributions are approximately normally –Have large sample sizes –Graph BOTH sets of data  ’s  ’s unknown

9. Hypothesis Statements: H 0 :  1 -  2 = 0 H a :  1 -  2 < 0 H a :  1 -  2 > 0 H a :  1 -  2 ≠ 0 H 0 :  1 =  2 H a :  1 <  2 H a :  1 >  2 H a :  1 ≠  2 BOTH Be sure to define BOTH  1 and  2 !

10. Formulas NOT Since in real-life, we will NOT know both  ’s, we will do t-procedures.

11. Hypothesis Test: Since we usually assume H 0 is true, then this equals 0 – so we can usually leave it out

12. Degrees of Freedom Option 1: use the smaller of the two values n 1 – 1 and n 2 – 1 This will produce conservative results – higher p-values & lower confidence. Option 2: approximation used by technology Calculator does this automatically!

13. Dr. Phil’s Survey VS If there were such a thing as a personality test, do you think that the guys’ personalities would be different from the girls?

14. Two competing headache remedies claim to give fast- acting relief. An experiment was performed to compare the mean lengths of time required for bodily absorption of brand A and brand B. Assume the absorption time is normally distributed. Twelve people were randomly selected and given an oral dosage of brand A. Another 12 were randomly selected and given an equal dosage of brand B. The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow: meanSDn Brand A20.18.712 Brand B18.97.512 Is there sufficient evidence that these drugs differ in the speed at which they enter the blood stream?

15. Have 2 independent randomly assigned treatments Given the absorption rate is normally distributed  ’s unknown Since p-value > a, I fail to reject H 0. There is not sufficient evidence to suggest that these drugs differ in the speed at which they enter the blood stream. State assumptions! Formula & calculations Conclusion in context H 0 :  A =  B H a :  A =  B Where  A is the true mean absorption time for Brand A &  B is the true mean absorption time for Brand B Hypotheses & define variables!

16. Suppose that the sample mean of Brand B is 16.5, then is Brand B faster? No, I would still fail to reject the null hypothesis.

17. A modification has been made to the process for producing a certain type of time-zero film (film that begins to develop as soon as the picture is taken). Because the modification involves extra cost, it will be incorporated only if sample data indicate that the modification decreases true average development time by more than 1 second. Should the company incorporate the modification? Original8.65.14.55.46.36.65.78.5 Modified5.54.03.86.05.84.97.05.7

18. Assume we have 2 independent SRS of film Both distributions are approximately normal due to approximately symmetrical boxplots  ’s unknown H 0 :  O -  M = 1 H a :  O -  M > 1 Where  O is the true mean developing time for original film &  M is the true mean developing time for modified film Since p-value > , I fail to reject H 0. There is not sufficient evidence to suggest that the company incorporate the modification.

19. Confidence intervals: Called standard error

20. Two competing headache remedies claim to give fast- acting relief. An experiment was performed to compare the mean lengths of time required for bodily absorption of brand A and brand B. Assume the absorption time is normally distributed. Twelve people were randomly selected and given an oral dosage of brand A. Another 12 were randomly selected and given an equal dosage of brand B. The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow: meanSDn Brand A20.18.712 Brand B18.97.512 Find a 95% confidence interval difference in mean lengths of time required for bodily absorption of each brand.

21. Assumptions: Have 2 independent randomly assigned treatments Given the absorption rate is normally distributed  ’s unknown We are 95% confident that the true difference in mean lengths of time required for bodily absorption of each brand is between –5.685 minutes and 8.085 minutes. State assumptions! Formula & calculations Conclusion in context From calculator df = 21.53, use t* for df = 21 & 95% confidence level Think “Price is Right”! Closest without going over

22. Brand A Brand B 517, 495, 503, 491493, 508, 513, 521 503, 493, 505, 495541, 533, 500, 515 498, 481, 499, 494536, 498, 515, 515 Compute a 95% confidence interval for the mean difference in amount of acetaminophen in Brand A and Brand B. In an attempt to determine if two competing brands of cold medicine contain, on the average, the same amount of acetaminophen, twelve different tablets from each of the two competing brands were randomly selected and tested for the amount of acetaminophen each contains. The results (in milligrams) follow.

23. Input Brand A data into L 1 and Brand B data into L 2. I am computing a 2-sample T interval for means at a 95% confidence level. Confidence level: (-28.48, -7.189) with 17 df I am 95% confident that the true difference in the mean amount of acetaminophen in Brand A is between 28.5 and 7.2 mg lower than in Brand B.

24. Note: confidence interval statements “mean difference”Matched pairs – refer to “mean difference” “difference of means”Two-Sample – refer to “difference of means”

25. Pooled procedures: sameUsed for two populations with the same variance When you pool, you average the two-sample variances to estimate the common population variance. DO NOT use on AP Exam!!!!! We do NOT know the variances of the population, so ALWAYS tell the calculator NO for pooling!

26. Robustness: more robustTwo-sample procedures are more robust than one-sample procedures BESTBEST to have equal sample sizes! (but not necessary)