1. Chapter 10 AP Statistics St. Francis High School Fr. Chris

2. Introduction to Inference 1. Estimating with Confidence 2. Tests of Significance 3. Using Significance Tests 4. Inference as a Decision.

3. Confidence Interval Two parts: the estimate and the confidence level Estimate±Margin of Error If you have an SRS, you can assume that the mean is normally distributed If you can estimate the spread, you can find the boundaries where 90%, 95% or 99% of the area under this normal curve lie.

4. A Pharmaceutical Manufacturer To establish the concentration of an active ingredient of a drug, the same specimen of a particular batch is measured three times: 0.8403, 0.8363, and 0.8447. The standard deviation is known to be 0.0068 grams per liter. What is the 95% confidence interval? So.8404 is a good estimate of the actual amount, but how big a margin of error do we need to be 95% confident the actual measurement is in our interval?

5. The 95% Confidence Interval 95% On a standard normal curve (N[0,1]), 95% is between -1.96 and +1.96 standard deviations. -1.96 +1.96 Since we know sigma, we convert the drug’s distribution to the standard normal so the interval is Where z = 1.96 (95% Confidence)

6. Substituting the numbers... So we can be 95% sure that the actual measurement of the sample is between.8303 and.8505 We would have a more narrow interval if we are willing to drop our confidence interval to 90%. On a standard normal curve, 90% of the area is between -1.645 and +1.65... Or (.8339,.8469)

7. Caution!!!!! Data must be SRS from the population Formulas for more complex designs are available Fancy formulas cannot rescue badly produced data Outliers can influence x-bar If n is small (<15), population needs to be normal Need to know the sigma of the population (If sample is huge, you can use s, otherwise, wait for more methods next chapter!)

8. 10.2 Tests of Significance H 0 and H a One Sided vs. Two Sided Alternative P-Value  Critical values and the z Statistic

9. Blood Pressure Males age 33-44 have a systolic blood pressure of 128 with a  =15. A researcher looks at the records of 72 male ATC’s in this age group and found their average systolic blood pressure to be 131.07. Obviously this is higher than the average. It is likely that this is due to chance variation, or that ATC’s have a significantly higher blood pressure? It is SIGNIFICANT if the result is unlikely to be caused by chance.

10. The Null Hypothesis Males age 33-44 have a systolic blood pressure of 128 with a  =15. 72 of our ATC’s average at over 131. The null Hypothesis is to say There is no difference between ATC’s and males in general. H 0 :  =  0 H a :  ≠  0 One way to state the alternative is to say that the average of ATC’s is different, so the alternative hypothesis is This is a “Two-tail” test, since the alternative includes two possibilities: Either the ATC’s have a higher blood pressure or they have a lower blood pressure.

11. Let’s get graphic A two tailed test means that the sample mean falling on either extreme will be considered grounds to reject the H0H0 Knowing what we know about the Normal Distribution, we know the critical z score that partitions the inner 95% is where the cumulative area is at 2.5% on both ends (thus our  =.05)

12. Are ATC’s average on the fringe? If they belong to the same distribution as anyone else, we would expect the  of the sample mean of 72 randomly selected males to be So to convert to the Standard Normal Curve...

13. Fail to reject the Null Hypothesis Since 1.7366 is less than 1.96, we are in the 95% of sample means, and we cannot reject the H0 H0 … But what if we only consider the alternative that ATC’s, due to the stressful nature of their work, would have a HIGHER blood pressure than other males...

14. The One Tailed Test The one tailed test has the same H0 H0 :  =0 =0 but a different alternative hypothesis, viz., Ha Ha :  >0>0 The critical value is different, since 95% is on the left, and 5% is on the upper end.

15. Rejecting the Null Hypothesis Since our critical value is now 1.645, and the sample mean of 131.07 has a z value of 1.7366, it is in the extreme upper 5%, so we can reject the null hypothesis in favor of the alternative hypothesis: The mean of ATC’s systolic blood pressure is higher than other males of this age group.

16. Review of terms Notice our  is still.05, but only when we consider the one-tailed test (which is appropriate in this case), are we able to reject the null hypothesis. Computers and statistical calculators can compute a p-value (probability) which allows to see how unlikely the average is if the null hypothesis is true.

17. Using your calculator Select STAT, then TESTS Select “1:Z-Test” Select “Stats” since we don’t have the raw data (72 sbp’s) Notice that it shows the z-score, and the p-value (.08 is bigger than.05). So we cannot reject the null hypothesis. Go back and instead of selecting “Calculate”, select “Draw” First let’s select the two tailed alternative hypothesis then press Calculate

18. Using your calculator for One Tail Test Select STAT, then TESTS Select “1:Z-Test” Select “Stats” since we don’t have the raw data (72 sbp’s) First let’s select the two tailed alternative hypothesis  >  0 then press Calculate Notice that it shows the z-score, and the p-value (.04 is less than.05). So we can reject the null hypothesis. Go back and instead of selecting “Calculate”, select “Draw”

19. 10.3 Using Significance Tests Choosing a level of significance When is Statistical Inference Valid Examples

20. Choosing a level P-values are more informative than Traditional  levels Small effects can be highly significant Lack of signifcance doesn’t imply H 0 is true

21. Validity Faulty data collection Outliers Testing a hypothesis on the same data that suggests a hypothesis can invalidate the test Many tests at once can produce significance due to chance alone, even if the H 0 ’s are true

22. Examples Hawthorne Effect: Will background music increase productivity? Any short term change can effect, regard less of what it is (See Example 10.19, p 590) A researcher looking for evidence of ESP tests 500 subjects. Four do significantly better (P<.01) than random guessing. Can we conclude these 4 have ESP? What should the researcher do to test whether these 4 have ESP? (10.61 p. 592)

23. Kidney Failure Patients with kidney failure are often treated by dialysis, and can also cause the retention of phosphorous that must be corrected by changes in the diet. One patient in a study had these 6 measurements over time: 5.6, 5.3, 4.6, 4.8, 5.7, 6.4 (These are separated over time and can be considered an SRS of the patient’s blood phosphorous level). If this level varies normally with  = 0.9 mg/dl, give a 90% confidence interval for the mean blood phosphorus level. (4.7956, 6.0044)

24. Conclusions The normal range of phosphorous in the blood is considered to be 2.6 to 4.8 mg/dl. Is there strong evidence that the patient has a mean phosphorous level that exceeds 4.8? H 0 :  =4.8 H a :  >4.8 Z=1.633 P-value is.051235 There is evidence that the patient’s mean phosphorous level is higher than 4.8 mg/dl, but it is not strong. Whether this is sufficient evidence is a judgement call..05 is not sacred What if the inference is use to determine a change in diet? Surgury? What are the hypotheses? What is the z-score/P-value? Conclusion?