Significance Tests. Review Confidence Intervals. The Gauss Model. Genetics

Transcription

1 15.0 Significance Tests Review Confidence Intervals The Gauss Model Genetics Significance Tests 1

2 15.1 CI Review The general formula for a two-sided C% confidence interval is: L, U = pe ± se cv (1 C)/2 where cv (1 C)/2 is the value from a table that has area (1 C)/2 under the curve and to the right. Often the critical value is z from a normal table it is the value that has area C in the middle. If n is large compared to the population, use se F P CF in place of the se The se is sd/ n for an average, sd n for a sum, and ˆp(1 ˆp)/n for a proportion. 2

3 Where do CIs come from? The CLT says: so z = X EV sd/ n N(0, 1) P[ z (1 C)/2 X EV sd/ n z (1 C)/2] C/100 where z (1 C)/2 is the value from a normal table that has area C shaded in the middle. We can use ordinary algebra to manipulate the terms inside the probability statement to solve for L and U. 3

4 C 100 P[ z (1 C)/2 X EV sd/ n z (1 C)/2] = P[ sd n z (1 C)/2 X EV sd n z (1 C)/2 ] = P[ sd n z (1 C)/2 X EV sd n z (1 C)/2 X] = P[ sd n z (1 C)/2 + X EV X sd n z (1 C)/2 ] So L = X sd n z (1 C)/2 U = X + sd n z (1 C)/2 4

5 15.2 The Gauss Model The Gauss model was first used (by Gauss) to find the orbit of Ceres, the first asteroid to be discovered. But before its observation could be confirmed by other astronomers, it disappeared behind the sun. Gauss used his model, and an incredibly complex calculation involving nine coordinate systems, to predict where and when astronomers should look to recapture it when its orbit carried it from behind the sun. 5

6 The Gauss model says that measurement errors in an observation are like draws from a box. The measurement is the true value plus whatever error is marked on the ticket: measurement = true value + random error. The EV of the box is zero (otherwise, the measurement process is biased). The standard deviations of the box is usually, but not always, unknown. 6

7 15.3 Genetics Gregor Mendel was an Augustinian monk in charge of the monastery s truck garden. He noted that several traits in pea plants, e.g.: color height wrinkled pods seemed to be inherited from the parent plants in a predictable way. To study color Mendel got inbred strains, whose progeny were always yellow or always green. Then he did experiments in which those inbred strains were crossed, and he observed the colors of the offspring. 7

8 Recall from biology: Mendelian theory says that each plant has two genes for color, and each parent contributes one of those genes, at random, to the progeny. Thus: GG GG GG Y Y Y Y Y Y Y Y GG Y G Y Y Y G Y Y, Y G GG Y G GG, GY Y G Y G GG, Y Y, Y G, GY Yellow is dominant. Any plant that has a yellow gene provides only yellow peas. 8

9 The inbred plants were GG or Y Y. When crossing these, the first generation all had yellow peas (because of dominance) even the though genetic composition of each plant was Y G. The second generation was formed by crossing the first generation plants. This cross was: Y G Y G GG, Y Y, Y G, GY and it gave plants such that 3/4 had yellow peas, 1/4 had green peas. 9

10 Mendel slowly developed his theory that each plant had two genes, contributed at random. He could predict that among, say, 100 second generation offspring, about 25 should bear green peas. So Mendel made many such crosses and found that his predicted numbers were close to those observed. But how can Mendel prove his theory? He had no statistical way to show that his observed counts of yellow and green pea plants matched well to the predictions from his model. All he could do was present his predictions, his counts, and wave his hands. So he (probably) faked his data in order to get better agreement and thus to present a stronger case. We believe this because his reported counts were too good to be true they were closer to his predictions than could happen under his model. In fact, we will clarify this when we talk about chi square tests. 10

11 15.4 Significance Tests A significance test is a way to decide whether the data strongly support point of view or another. There are many kinds of significance tests, but all involve: a null and alternative hypothesis a test statistic a significance probability (P ). The following gives an overview of most of the different kinds of significance tests. 11

12 Step 1: Pick the null and alternative hypotheses. If possible, put what you want to prove as the alternative. Or make the alternative hypothesis the one that leads to new action. Example 1: H 0 : The mean lifetime with the new drug 72 years. H A : The mean lifetime with the new drug > 72 years. Example 2: H 0 : The mean annual global temperature = 56 F. H A : The mean annual global temperature 56 F. 12

13 Step 2: Calculate the test statistic. The test statisic is a one-number summary of all the information in the sample regarding the correctness of the alternative hypothesis. Different kinds of hypothesis tests (e.g., about means, proportions, differences of means, differences of proportions, etc.) require different test statistics. Soon we shall list many standard cases. Step 3: Find the P -value (or significance probability). Use a table to find the P -value. This is the probability of observing data that is as or more supportive of the alternative hypothesis when the null hypothesis is correct. This interpretation of the P -value is a bit subtle you will need to think about it. 13

14 I The Three Possible Pairs of Null and Alternative Hypotheses 1. H o : θ = θ o versus H A : θ θ o 2. H o : θ θ o versus H A : θ > θ o 3. H o : θ θ o versus H A : θ < θ o Here θ represents a generic parameter. It could be a population mean, a population proportion, the difference of two population means, or many other things. And θ 0 is read theta-nought. It represents some specific value that is being tested. 14

15 II Possible Test Statistics a. For a test on the population mean we take θ to be the population mean µ. If you know the population SD, or for n > 26 you use the sample SD as an estimate of the SD, then you get the significance probability from a z-table and the test statistic is: ts = ( X µ o )/(SD/ n). b. For the previous case, if you have a sample of size n 26 and use the sample SD to estimate the population SD, then the significance probability comes from a t n 1 table and the test statistic is: ts = ( X µ o )/(SD/ n 1). 15

16 c. For a test about a proportion, θ = p. The significance probability comes from a z-table and the test statistic is: po (1 p o ) ts = (ˆp p o )/. n d. For a test of the difference of two means, θ = µ 1 µ 2. Assuming that the sample sizes from each population satisfy n 1 > 26 and n 2 > 26, then the significance probability comes from a z-table and the test statistic is: ts = ( X 1 X 2 θ o )/ SD 2 1 n 1 + SD2 2 n 2. 16

17 e. For a test of the difference of two proportions, take θ = p 1 p 2. Use a z-table for the significance probability and the test statistic: ˆp 1 (1 ˆp 1 ) ts = (ˆp 1 ˆp 2 θ o )/ + ˆp 2(1 ˆp 2 ). n 1 n 2 f. For n 26, with θ = µ 1 µ 2, and n paired differences X i Y i, use t n 1 for the significance probability and the test statistic is: ts = ( X Ȳ θ o)/(sd d / n 1). Here SD d is the sample variance of the n differences. 17

18 III The significance probability of the test statistic depends upon the hypothesis chosen in Part I. For that choice, let W be a random variable with a z or t n 1 distribution, as indicated in Part II. Then 1. The significance probability is P[W ts ] + P[W ts ]. 2. The significance probability is P[W ts]. 3. The significance probability is P[W ts]. The significance probability is THE CHANCE OF OBSERVING DATA THAT SUPPORTS THE ALTERNATIVE HYPOTHESIS AS OR MORE STRONGLY THAN THE DATA YOU HAVE SEEN, WHEN THE NULL HYPOTHESIS IS CORRECT. 18

19 Example: Suppose you have a new drug that may extend life expectancy. You give it to 36 random people and find that their average lifespan is 76 years, and the standard deviation in their lifespan is 12. You know that the average U.S. lifespan is 72, and hope to show that your drug improves that. The first step is to choose the null and alternative hypotheses. For this problem, the following are the right ones. This is case I.2 of the handout: H 0 : The mean lifetime with the new drug 72 years. H A : The mean lifetime with the new drug > 72 years. The second step is to find the test statistic. We are in case II.a of the handout, so: ts = X µ 0 sd/ n = / 36 = 2 19

20 The third step finds the significance probability. If you use hypotheses I.2, then you use rule III.2. The significance probability, or P-value, is P[z > 2] = (1.9545)/2 = So if the null hypothesis is true and the drug does not help, then you have only about chances in 100 of observing the result in your experiment. This is pretty persuasive that your drug helps. If you had a larger sample, then you might get a significance probability that is even smaller, say In this case, there is only 1 chance in a million that you would get such strong support when the drug is not helpful. 20