Statistical_Hypothesis

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Unformatted text preview: Statistical Hypothesis Statistical Hypothesis Testing Chapter 8 What is the probability of finding a sample mean (X) ≥ 108, based on a sample of 9 drawn from a population with μ = 100, σ = 12? Find the standard error of the mean! σx = σ/√N = 12/√9 = 4.0 z = (X­ μ)/ σx = (108­100)/4.0 = 8/4 = 2.0 look up value in col c prob = .0228 Statistical Hypothesis Testing Statistical Hypothesis Testing Why are sampling distributions useful? Used to determine if a sample is representative of a population Did the sample come from a population with known μ? Statistical Hypothesis Statistical Hypothesis Statement of expectation or prediction Sets up a testable situation Establishes a situation in which the theoretical sampling distribution can be obtained H0 specifies Sampling Distribution Gives Probability density function against which to evaluate probability of observing X Test statistic zobs = X­ μ σX Null Hypothesis: H0 States the characteristics of the population from which the sample was believed to be drawn Assume we want to test the hypothesis that USC students have IQ = 100 H : μ = 100 0 Alternative Hypothesis: H1 Covers all other possibilities H1: μ ≠ 100 Example Know μ IQ is 100, σ = 20. We want to take a sample of 16 USC students and test the following Statistical hypothesis: H0 : μ = 100 H1 : μ ≠ 100 σX= σ/√N = 20/√16 = 5.0 ­2 ­1 μ=100 +1 +2 Z Test Z Test Zobs = X – μ σX = σ σX √N Is the IQ of USC students = 100? μ = 100 of US Population σ = 20 NUSC = 16 XUSC = 110 Z = (110­100) = 10 = 2.0 (20/√16) 5 Do USC students have Do USC students have an average IQ = 100? Seems unlikely Prob = .0028 Zobs = 2.0 μ = 100 ­2 z=0 +2 Do USC students have Do USC students have an average IQ = 100? Yes or no? Don’t know for sure Would have to measure everyone in population Statistical decisions by convention If probability of Zobs ≤ .05 we say NO. Decision Conventions Decision Conventions If probability of Zobs is ≤ .05 or ≤ .01 Two­tailed One­tailed H0 : μ = 100 H0 : μ ≤ 100 H1 : μ ≠ 100 .05 or .01 H1 : μ > 100 .05 or .01 Ho and H1 characteristics Ho and H1 characteristics 1) Must be mutually exclusive 2) Include all possible values b) H0 : μ ≤ 100 H1 : μ > 100 a) H0 : μ = 100 H1 : μ ≠ 100 Decision Criterion Decision Criterion Reject H0, accept H1 if probability of Zobs ≤ . 05 (.01) Do not reject H0, do not accept H1 if probability of Zobs > .05 (.01) Critical values of Zobs Two­tailed One­tailed α .05 .01 .05 .01 Zobs 1.960 2.576 1.645 2.326 Statistically significant Zobs falls in the rejection region of the sampling distribution as established by α (.05 or .01) and statistical hypotheses (one or two­tailed) Value of the test statistic is sufficiently rare (by conventional standards) to reject H0 and accept H1 Zobs did not fall in the rejection region Not rare (by convention) Statistically non­significant 5­Step Procedure 5­Step Procedure 1) E must state H0 & H1 2) E must pick a level of α 3) Set the decision stage a. choose a test statistic b. Locate rejection region in the sampling distribution of test­statistic c. identify critical limits d. clearly state value of obs test statistic necessary to reject H0 5­Step Procedure 5­Step Procedure 4) Take a random sample and compute the test statistic 5) Decide: Accept or Reject H0 Statistics Class Performance Statistics Class Performance National average μ = 75.0, σ = 10. USC: Spring 1997 X = 83.4 N = 120 5­Step Procedure 1) H0 : μ ≤ 75 H1 : μ > 75 2) α = .01 3) set decision stage A. use Z as test statistic. μ & σ are known! B. Rejection region C. critical limit 2.33 D. If Zobs ≥ 2.33 reject H 0 If Zobs < 2.33 do not reject H0 4) Take a random sample and compute the test statistic Class of 120 X = 83.4 Zobs = X ­ μ = 83.4 – 75 (σ/√N) (10/√120) Zobs = 8.4 = 8.4 = 9.23 (10/10.95) .91 5) Reject H accept H 0, 1 S97 class did better than National Average Hypothesis Testing and Error Hypothesis Testing and Error State of Nature H0 True Statistical Decision H0 True Correct Decision H1 True Type II β We fail to reject the null – but we should reject! H1 True Type I α We reject the null – but we shouldn’t have Correct Decision One Sample t­test One Sample t­test t­test ­ the procedure is the similar to the one we used for the z test THE BIG DIFFERENCE ­ μ is known but σ(s) estimated t = X – μ sX t = X – μ s/√N Sampling Distribution is the t family Not quite Normal df=2 T Family of Distributions T Family of Distributions ­5 ­4 ­3 ­2 ­1 0 1 2 3 4 5 df = 4 ­5 ­4 ­3 ­2 ­1 0 1 2 3 4 5 df = 60 ­5 ­4 ­3 ­2 ­1 0 1 2 3 4 5 Degrees of Freedom Degrees of Freedom How many pieces of information (scores) are independent of one another (free to vary) in the computation of a statistic. Why does the estimated pop. variance have df = N­1? When you figure out the variance from a sample, you first need to know the mean If you know the mean, one score automatically has to be fixed at a given value Example: X & s Example: X & s 2 X X­X 6 2 5 1 4 0 S2 = ∑(X­X)2 3 ­1 N­1 2 ­2 ∑X = 20 ∑(X­X) = 0 X df = 5 (N) X = 20/5 = 4.0 S2 df = 4 (N­1) What are the dfs? What are the dfs? For X df always equals the number of terms you have N df always equals the number of terms you have – 1 N­1 For s2 and s How to Consult the Family of ts How to Consult the Family of ts 1) Must know dfs: For one­sample t, df = N­1 2) Must know if one­ or two­tailed 3) Must know α for test. Two­tailed α One­Tailed .025 .05 .05 . 005 .01 .01 Critical limits of t Critical limits of t Table A­2 p.491 Two­tailed df .05 .01 2 4.30 9.93 4 2.78 4.60 10 2.23 3.17 60 2.00 2.66 ∞ 1.96 2.58 One­tailed df .05 .01 2 2.92 6.97 4 2.13 4.54 10 1.81 2.76 60 1.67 2.39 ∞ 1.64 2.33 Example of one­sample Example of one­sample Statistics Class Performance National average μ = 75.0, σ = ? USC: Spring 1997 X = 83.4 s=13.4 N = 120 5­Step Procedure 1) H0 : μ ≤ 75 H1 : μ > 75 2) α = .01 3) Set the decision stage A. use t­test because σ is unknown B. Rejection region (df = N­1) C. Critical value df=(120­1)=119 t119=+2.39 D. Reject H0 if t 0 obs ≥ 2.39 Do not reject H0 if tobs< 2.39 4) Collect sample data and compute t X = 83.4 s=13.4 N = 120 t = X – μ = 83.4­75 = 8.4 = 8.4 s/√N (13.4/√120) (13.4/10.95) 1.22 t(119) = 6.89 5) Reject H0, accept H1 S 97 class did better than national average ...
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