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Statistical Hypotheses A statistical hypothesis is an assertion or conjecture about a population, which may be expressed in terms of ◦ some parameter: mean is zero; ◦ some parameters: mean and median are identical; or ◦ some sampling distribution: this sample is normally distributed. Test problem - decide between two hypotheses ◦ the null hypothesis H0 and ◦ the alternative hypothesis Ha . Popperian approach to scientific theories ◦ Scientific theories are subject to falsification. ◦ It is impossible to verify a scientific theory. Null hypothesis H0 default (current) theory which we try to falsify Alternative hypothesis Ha alternative to adopt if null hypothesis is rejected Examples: ◦ Clinical study of new drug - H0 : drug has no effect ◦ Criminal case - H0 : suspect is not guilty ◦ Safety test of nuclear power station - H0 : power station is not safe ◦ Chances of new investment - H0 : project not profitable ◦ Testing for independence - H0 : random variables are independent Testing Hypotheses II, Feb 18, 2004 -1- Statistical Tests Example: Testing for pesticide in discharge water Suppose the Environmental Protection Agency takes 10 readings on the amount of pesticide in the discharge water of a chemical company. Question: Does the concentration cP of pesticide in the water exceed the allowed maximum concentration c0 ? ◦ Before taking action against the company, the agency must have some evidence that the concentration cP exceeds the allowed level. ◦ Without evidence the agency assumes that the pesticide concentration cP is within the limits of the law. Consequently, the null hypothesis of the agency is that the pesticide concentration cP does not exceed c0 . Thus the question corresponds to the test problem H0 : c P ≤ c 0 vs Ha : c P > c 0 . Suppose that the company regularly also runs tests on the amount of pesticide in the discharge water. Question: Does the concentration cP of pesticide in the water exceed the allowed maximum concentration c0 ? ◦ The aim of the company is to avoid fines for exceeding the allowed level. Thus the company wants to make sure that the concentration stays within the allowed limits. Thus, the null hypothesis of the company should be that the pesticide concentration cP exceeds c0 . The question now corresponds to the test problem H0 : c P ≥ c0 vs Testing Hypotheses II, Feb 18, 2004 Ha : c P < c 0 . -2- Six Steps of Conducting a Test Steps of a significance test 1. Determine null hypothesis H0 and alternative Ha . 2. Decide on probability of type I error, the significance level α. 3. Find an appropriate test statistic T . 4. Based on the sampling distribution of T , formulate a criterion for testing H0 against Ha . 5. Calculate value of the test statistic T . 6. Decide whether or not to reject the null hypothesis H0 . Example: Fair coin (contd) We want to decide from 100 tosses of a coin whether it is fair or not. Let θ be the probability of heads. 1. Test problem: H0 : θ = 21 vs Ha : θ 6= 1 2 2. Significance level: α = 0.05 (most commonly used significance level) 3. Test statistic: T =X (number of heads in 100 tosses of the coin) 4. Rejection criterion: reject H0 if T ∈ / [40, 60] 5. Observed value of test statistic: Suppose after 100 tosses we obtain t = 55 6. Decision: Since 55 does not lie in the rejection region, we do not reject H0 . Testing Hypotheses II, Feb 18, 2004 -3- One and Two-sided Hypotheses Example: Blood cholesterol after a heart attack Suppose we are interested in whether the blood cholesterol level two days after a heart attack differs from the average cholesterol level in the (general) population (µ0 = 193). Two cases: ◦ We are interested in any difference from the population mean µ0 . Then we have a two-sided test problem H0 : µY1 = µ0 vs H0 : µY1 6= µ0 . ◦ We suspect that the cholesterol level after a heart attack might me higher than in the general population. In this case, we have a one-sided test problem H0 : µY1 = µ0 vs H0 : µY1 > µ0 . Remark: ◦ More generally, we might be interested in one-sided test problems of the form H0 : µY1 ≤ µ0 vs H0 : µY1 > µ0 , which accounts for the possibility that µ might be smaller than µ0 . ◦ For all common test situations (in particular those discussed in this course), the form of the test does not depend on the form of H0 , but only on the parameter value in H0 that is closest to Ha , that is µ0 . Testing Hypotheses II, Feb 18, 2004 -4- Test Statistic Let θ be the parameter of interest. Two-sided test problem H0 : θ = θ0 against Ha : θ 6= θ0 One-sided test problem H0 : θ = θ0 against Ha : θ > θ0 (or Ha : θ < θ0 ) Suppose that θ̂ is an estimate for θ. ◦ If θ = θ0 (null hypothesis), we expect the estimate θ̂ to take a value near θ0 . ◦ Large deviations from θ0 are evidence against H0 . This suggests the following decision rules: ◦ Ha : θ > θ0 : reject H0 if θ̂ − θ0 is much larger than zero ◦ Ha : θ < θ0 : reject H0 if θ̂ − θ0 is much smaller than zero ◦ Ha : θ 6= θ0 : reject H0 if |θ̂ − θ0 | is much larger than zero Problem: Often the sampling distribution of the estimate θ̂ depends on the unknown parameter θ. Definition (Test statistic) A test statistic is a random variable ◦ that measures the compatibility between the null hypothesis and the data and ◦ has a sampling distribution which we know (under H0 ). Testing Hypotheses II, Feb 18, 2004 -5- Test Statistic Example: Blood cholesterol after a heart attack Data: X1 , . . . , X28 ◦ blood cholesterol level of 28 patients two days after a heart attack 2 ◦ assumed to be normally distributed with mean µX and variance σX The parameter µ can be estimated by the sample mean 28 2 1 P σX X̄ = Xi ∼ N µX , . 28 i=1 28 This suggests to the standardized sample mean as a test statistic X̄ − µ0 √ ∼ N (0, 1) σ/ 28 (under H0 ). Test H0 : µ ≤ 193 vs Ha : µ > 193 at significance level α = 0.05 ◦ Test statistic: Assume σ = 47.7 to be known. T = X̄ − µ0 √ σ/ 28 ◦ Rejection criterion: Reject H0 if T > z0.05 = 1.645 ◦ Outcome of test: Since the observed value of T is t= 253.9 − 193 √ = 6.76, 47.7/ 28 we reject the null hypothesis that µ = 193. Testing Hypotheses II, Feb 18, 2004 -6- Tests for the Mean Tests for the mean µ (σ 2 known): ◦ Test statistic: T = X̄ − µ0 √ σ/ n ◦ Two sided test: H0 : µ = µ0 against Ha : µ 6= µ0 reject H0 if |T | > zα/2 ◦ One sided tests: H0 : µ = µ0 against Ha : µ > µ0 reject H0 if T > zα (µ < µ0 ) (T < −zα ) Tests for the mean µ (σ 2 unknown): ◦ Test statistic: T = X̄ − µ0 √ s/ n ◦ Two sided test: H0 : µ = µ0 against Ha : µ 6= µ0 reject H0 if |T | > tn−1,α/2 ◦ One sided tests: H0 : µ = µ0 against Ha : µ > µ0 reject H0 if T > tn−1,α (µ < µ0 ) (T < −tn−1,α ) Example: Blood cholesterol after a heart attack Estimating the standard deviation from the data, we obtain the test statistic T = X̄ − µ0 √ ∼ t27 . s/ 28 Noting that t27,0.05 = 1.703 and t = 6.76, we still reject H0 . Testing Hypotheses II, Feb 18, 2004 -7- Tests and Confidence Intervals Consider level α significance test for the two-sided test problem H0 : θ = θ0 vs Ha : θ 6= θ0 . Let ◦ T = Tθ0 (X) be the test statistic of the test (depends on θ0 ) ◦ R be the critical region of the test Then C(X) = {θ : Tθ (X) ∈ / R} is a (1 − α) confidence interval for θ: If θ is the true parameter, then Pθ θ ∈ C(X) = Pθ Tθ (X) ∈/ R = 1 − Pθ Tθ (X) ∈ R = 1 − α. We have θ0 ∈ C(X) ⇔ Tθ0 (X) ∈ / R ⇔ H0 is not rejected Result A level α two-sided significance test rejects the null hypothesis H0 : θ = θ0 if and only if the parameter θ0 falls outside a (1 − α) confidence interval for θ. Example: Normal distribution iid Let X1 , . . . , Xn ∼ N (µ, σ 2 ). We reject H0 : µ = µ0 if X̄ − µ0 √ > tn−1,α/2 s/ n or equivalently X̄ − µ0 > tn−1,α/2 √s n Rearranging terms, we find that we reject if h i s s µ0 ∈ / X̄ − tn−1,α/2 √ , X̄ + tn−1,α/2 √ . n Testing Hypotheses II, Feb 18, 2004 n -8- The P -value Definition (P -value) The probability that under the null hypothesis H0 the test statistic would take a value as extreme or more extreme that that actually observed is called the P -value of the test. The P -value is often interpreted a measure for the strength of evidence against the null hypothesis: the smaller the P -value, the stronger the evidence. However: ◦ The P -value is a random variable (under H0 uniformly distr. on [0, 1]). ◦ Without a measure of its variability it is not safe to interpret the actually observed P -value. ◦ If the P -value is smaller than the chosen significance level α, we reject the null hypothesis H0 . Three approaches to deciding on test problem: ◦ reject if θ0 ∈ / C(X) ◦ reject if T (X) ∈ R ◦ reject if P -value p ≤ α Example: Blood cholesterol after a heart attack The observed value for the test statistic T = X̄ − µ0 √ ∼ t27 . s/ 28 is t = 6.76. The corresponding P -value is P(T > 6.76) = 1.47 · 10−07. We thus reject the null hypothesis. Equivalently, the confidence interval for µ is [235.43, 272.42]. Since it does not contain µ0 = 193 we reject H0 (for the third and last time!). Testing Hypotheses II, Feb 18, 2004 -9- Example Data: Banks’ net income ◦ percent change in net income between first half of last year and first half of this year ◦ sample mean x̄ = 8.1% ◦ sample standard deviation s = 26.4% Test problem: H0 : µ = 0 against Ha : µ 6= 0 . ttesti 110 8.1 26.4 0 One-sample t test -----------------------------------------------------------------| Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ----+------------------------------------------------------------x | 110 8.1 2.517141 26.4 3.111108 13.08889 -----------------------------------------------------------------Degrees of freedom: 109 Ho: mean(x) = 0 Ha: mean < 0 t = 3.2179 P < t = 0.9991 Ha: mean != 0 t = 3.2179 P > |t| = 0.0017 Ha: mean > 0 t = 3.2179 P > t = 0.0009 Critical value of t distribution with 109 degrees of freedom: t109,0.025 = 1.982 Result: ◦ |t| > t109,0.025 , therefore the test rejects H0 at significance level α = 0.05. ◦ Equivalently, µ0 = 0 ∈ / [3.11, 13.09] and thus the test rejects H0 . ◦ Equivalently, P -value is less than α = 0.05 and thus the test rejects H0 . Testing Hypotheses II, Feb 18, 2004 - 10 -