* A statistical hypothesis is a claim about the value. of a single population characteristic. * Usually in any hypothesis testing problem,

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1 STAT 215 Fall 2006 L20 TESTING HYPOTHESIS * A statistical hypothesis is a claim about the value of a single population characteristic. * Usually in any hypothesis testing problem, there are two contradictory hypothesis under consideration * The claim or research hypothesis that we wish to establish is called the alternative hypothesis H 1 1

2 * The opposite statement, one that nullifies the research hypothesis H 1, is called the null hypothesis H 0 * A test procedure is a rule, based on the sample data, for deciding whether to reject H 0. The test procedure is specified by 1). a test statistic, a function of the sample data on which the decision (reject H 0 or do not reject H 0 ) is to be based. 2). a rejection region, the set of all test statistic values for which H 0 will be rejected. * The null hypothsis H 0 will be rejected 2

3 if and only if the observed or computed test statistic value falls in the rejection region. Example 1. claims Suppose a cigarette manufacturer that the average nicotine content µ of a brand B cigarettes is at most 1.5 mg. So that, the alternative hypothesis H 1 : µ < 1.5 and the null hypothesis is H 0 : µ = ). Errors in hypothesis testing * A Type I error consists of rejecting the 3

4 null hypothesis H 0 when it is true. * A Type II error involves not rejecting the null hypothesis H 0 when it is false. 4). Level of significance α = Probability of making a Type I error is called the level of significance. Usually, the specified level of significance α =.05 or.01 Power of a Test: 1 - P (Typer II error) = 1 - β Problems: 1. Stated here are some claims or research hypotheses that are to be substantiated by sample data. 4

5 In each case, identify the null hypothesis H 0 and the alternative hypothesis H 1 in terms of the population mean µ. 5

6 Examples: (a) The average mathematics score of the college-bound students in Milwaukee who participated in the American College Testing (ACT) program in 1996 was highter than (b) The meantime for an airline passenger to obtain his or her luggage, once luggage starts coming out on the conveyer belt, is less than 210 seconds. (c) The content of fat in a name-brand chocolate ice cream is more than 4%, the amount printed on the label. 6

7 (d) The average weight of a brand of motors is different from the manufaturer s target of 6 pounds. LARGE SAMPLE CASE. Assume that we have some population with unknown mean = µ and sd = σ. Suppose X 1,..., X n is a random sample from this population with n - sample size. Let us consider the case when n 30. To test the hypothesis H 0 : µ = µ 0 versus the alternative H 1 : µ < µ 0, ( or µ > µ 0 7

8 or µ µ 0 ) consider the standartized random variable here again Z = X µ 0 S/ n, and X = 1 n n X i i=1 S = 1 n 1 n i=1 (X i X) 2. 8

9 Definitions and statistical concepts: 1. Null Hypothesis H 0 : µ = µ 0, the alternative hypothesis H 1 : µ < µ 0 ( H 1 : µ > µ 0 or, for example, H 1 : µ µ 0 ) 2. Type I and Type II errors: Type I error: if H 0 is true and we reject H 0 Type II error: if H 1 is true and we accept H 0 3. The Level of significance: α = P (Type I error) 4. The Power of a Test: 1 - P (Type II error) = 1 - β 9

10 5. Rejection Region: Z z α ( or Z z α ), where Z is a test statistic: Z = X µ 0 S/ n and z α is the α -upper point of standard normal distribution. Our general aim in Hypothesis Testing is to use statistics (Tests) that make α and β as small as possible. These actions are contradictory. Instead, our general strategy: to fix α at some specific level, for example, α =.05,.01, and to 10

11 use the test that max the power of the Test. When H 0 is true and n 30 then statistic Z has approximately standard normal distribution, z-distribution. From the corresponding Table we can determine the critical value z α : P (Z z α ) = α. 11

12 The following sequence of steps is recommended in the Hypothesis testing analysis: 1. Identify the parameter of interest and describe it in the context of the problem situation; 2. Determine the null value and state the null hypothesis H 0 ; 3. State the appropriate alternative hypothesis H 1 ; 4. Give the formula for the computation 12

13 the value of the Test statistic ( T or Z statistics); 5. State the rejection region R for the specified significance level α ( with specified values z α or t α ); 6. Compute any necessary sample quantities ( sample mean, sample deviation), substitute them into the formula for the test statistics and compute that value; 7. Decide whether H 0 should be rejected and state this conclusion in the problem context. 13

14 Testing of statistical hypotheses about population mean µ. Small samples Assume that the population distribution is normal with unknown mean = µ and sd = σ. Suppose X 1,..., X n is a random sample from this population with n 30. To test the hypothesis H 0 : µ = µ 0 versus the alternative H 1 : µ > µ 0, consider the standartized random variable here again T = X µ 0 S/ n, 14

15 and X = 1 n n X i i=1 1 S = (X i n 1 X) 2. i=1 When H 0 is true, T statistic has t-distribution (Student s distribution) with degree of freedom (d.f.) = n - 1. This knowledge allows us to construct the rejection region R : T t α, such that n P (type I error) = P (H 0 is rejected when it is true) = P (T t α ) = Here α is the specified level of significance (usually =.05 ;.01). Remark. The test statistic is the same as in the 15

16 large sample case but is labeled by T to emphasis that its distribution is t-distribution with d.f. n 1 rather then standard normal z - distribution. Examples: 1. A random sample of size 20 from a normal population has x = 182 and s = 2.3. To test H 0 : µ = 181 against H 1 : µ > 181 with α =.05. The null hypothesis H 0 : µ = 181, the alternative hypothesis H 1 : µ > 181. The test statistic T = X 181 S/ 20 Since the sample size n = 20, the statistic T 16

17 has t-distribution with d.f. = 19. The specified level of significance is α =.05. The upper.05 point of t-distribution with d.f. = 19 is t.05 = 1.729, so that the rejection region is R : T Now let us compare the calculated value of T with the value t.05 = 1.729, we get T = X 181 S/ = / = 1.94 > i.e. the value of T lies in the rejection region R. So that we reject the null hypothesis H 0 in favor of the alternative hypothesis H 1 : µ >