HYPOTHESIS TESTING

Chapter 7 HYPOTHESIS TESTING

  A statistical hypothesis is an assertion or conjecture, denoted by H, about a distribution of one or more random variables. If the statistical hypothesis completely specifies the distribution is simple, otherwise is composite.

  Example Let X 1 ,X 2 , ..., X n be a random sample from f (x; θ) = φ μ,25 (x) .

  The statistical hypothesis that the mean of the normal population is less or equal to 17 is denoted by: H : θ ≤ 17. Such a hypothesis is composite, as it does not

  completely specify the distribution. On the other hand, the hypothesis H : θ ≤ 17 is simple since it completely specifies the distribution.

  A test of statistical hypothesis H is a rule or procedure for deciding whether

  to reject H.

  Example Let X 1 ,X 2 , ..., X n be a random sample from f (x; θ) = φ μ,25 (x) .

  Consider H : θ ≤ 17. One possible test Y is as follows: Reject H if and only if

  √

  X > 17 + 5 n .

  In many hypotheses-testing problems two hypotheses are discussed. The first,

  the hypothesis being testing, is called the null hypothesis, denoted by H 0 , and the second is called the alternative hypothesis denoted by H 1 . We say that H 0 is

  tested against or versus H 1 . The thinking is that if the null hypothesis is wrong the

  alternative hypothesis is true, and vice versa. We can make two types of errors:

  Rejection of H 0 when H 0 is true is called a Type I error, and acceptance of

  H 0 when H 0 is false is called a Type II error. The size of Type I error is defined

  Hypothesis testing

  to be the probability that a Type I error is made, and similarly the size of a Type

  II error is defined to be the probability that a Type II error is made.

  Significance level or size of a test, denoted by α, is the supremum of the

  probability of rejecting H 0 when H 0 is correct, i.e. it is the supremum of the Type I

  error. In general to perform a test we fix the size to a prespecified value in general

  10, 5 or 1.

  Example Let X 1 ,X 2 , ..., X n be a random sample from f (x; θ) = φ μ,25 (x) .

  √ Consider H 0 :θ ≤ 17 and the test Y: Reject H 0 if and only if X > 17 + 5 n . Then

  of the test is

  h 17+5 √ i

  1 17+5 −P n−θ X−θ

  = sup 1 −Φ n−θ 5 √ n =1 − Φ(1) = 0.159

  θ ≤17

  7.1 Testing Procedure Let us establish a test procedure via an example. Assume that n = 64, X = 9.8 and

  σ 2 = 0.04 . We would like to test the hypothesis that μ = 10.

  1. Formulate the null hypothesis:

  H 0 : μ = 10

  2. Formulate the alternative:

  H 1 :μ 6= 10

  3. select the level of significance: α = 0.01 From tables find the critical values for Z, denoted by c Z = 2.58 .

  4. Establish the rejection limits:

  Reject H 0 if Z < −2.58 or Z > 2.58.

  5. Calculate Z:

  Z= X−μ σ 0 √ 9.8−10 =

  0.2 √ 64 = −8

  n

  6. Make the decision:

  Testing Proportions

  Since Z is less than −2.58, reject H 0 .

  To find the appropriate test for the mean we have to consider the following

  cases:

  1. Normal population and known population variance (or standard deviation). In this case the statistic we use is:

  X −μ 0

  Z=

  v N(0, 1)

  √ σ

  n

  2. Large samples in order to use the central limit theorem. In this case the statistic we use is:

  X −μ 0

  Z=

  v N(0, 1)

  √ n

  S

  3. Small samples from a normal population where the population variance (or

  standard deviation) is unknown.

  In this case the statistic we use is:

  7.2 Testing Proportions The null hypothesis will be of the form:

  H 0 :π=π 0

  an the three possible alternatives are:

  (1) H 1 :π 6= π 0 two sided test, (2) H 1 :π<π 0 one sided, (3) H 1 :π>π 0 one

  sided. The appropriate statistic is based on the central limit theorem and is:

  p −π 0 2

  Z= S v N(0, 1) where S =π 0 (1 −π 0 )

  √ n

  Example: Mr. X believes that he will get more 60 of the votes. However, in a sample of 400 voters 252 indicate that they will vote for X. At a significance level of 5 test Mr. X belief.

  Hypothesis testing

  p= 252

  400 = 0.63 ,S 2 = 0.6(1 − 0.6) = 0.24. The H 0 :π=π 0 and the alternative is H 1 :π>π 0 . The critical value is 1.64. Now Z = p−π 0 √ S = 0.63−0.6 0.489 √

  n

  Consequently, the null is not rejected as Z < 1.64. Thus Mr. X belief is wrong.

  If fact we have the following possible outcomes when testing hypotheses:

  H 0 is accepted

  H 1 is accepted

  H 0 is correct Correct decision (1 − α)

  Type I error (α)

  H 1 is correct

  Type II error (β)

  Correct decision (1 − β)

  An operating characteristic curve presents the probability of accepting a null hypothesis for various values of the population parameter at a given significance level α using a particular sample size. The power of the test is the inverse function of the operating characteristic curve, i.e. it is the probability of rejecting the null hypothesis for various possible values of the population parameter.

  Part III

  Asymptotic Theory