Introduction to Statistics

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Christine Gibbs
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1 MTH4106 Introduction to Statistics Notes 15 Spring 2013 Testing hypotheses about the mean Earlier, we saw how to test hypotheses about a proportion, using properties of the Binomial distribution It is quite tricky Testing hypotheses is actually easier when the underlying distribution is continuous, because then we can choose the rejection region to have any specified significance level First of all we will revisit the snowy day example Example The time required by workers to complete an assembly job usually has a mean of 50 minutes and a standard deviation of minutes On a snowy day, the supervisor wonders if they are working at the same speed as usual He intends to record the times that 60 workers take to complete one assembly job apiece Suppose that the supervisor asks Has the mean time changed? The null hypothesis is one of no change, the alternative is there is a change so Null hypothesis: µ = 50 Alternative hypothesis: µ 50 We know that the distribution of X is approximately normal with mean µ and variance 2 60, since σ2 = 2 and n = 60 ie X N(µ, 2 60 ) It follows, taking off the mean and dividing by the standard deviation, that Z = X µ 60 = ( X µ) 60 N(0, 1) 1

2 So if the null hypothesis is true the test statistic Z = ( X 50) 60 will have a standard normal distribution Large values of Z will give evidence against the null hypothesis H 0 Suppose the supervisor uses a significance level of 5% Remember this means the probability of a Type I error is 5%, that is the probability of rejecting H 0 when it is true is 5% The rejection region is {z : z < 196 or z > 196} which we can also write as {z : z > 196} If the supervisor observes x = 5267, the observed value of the test statistic is z = ( ) 60 = 255 Since 255 > 196 this value lies in the rejection region and we can reject H 0 at the 5% significance level Note that the probability of a Type II error, ie accepting the null hypothesis H 0 when it is false, is not usually calculated Firstly, it needs the supposed true vale of µ Secondly, it is usually used to compare different tests with equal significance levels to see which is better and here we have only one test But we can say that the probability of a Type II error will go down the larger n is Suppose now that the supervisor actually suspects that they have got slower, ie they are taking longer to complete the job, and wants to find out if he is right The null hypothesis is one of no change, the alternative is there is an increase Null hypothesis: µ = 50 Alternative hypothesis: µ > 50 Note we could also write the null hypothesis as µ 50 in this one-sided test 2

3 As before we know that the distribution of X is approximately It follows that N(µ, 2 60 ) Z = X µ 60 = ( X µ) 60 N(0, 1) So if the null hypothesis is true the test statistic Z = ( X 50) 60 will have a standard normal distribution Now large values of Z will give evidence against the null hypothesis H 0 Suppose the supervisor uses a significance level of 5% Remember this means the probability of a Type I error is 5%, that is the probability of rejecting H 0 when it is true is 5% The rejection region is {z > 16449} 0 5% If the supervisor observes x = 5267, the observed value of the test statistic is z = ( ) 60 = 255 Since 255 > this value lies in the rejection region and we can reject H 0 at the 5% significance level 3

4 P values We discussed the use of P values They can be useful as a hypothesis test with a fixed significance level α does not give any indication of the strength of evidence against the null hypothesis The P value depends on whether we have a one-sided or two-sided test Consider a one sided test of H 0 : µ = µ 0 versus H 1 : µ > µ 0 Assuming the population variance is known the P value is P(Z > z obs ) where z obs is the observed value of Z For H 1 : µ < µ 0 the P value is P(Z < z obs ) For a two sided test with H 1 : µ µ 0 it is P(Z > z obs Z < z obs ) = 2P(Z > z obs ) In each case we can think of the P value as the probability of obtaining a value of the test statistic more extreme than we did observe assuming that H 0 is true What is regarded as more extreme depends on the alternative hypothesis If the P value is small that is evidence that H 0 may not be true It is useful to have a scale of evidence to help us interpret the size of the P value There is no agreed scale but the following may be useful as a first indication: P value Interpretation P > 010 No evidence against H < P < 010 Weak evidence against H < P < 005 Moderate evidence against H < P < 001 Strong evidence against H 0 P < 0001 Very strong or overwhelming evidence against H 0 Note that the P value is the smallest level of significance that would lead to rejection of the null hypothesis Example Consider the snowy day example again For the two sided test the P-value is 2 P( X 5267) if µ = 50 This is equal to ( ) P Z / = 2 P(Z 255) = This is very small and there is very strong evidence against H 1 Similarly for the one-sided test the P-value is P(Z 255) =

5 Example Drills being manufactured are supposed to have a mean length of 4cm From past experience we know the standard deviation is equal to 1cm and the lengths are normally distributed A random sample of 10 drills had a mean of 45cm Test the hypothesis that the mean is 40 with significance level α = 005 and find the P value We have H 0 : µ = 40 versus H 1 : µ 40 We know that so if H 0 is true ( X N µ, 1 ) 10 Z = X 4 1/10 N(0,1) The observed value of Z is /10 = 15 For a 2 sided test with α = 005 the rejection region is {z : z > 196} Since z = 15 we do not reject H 0 at the 5% level The P value is 2 P(Z > 15) = 2 (1 Φ(15)) = 2( ) = and so there is no evidence against H 0 Example An advertisement for a certain brand of cigarettes claimed that on average there is no more than 1mg of nicotine per cigarette A test of 12 cigarettes gave a sample mean of x = 191 Assuming σ 2 = 4 and the distribution of the amount of nicotine is normally distributed test this claim with a significance level of α = 005 and find the P value We have H 0 : µ = 10 versus H 1 : µ > 10 We know that so if H 0 is true ( X N µ, 4 ) 12 Z = X 1 1/3 N(0,1) The observed value of Z is /3 = 1905 For a 1 sided test with α = 005 the rejection region is {z : z > 16449} Since z = we can reject H 0 at the 5% level 5

6 Note that if we had chosen α = 002 then z α = we we would fail to reject H 0 at the 2% significance level The P value is P(Z > 1905) = (1 Φ(1905)) = = 0024 Thus we have moderate evidence against H 0 General procedure for hypothesis tests of the mean value Suppose that µ is the unknown mean of some large population with variance σ 2 ; that H 0 and H 1 are phrased in terms of µ; that X is the mean of a random sample of size n; and that we can assume that X is (approximately) normal Remember that E( X) = µ and Var( X) = σ 2 /n 1 Population variance σ 2 known (a) One-sided tests (also called one-tailed tests) (i) H 0 : µ µ 0, where µ 0 is known; H 1 : µ > µ 0 We know that X µ 0 σ/ n N(0,1) under H 0, so we take X µ 0 σ/ as the test statistic For significance level α, the n rejection region is {z : z > z α } α 0 z α Given the sample mean x of the data, reject H 0 if x µ 0 σ/ n z α (ii) H 0 : µ µ 0, where µ 0 is known; H 1 : µ < µ 0 By a similar argument, reject H 0 if x µ 0 σ/ n z α 6

7 (b) Two-sided tests (also called two-tailed tests) H 0 : µ = µ 0, where µ 0 is known; H 1 : µ µ 0 Use the same test statistic X µ 0 σ/ but with the symmetric rejection region n {z : z < z α/2 z > z α/2 } α/2 α/2 z α/2 0 z α/2 Reject H 0 if x µ 0 σ/ n z α/2 or x µ 0 σ/ n z α/2 2 σ 2 unknown, but n large (n 30) As Case 1, but replace the known σ by the sample standard deviation s Thus the test statistic is X µ 0 S/ n but the rejection region does not change These tests are all called one-sample z-tests Relationship between hypothesis tests and confidence intervals If we do a two-sided z-test with H 0 : µ = µ 0, test statistic Z = ( X µ 0 ) n σ and significance level α, z Rejection Region z z α/2 or z z α/2 ( x µ 0) n z σ α/2 or ( x µ 0 ) n σ z α/2 7

8 x µ 0 z α/2 σ n or x µ 0 + z α/2 σ n σ σ µ 0 x + z α/2 n or µ 0 x z α/2 n ( ) σ σ µ 0 / x z α/2 n, x + z α/2 n µ 0 / (1 α)-confidence interval for µ So a 100(1 α)% confidence interval is all the values µ 0 which would not be rejected using a two-sided z-test with significance level α Tests of a Poisson mean Suppose X 1, X 2, X n are independent random variables with the Poisson distribution with mean λ If nλ > 9 then we can use the normal approximation, which is justified by the Central Limit Theorem So the sample mean X will be approximately N(λ, λ/n) So we can test a hypothesis about λ, H 0 : λ = λ 0 by using the test statistics Z = ( X λ 0 ) λ 0 n Example X 1,,X n is a random sample from a Poisson distribution with mean λ Suppose n = 10 and x = 1 test the null hypothesis H 0 : λ = 3 against a two-sided alternative using a significance level of 5% We use the test statistic Z = ( X 3) 3 10 and Z N(0,1) if H 0 is true For a two-sided test at the 5% significance level the rejection region is {z : Z > 196} The observed value of the test statistic is z = (1 3) 3 10 = 219 which is in the rejection region and so we reject H 0 at the 5% significance level For the Poisson distribution, if we know the mean then we know the variance So there is no longer a simple relationship between two-sided hypothesis test and confidence intervals

CHAPTER EIGHT TESTS OF HYPOTHESES

CHAPTER EIGHT TESTS OF HYPOTHESES 11/18/213 CAPTER EIGT TESTS OF YPOTESES (8.1) Definition: A statistical hypothesis is a statement concerning one population or more. 1 11/18/213 8.1.1 The Null and The Alternative ypotheses: The structure