Hypothesis for Means and Proportions

Transcription

1 November 14, 2012

2 Hypothesis Tests - Basic Ideas Often we are interested not in estimating an unknown parameter but in testing some claim or hypothesis concerning a population. For example we may wish to test a manufacturer s claim concerning the mean lifetime of a product, or test a claim that one drug is more effective than another in curing a disease. As with estimation, we will use sample data to test claims or hypotheses concerning population parameters. The general procedure for doing this is known as hypothesis testing.

3 Hypothesis Tests - Basic Ideas To illustrate the general idea let s consider an example. A manufacturer of AA batteries claims that the mean lifetime of their batteries is 800 hours. We wish to test this claim and decide whether to accept or reject it. In general, we set up a hypothesis test as follows: Formulate a null hypothesis H 0. In this case H 0 : µ = 800. Formulate an alternative hypothesis H 1. If we are only interested in whether the mean is equal to 800 or not equal to 800, we would choose H 1 : µ 800. This is a two-sided alternative. If we are testing a claim that the mean lifetime is at least 800 hours, then a one-sided alternative H 1 : µ < 800 is appropriate.

4 Hypothesis Tests - Basic Ideas Determine a decision procedure or criterion for rejecting the null hypothesis. For instance, we may choose a sample of 50 batteries and calculate the mean liftime, x of the batteries in the sample. We may decide that the null hypothesis is to be rejected if x < 790 or x > 810 (two-sided) or to reject H 0 if x < 790 (one-sided).

5 Hypothesis Tests - Basic Ideas We usually construct a decision criterion in the following way. First we choose a random sample and calculate a test statistic (such as X µ σ/ n ). We then determine critical values for this statistic from its sampling distribution. If the value of the test statistic lies outside a region determined by the critical values, then we reject the null hypothesis.

6 Type I and Type II Errors It is possible to make two different types of error in hypothesis testing. 1 We can reject a true null hypothesis - this is known as a Type I Error. 2 Alternatively we can fail to reject a false null hypothesis - this is known as a Type II Error. Decision H 0 True H 0 False Fail to reject H 0 No Error Type II Error Reject H 0 Type I Error No Error

7 Type I and II errors Example Recall our example of testing the claim about mean lifetimes of AA batteries. H 0 : µ = 800 H 1 : µ 800 Suppose we apply the decision procedure to reject H 0 if the mean of a random sample of size n = 40 lies in the critical region x > 810 or x < 790. If the standard deviation of the population is known to be σ = 22hours, what is the probability of a type I error?

8 Type I and II errors If the null hypothesis is true, then µ = 800 and the probability of making a type I error is given by P( X < 790) + P( X > 810) = P(Z < 22/ ) + P(Z > 40 22/ 40 ) = P(Z < 2.87) + P(Z > 2.87) =

9 Type I and II errors To calculate the probability of a type II error, we require a more specific alternative hypothesis. For instance, we can calculate the probability of a type II error in this example if the true mean is actually 808 hours. In this case, the probability of a type II error is P(790 X 810) = P( 22/ Z 40 22/ 40 ) = P( 5.17 Z 0.57) =

10 Type I and II errors The probability of making a type I error is usually referred to as the significance level of a test. Typically, we decide on the significance level of a test before determining the critical region. In fact, the general procedure for hypothesis testing is as follows: formulate the null and alternative hypotheses; specify a significance level α; choose a test statistic; use the sampling distribution of the test statistic to determine the critical region; calculate a value for the test statistic based on sample data and determine whether or not this lies in the critical region. If it does, then we reject H 0 ; if it does not, then we fail to reject H 0.

11 Large Sample Hypothesis Tests on the Mean We now consider the problem of testing hypotheses concerning the mean of a population. To begin with we consider large-sample tests. If the sample size is large and σ is known then to test the hypothesis H 0 : µ = µ 0, we use as a test statistic. Z 0 = X µ 0 σ/ n Z 0 follows a standard normal distribution.

12 Large Sample Tests on the Mean If we choose a significance level α, then our criterion for rejecting H 0 is: Z 0 > z α/2 or Z 0 < z α/2 if the test is two-sided - that is if the alternative hypothesis is H 1 : µ µ 0 ; Z 0 < z α if the test is one-sided and the alternative hypothesis H 1 : µ < µ 0 ; Z 0 > z α if the test is one-sided and the alternative hypothesis H 1 : µ > µ 0.

13 Large Sample Hypothesis Tests on the Mean Example A manufacturer of steel cables claims that the mean diameter of its cables is 2.2cm. To test this claim, a random sample of 35 cables from this manufacturer is selected and the mean diameter of the cables in the sample is 2.05cm. The standard deviation of the diameters of cables from this manufacturer is known to be 0.3cm. Perform a hypothesis test on the manufacturer s claim at both 5% and 1% levels of significance and state clearly your conclusions.

14 Large Sample Hypothesis Tests on the Mean 1 Null Hypothesis is H 0 : µ = Alternative Hypothesis is H 1 : µ Significance level is α = Test Statistic: X / Reject H 0 if Z 0 > 1.96 or Z 0 < The value of Z 0 is / 35 = As the value of the test statistic lies in the rejection region, we reject the manufacturer s claim at at 5% level of significance. The procedure for 1% level of significance is the same only the rejection region would be Z 0 < 2.58 or Z 0 > In the current example, we would also reject H 0 at the 1% level of significance.

15 Large Sample Hypothesis Tests on the Mean Example For safety reasons, it is important the the mean concentration of a chemical used in a make of cough syrup does not exceed 8mg/l. A random sample of 34 bottles of the syrup is selected and the mean concentration for the bottles in the sample is 8.25mg/l. From past experience it is known that the standard deviation of the concentrations of this chemical in bottles of the syrup is 0.9mg/l. Perform a hypothesis test at a significance level of 5% on whether or not this cough syrup conforms to safety requirements, stating your conclusions clearly.

16 Large Sample Hypothesis Tests on the Mean 1 Null Hypothesis is H 0 : µ = 8. 2 Alternative Hypothesis is H 1 : µ > 8. 3 Significance level is α = Test Statistic: 5 Reject H 0 if Z 0 > X 8 0.9/ The value of Z 0 is / 34 = As the value of the test statistic does not lie in the rejection region, we cannot reject the null hypothesis that the mean concentration does not exceed 8mg/l at a 5% level of significance.

17 One-tailed Tests - Comments It is always more difficult to reject a null hypothesis. Our choice of alternative hypothesis in one-tailed tests can influence our recommendations. In the above example, our choice of alternative hypothesis makes it difficult to conclude that the mean concentration exceeds 8mg/l. The previous example could also be approached in the following way.

18 1-tailed Tests - Comments We could test the null hypothesis H 0 : µ = 8 against the alternative hypothesis H 1 : µ < 8. In this case, the syrup is deemed unsafe unless we can reject the null hypothesis. For the above data we would be unable to reject this null hypothesis so we would deem the syrup unsafe if we took this approach. How we set up a one-tailed test influences our recommendations and is determined by where we wish the burden of proof to lie.

19 Small Sample Hypothesis Testing on the Mean - Two-tailed tests Consider the following situation. 1 We have a small sample (n < 30) with σ unknown. 2 The population being studied is normal. 3 We wish to test the hypothesis H 0 : µ = µ 0 against the alternative hypothesis H 1 : µ µ 0 at a significance level of α.

20 Small sample tests on the mean In this case, we use the test statistic t = X µ 0 S/ n which has a t-distribution with n 1 degrees of freedom. The critical values for this test are t n 1,α/2 and t n 1,α/2. 1 We calculate the sample mean x and the sample standard deviation s from our sample data and then compute the value of the test statistic: t 0 = x µ 0 s/ n. 2 If the value of t 0 > t n 1,α/2 or t 0 < t n 1,α/2 then we reject the null hypothesis. Otherwise we fail to reject the null hypothesis.

21 Small sample tests on the mean Example A manufacturer of ball bearings claims that the average mass of of their bearings is 35g. To test this claim, a random sample of 14 bearings is selected. The mean and standard deviation of the masses of the bearings in the sample were 35.8g and 1.7g respectively. Perform a hypothesis test on the manufacturer s claim at a 5% level of significance and explain your results.

22 Small sample tests on the mean 1 Null Hypothesis is H 0 : µ = Alternative Hypothesis is H 1 : µ Significance level is α = Test Statistic: t 0 = X / Reject H 0 if t 0 > t 13,0.025 = 2.16 or t 0 < t 13,0.025 = The value of t 0 is / 14 = As the value of the test statistic does not lie in the rejection region, we cannot reject the manufacturer s claim at a 5% level of significance.

23 Small Sample Hypothesis Testing on the Mean - One-tailed tests Consider the following situation. 1 We have a small sample (n < 30) with σ unknown. 2 The population being studied is normal. 3 We wish to test the hypothesis H 0 : µ = µ 0 against the alternative hypothesis H 1 : µ < µ 0 at a significance level of α. As above, we use the test statistic t = X µ 0 S/ n which has a t-distribution with n 1 degrees of freedom. The critical value for this test is t n 1,α.

24 Small sample 1-tailed tests on the mean 1 We calculate the sample mean x and the sample standard deviation s from our sample data and then compute the value of the test statistic: t 0 = x µ 0 s/ n. 2 If the value of t 0 < t n 1,α then we reject the null hypothesis. Otherwise we fail to reject the null hypothesis. If we wish to test the null hypothesis H 0 : µ = µ 0 against the alternative hypothesis H 1 : µ > µ 0 at a significance level of α, the procedure is largely the same. In this case we would reject the null hypothesis if t 0 > t n 1,α.

25 Small sample 1-tailed tests on the mean Example A producer of car windscreens wishes to test whether the mean thickness of their windscreens is at least 12mm. To do so, a random sample of 8 windscreens is selected. The thicknesses of the screens (in mm) in the sample are listed below: 11.6, 11.5, 12.3, 12.4, 12.2, 11.4, 12.3, Perform a hypothesis test on these data at a 5% level of significance and explain your conclusions.

26 Small sample 1-tailed tests on the mean 1 Null Hypothesis is H 0 : µ = Alternative Hypothesis is H 1 : µ < Significance level is α = From the sample data: x = , s = Test Statistic: t 0 = X / 8. 6 Reject H 0 if t 0 < t 7,0.05 = The value of t 0 is / 8 = As the value of the test statistic does not lie in the rejection region, we cannot reject the producer s claim at a 5% level of significance.

27 Hypothesis Tests on Proportions In many practical situations, we are interested in testing a claim about a population proportion. For instance, a manufacturer s claim concerning the proportion of their products that fail to meet some customer specification, or the claim of a drug company about the proportion of people for whom a treatment is effective. As usual we let p denote the proportion of successes in the population. As with hypothesis tests on the mean, tests on proportions can be one-sided or two-sided.

28 Two-tailed tests on proportions 1 We wish to test the hypothesis H 0 : p = p 0 against the alternative hypothesis H 1 : p p 0. 2 If ˆP denotes the sample proportion the appropriate test statistic is ˆP p 0 Z 0 = p0 (1 p 0 )/n which approximately follows a standard normal distribution. To test the hypothesis H 0 : p = p 0 against the alternative H 1 : p p 0, we proceed as follows.

29 Two-tailed tests on proportions 1 Decide on a significance level α. 2 Take a sample of size n and compute the sample proportion ˆp for the sample. 3 Compute the value of the test statistic z 0 = ˆp p 0 p0 (1 p 0 )/n 4 The critical values for the two-tailed test are z α/2 and z α/2. 5 If z 0 < z α/2 or z 0 > z α/2, we reject H 0. Otherwise we fail to reject H 0.

30 Two-tailed tests on proportions Example A government agency claims that the proportion of the population who have access to wireless broadband at home is 70%. To test this claim, a random sample of 500 people is selected and 370 of these had access to broadband at home. Test the agency s claim at a 5% level of significance and explain your conclusions clearly.

31 Two-tailed tests on proportions 1 Null Hypothesis is H 0 : p = Alternative Hypothesis is H 1 : p Significance level is α = From the sample data: ˆp = = Test Statistic: ˆp 0.7 Z 0 =. (0.7)(0.3)/500 6 Reject H 0 if Z 0 < z = 1.96 or Z 0 > z = The value of z 0 for our sample is = (0.7)(0.3)/500 8 As the value of the test statistic does not lie in the rejection region, we cannot reject the producer s claim at a 5% level of significance.

32 One-tailed tests on proportions 1 We wish to test the hypothesis H 0 : p = p 0 against the alternative hypothesis H 1 : p < p 0. 2 Decide on a significance level α. 3 Take a sample of size n and compute the sample proportion ˆp for the sample. 4 Compute the value of the test statistic z 0 = ˆp p 0 p0 (1 p 0 )/n 5 The critical value for the one-tailed test is z α. 6 If z 0 < z α, we reject H 0. Otherwise we fail to reject H 0.

33 1-tailed tests on proportions If we wish to test the null hypothesis H 0 : p = p 0 against the alternative hypothesis H 1 : p > p 0 at a significance level of α, the procedure is largely the same. In this case we would reject the null hypothesis if z 0 > z α. Example A manufacturer of automatic gearboxes claims that at most 5% of their products fail to meet the specifications required by a major customer. The customer tests this claim by choosing a random sample of 400 gearboxes from the manufacturer. 24 gearboxes in the sample failed to meet specifications. Perform a hypothesis test on the manufacturer s claim at a 5% level of significance and state your conclusions clearly.

34 1-tailed tests on proportions 1 Null Hypothesis is H 0 : p = Alternative Hypothesis is H 1 : p > Significance level is α = From the sample data: ˆp = = Test Statistic: Z 0 = ˆp 0.05 (0.05)(0.95)/ Reject H 0 if Z 0 > z 0.05 = The value of z 0 for our sample is = (0.05)(0.95)/400 8 As the value of the test statistic does not lie in the rejection region, we cannot reject the producer s claim at a 5% level of significance.

35 1-tailed tests on proportions As with one-tailed tests on the mean, the customer could demand that the evidence is more strongly on the side of the proportion of non-conforming gearboxes being less than In this case, we would choose the alternative hypothesis H 1 : p < 0.05 and only accept the manufacturer s claim if we can reject the null hypothesis. Clearly, in the above example we would not be able to reject the null hypothesis in this way so we would not accept the manufacturer s claim.