1. Economics 173 Business Statistics Lecture 4 Fall, 2001 Professor J. Petry http://www.cba.uiuc.edu/jpetry/Econ_173_fa01/

2. Introduction to Hypothesis Testing Chapter 10

3. 3 10.1 Introduction The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief about a parameter. Examples –Is there statistical evidence in a random sample of potential customers, that support the hypothesis that more than p% of the potential customers will purchase a new product? –Is a new drug effective in curing a certain disease? A sample of patients is randomly selected. Half of them are given the drug and half are given a placebo. The improvement in the patients conditions is then measured and compared.

4. 4 10.2 Concepts of hypothesis testing –There are two hypotheses (about a population parameter(s)) H 0 - the null hypothesis [ for example  = 5] H 1 - the alternative hypothesis [  > 5] –The null always contains =, and may contain ≥, ≤ –The alternative hypothesis is most important, it is what you are trying to prove. The alternative can involve >, < or ≠ The alternative establishes whether the test is one-tailed or two-tailed. The alternative establishes the location of the rejection region

5. 5 –We always assume the null hypothesis is true. Calculate a statistic related to the parameter hypothesized. Pose the question: How probable is it to obtain a statistic value at least as extreme as the one observed from the sample? If the sample statistic is in an extreme location of the sampling distribution, you are going to be likely to reject the null. On the other hand, if it is reasonably close to the center of the sampling distribution you will not reject the null.  = 5

6. 6 –Make one of the following two conclusions (based on the test): Reject the null hypothesis in favor of the alternative hypothesis. »There is enough evidence to infer that the alternative is true Do not reject the null hypothesis in favor of the alternative hypothesis. »There is not enough evidence to infer that the alternative is true –Two types of errors are possible when making the decision whether to reject H 0 Type I error - reject H 0 when it is true. Type II error - do not reject H 0 when it is false. –Resulting in one of the following four situations...

7. 7 –Terminology Null hypothesis, Alternative hypothesis Test statistic, standardized test statistic, p-value Critical value, standardized critical value, significance level –Analogy: Hypothesis testing is similar to a jury trial Assume innocent until proven guilty –Assume H 0 is true until proven otherwise Level of proof required to establish guilty verdict? What if you convict an innocent person? –Identical to establishing significance level of test. Type I error is equivalent to convicting an innocent person. –“Beyond a reasonable doubt” is court of law norm –Standard in statistics varies depending upon the issue at stake: »Overwhelming evidence = 1% significance level »Strong evidence = 1.001-5% significance level »Weak evidence = 5.001-10% significance level »No statistical evidence = 10.001% or higher significance level

8. 8 10.3 Testing the Population Mean When the Population Standard Deviation is Known Example 10.1 –A new billing system for a department store will be cost- effective only if the mean monthly account is more than $170. –A sample of 400 monthly accounts has a mean of $178. –If the account are approximately normally distributed with  = $65, can we conclude that the new system will be cost effective?

9. 9 Solution –The population of interest is the credit accounts at the store. –We want to show that the mean account for all customers is greater than $170. H 1 :  > 170 –The null hypothesis must specify a single value of the parameter  H 0 :  = 170

10. 10 Is a sample mean of 178 sufficiently greater than 170 to infer that the population mean is greater than 170? 178 If  is really equal to 170, then. The distribution of the sample mean should look like this. Is it likely to have under the null hypothesis (  = 170)?

11. 11 –Instead of using the statistic, we can use the standardized value z. –Then, the rejection region becomes One tail test The standardized test statistic

12. 12 Example 10.1 - continued H 0 :  = 170 H 1 :  > 170 –Test statistic: –Rejection region: z > z.05  1.645. –Conclusion: Since 2.46 > 1.645, reject the null hypothesis in favor of the alternative hypothesis. –2.46 is test statistic value, p-value is probability associated w/ the test statistic value (see below). –1.645 is the critical value. Significance level of the test is the probability associated with this value.

13. 13 –The p - value provides information about the amount of statistical evidence that supports the alternative hypothesis. – The p-value of a test is the probability of observing a test statistic at least as extreme as the one computed, given that the null hypothesis is true. – Illustrating with example 10.1... P-value method

14. 14 The probability of observing a test statistic at least as extreme as 178, given that the null hypothesis is true is: The p-value

15. 15 Interpreting the p-value –Because the probability that the sample mean will assume a value of more than 178 when  = 170 is so small (.0069), there are reasons to believe that  > 170. …it becomes more probable under H 1, when Note how the event is rare under H 0 when but... We can conclude that the smaller the p-value the more statistical evidence exists to support the alternative hypothesis. We can conclude that the smaller the p-value the more statistical evidence exists to support the alternative hypothesis.

16. 16 Describing the p-value –If the p-value is less than 1%, there is overwhelming evidence that support the alternative hypothesis. –If the p-value is between 1% and 5%, there is a strong evidence that supports the alternative hypothesis. –If the p-value is between 5% and 10% there is a weak evidence that supports the alternative hypothesis. –If the p-value exceeds 10%, there is no evidence that supports of the alternative hypothesis.

17. 17 The p-value and rejection region methods –The p-value can be used when making decisions based on rejection region methods as follows: Define the hypotheses to test, and the required significance level  Perform the sampling procedure, calculate the test statistic and the p-value associated with it. Compare the p-value to  Reject the null hypothesis only if p <  ; otherwise, do not reject the null hypothesis.  = 0.05 P-value = 0.0069

18. 18 Example 10.2 –A government inspector samples 25 bottles of catsup labeled “Net weight: 16 ounces”, and records their weights. –From previous experience it is known that the weights are normally distributed with a standard deviation of 0.4 ounces. –Can the inspector conclude that the product label is unacceptable?

19. 19 Solution –We need to draw a conclusion about the mean weights of all the catsup bottles. –We investigate whether the mean weight is less than 16 ounces (bottle label is unacceptable). H 1 :  < 16 H 0 :  = 16 –The test statistic is – Select a significance level:  = 0.05 – Define the rejection region z < - z   1.645 Then One tail test

20. 20 we want this mistake to happen not more than 5% of the time. 16  0.05 A sample mean far below 16, should be a rare event if  = 16. So, if in reality  =16, but we reject this hypothesis in favor of  < 16 because was very small, Rejection region -1.25  0.05 0 -z   = -1.645

21. 21 0 -z   = -1.645  0.05 Rejection region -1.25 Since the value of the test statistic does not fall in the rejection region, we do not reject the null hypothesis in favor of the alternative hypothesis. There is insufficient evidence to infer that the mean is less than 16 ounces. The p-value = P(Z.05

22. 22 Example 10.3 –The amount of time required to complete a critical part of a production process on an assembly line is normally distributed. The mean was believed to be 130 seconds. –To test if this belief is correct, a sample of 100 randomly selected assemblies was drawn, and the processing time recorded. The sample mean was 126.8 seconds. –If the process time is really normal with a standard deviation of 15 seconds, can we conclude that the belief regarding the mean is incorrect ?

23. 23 Solution –Is the mean different than 130? H 0 :  = 130 Then – Define the rejection region z z  /2

24. 24 130 0 A sample mean far below 130 or far above 130, should be a rare event if  = 130. we want this mistake to happen not more than 5% of the time. So, if in reality  =130, but we mistakenly reject this hypothesis in favor of because was very small or very large,  2  0.025 -z   = -1.96 z   = 1.96 Rejection region

25. 25 0  2  0.025 -z   = -1.96 z   = 1.96 Since the value of the test statistic falls in the rejection region, we reject the null hypothesis in favor of the alternative hypothesis. There is sufficient evidence to infer that the mean is not 130. -2.13 The p-value = P(Z 2.13) = 2(.0166) =.0332 <.05 2.13

26. 26 –Interval estimators can be used to test hypotheses. –Calculate the 1 -  confidence level interval estimator, then if the hypothesized parameter value falls within the interval, do not reject the null hypothesis, while if the hypothesized parameter value falls outside the interval, conclude that the null hypothesis can be rejected (  is not equal to the hypothesized value). Testing hypotheses and intervals estimators

27. 27 Drawbacks –Two-tail interval estimators may not provide the right answer to the question posed in one-tail hypothesis tests. –The interval estimator does not yield a p-value. There are cases where only tests produce the information needed to make decisions.

28. 28 Calculating the Probability of a Type II Error To properly interpret the results of a test of hypothesis, we need to –specify an appropriate significance level or judge the p-value of a test; –understand the relationship between Type I and Type II errors. –How do we compute a type II error?

29. 29 Calculation of a type II error requires that –the rejection region be expressed directly, in terms of the parameter hypothesized (not standardized). –the alternative value (under H 1 ) be specified. H 0 :  0 H 1 :  1 (  0 is not equal to  1 )  0  1 

30. 30 Revisiting example 10.1 –The rejection region was with  =.05. –A type II error occurs when a false H 0 is not rejected.    170    180 .05 175.34 …but H 0 is false Do not reject H 0

31. 31 Effects on  of changing  –Decreasing the significance level  increases the the value of  and vice versa       

32. 32 Judging the test –A hypothesis test is effectively defined by the significance level  and by the the sample size n. –If the probability of a type II error  is judged to be too large, we can reduce it by increasing , and/or increasing the sample size.

33. 33     By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, decreases. As a result  decreases –In example 10.1, suppose n increases from 400 to 1000.

34. 34 In summary, –By increasing the sample size, we reduce the probability of type II error. –Hence, we shall accept the null hypothesis when it is false less frequently. Power of a test –The power of a test is defined as 1 -  –It represents the probability to reject the null hypothesis when it is false.