Hypothesis Testing PDF

CHAPTER 8: HYPOTHESIS TESTING  Simply put, we want to test the probability that a given hypothesis/theory/claim is true or not. The general procedure is to choose a specific hypothesis to be tested -> pick an appropriate sample -> use measurements/calculations from the sample -> then determi...

CHAPTER 8: HYPOTHESIS TESTING  Simply put, we want to test the probability that a given hypothesis/theory/claim is true or not. The general procedure is to choose a specific hypothesis to be tested -> pick an appropriate sample -> use measurements/calculations from the sample -> then determine the likelihood of that hypothesis.  The process of hypothesis testing consists of four steps: 1) Define H0 and H1  In an hypothesis test, there are always two hypotheses called a null hypothesis (H0) and an alternative hypothesis (H1).  We begin the test assuming the null hypothesis (H 0) is true, which is always equal to a specific value. This is the theory/claim to be tested.  The main goal of this testing procedure is to determine whether there is enough evidence to infer/conclude that the alternative hypothesis (H1) is true.  So we are testing the alternative hypothesis against the null hypothesis. The alternative hypothesis can then be determined using three different inequalities (>; H1: < H1: ≠ - More than or bigger than… - Smaller than or less than… - Different or not equal to… - One tail test to the right =the rejection region is - One tail test to the left =the rejection region is in - Two tail test =the rejection region is in both the in the right tail. the left tail. left and the right tails. - α is NOT divided by 2 since we only concentrate - α is NOT divided by 2 since we only concentrate - α IS divided by 2 because both tails are being on the one tail on the right. on the one tail on the left. used.  Type I error = rejecting the null hypothesis (H0) when it is in fact true and it should not have been rejected. 2  Type II error = not rejecting the null hypothesis (H0)when it is in fact false and it should have been rejected. 3) Calculate the test statistic  Depending on the size of the sample (randomly selected from the population), either a Z test statistic (n 30) or T test statistic (n 30) must be calculated.  These test statistics are different for each scenario under consideration: one population mean; the difference between two population means; one population proportion; and the difference between two population proportions. 4) Make a conclusion  The decision to reject H0 or not reject H0 is based on where the test statistic (in step 3) is calculated to be in comparison with the critical value (in step 2).  There are two possible decisions: 1. Reject the null hypothesis H0 and accept the alternative hypothesis H1. If the test statistic falls in the rejection region then we reject the null hypothesis. As a result, we conclude that there is enough statistical evidence to infer that the alternative hypothesis is true. 2. Do not reject the null hypothesis H0. If the test statistic does not fall in the rejection region (and falls in the acceptance region) then we do not reject the null hypothesis. We conclude that there is not enough statistical evidence to infer that the alternative hypothesis is true. 3  So our conclusion is always in reference to the null hypothesis even though we are testing the alternative hypothesis. 4 E.g. 8.1; Pg. 5: (Population standard deviation is known) The manager of a department store is thinking about establishing a new payment system for the store’s credit customers. After a thorough financial analysis, she determines that the new system will only be cost effective if the mean monthly installment is more than R170. A random sample of 400 monthly accounts is drawn, for which the sampling mean is R178. The manager knows that the accounts are approximately normally distributed with a population standard deviation of R65. Can the manager conclude from this that the new system will be cost effective? Solution: Step 1: Define H0 and H1  Null hypothesis is always an equality H0: = 170  Alternative hypothesis is either >, 170 (mean monthly installment is more than R170) Step 2: Define CV and RR  Based on the H1 we selected in step 1 (>) this means that we have a one tail test to the right (positive side) and therefore stays as is. By default, the level of significance is 5% (when not stated in the question) therefore = 0.05.  Our random sample n = 400 which is larger than 30, so we definitely use Z-distribution  CV: Use Z-tables on pg. A6 to find the z-score: = = 1.645 (separates the acceptance region from the rejection region)  RR: Therefore, we reject H0 if our Z statistic (step 3) is greater than 1.645. If it is smaller than 1.645, then we do not reject H0. Step 3: Calculate the test statistic  Use the Z-test statistic when is known (very rare): Z = = = = 2.4615 Step 4: Make a conclusion  The Z-test statistic = 2.4615 is more than the critical value of 1.645, thus falls in the rejection region. Therefore, H 0 is 5 rejected in favour of the alternative hypothesis H1. Thus, we have enough statistical evidence to infer that the mean monthly instalments are more than R170. E.g. 8.2; Pg. 7: (Population standard deviation unknown and n 30) The manufacturers of ‘Road King’ tyres claim that their tyres last an average of 70 000km or more. The Consumer Board suspects that this claim may be false (they have received some complaints) and decides to test the manufacturer’s claim. The Consumer Board took a random sample of 500 tyres from the population. They tested the tyres until they failed to meet requirements for tread depth. The sampling mean is = 69914 and the standard deviation is s = 2131. Solution: Step 1: Define H0 and H1  Null hypothesis is always an equality H0: = 70000  Alternative hypothesis is either >, ) this means that we have a one tail test (positive side) and therefore must stay as is. Thus = 1% = 0.01.  Both samples are larger than 30, so we must use the Z-distribution  CV: Use Z-tables on pg. A6 to find the z-score = = 2.325 (separates the acceptance region from the rejection region)  RR: Therefore, we reject H0 if our Z statistic (step 3) is bigger than 2.325. If it is smaller than 2.325, we do not reject H 0. Step 3: Calculate the test statistic  Use the Z-test statistic: Z = = = = = 3.4816 Step 4: Make a conclusion 9  The Z-test statistic = 3.4816 which is greater that the critical value of 2.325. Therefore Z-test statistic falls in the rejection region. Thus the null hypothesis H0 is rejected in favour of the alternative H 1. There is sufficient evidence to infer that the average salary of privately employed pharmacists does exceed the average salary of pharmacists employed by the state by more than R6 000 p.a. nssX2 E.g. 8.5; Pg. 13: (Population standard deviations and unknown with < 30 or < 30 or both < 30) University 1 University 2 Two randomly selected samples of students from different universities were Sampling mean ( 18.8 10.3333 Sample size n 5 6 given a statistics test out of 20. Determine whether there is a significant Sample variance 1.7 2.2667 difference between the two universities. Sample standard deviation s 1.3038 1.5055 Solution: Step 1: Define H0 and H1  Null hypothesis is always an equality H0: - = 0 OR H0: = (significant difference between the two)  Alternative hypothesis is either >, ) we have a one tail test to the right (positive side) and so 1 13 stays as is = 0.05. When working with proportions, we always use the Z-distribution  CV: Use Z-tables on pg. A6 to find the z-score: = = 1.645 (separates the acceptance region from the rejection region)  RR: Therefore, we reject H0 if our Z statistic (step 3) is bigger than 1.645. If it is smaller than 1.645, then we do not reject H 0. Step 3: Calculate the Z-test statistic  sample proportions are: = = 0.31 = = 0.24 = = = = 0.275 Z= = = = 1.11 Step 4: Conclusion  The Z-test statistic (equal to 1.11) is less than the critical value of 1.645. Hence the calculated Z-test statistic falls in the acceptance region. We therefore do not reject the null hypothesis. There is no significant difference between the proportion of males and females that indicated a particular preference. 14