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1 Tests of Hypotheses: Chapter Large Samples Rejection region Acceptance region GOALS 2 TO DEFINE HYPOTHESES AND HYPOTHESIS TESTING. TO DESCRIBE THE HYPOTHESIS TESTING PROCEDURE. TO DISTINGUISH BETWEEN ONE-TAILED AND TWO-TAILED TEST OF HYPOTHESIS. TO CONDUCT A TEST FOR THE POPULATION MEAN OR PROPORTION. GOALS 3 TO CONDUCT A TEST FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS OR PROPORTIONS. TO DESCRIBE STATISTICAL ERRORS ASSOCIATED WITH HYPOTHESIS TESTING. WHAT IS A HYPOTHESIS? 4 Hypothesis: A statement about the value of a population parameter developed for the purpose of testing. Examples of hypotheses, or statements, made about a population parameter are: » The mean monthly income from all sources for systems analysts is $3,625. » Twenty percent of all juvenile offenders ultimately are caught and sentenced to prison. WHAT IS HYPOTHESIS TESTING? 5 Hypothesis testing: A procedure, based on sample evidence and probability theory, used to determine whether the hypothesis is a reasonable statement and should not be rejected, or is unreasonable and should be rejected. Following is a five-step procedure for testing a hypothesis. STEP 1 State null H0 and alternative 6 hypotheses H1 STEP 2 Select a level of significance STEP 3 Identify the test statistic STEP 4 Formulate a decision rule STEP 5 Take a sample , arrive at a decision Reject H0 and Do not reject H0 accept H1 Null Hypothesis H0: A statement about the value of a population parameter. 7 Alternative Hypothesis H1: A statement that is accepted if the sample data provide evidence that the null hypothesis is false. Level of Significance: The probability of rejecting the null hypothesis when it is actually true. Type I Error: Rejecting the null hypothesis, H0, when it is actually true. Type II Error: Accepting the null hypothesis, H0, when it is actually false. 8 Researcher Null hypothesis Accepts H0 Rejects H0 Correct Type I If H0 is true and Decision error Type II Correct If H0 is false and error Decision Test statistic: A value, determined from sample information, used to determine 9 whether or not to reject the null hypothesis. Critical value: The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected. ONE-TAILED TESTS OF SIGNIFICANCE 10 A test is one-tailed when the alternate hypothesis, H1, states a direction such as: H0: The mean income of the females is less than or equal to the mean income of the males. H1: The mean income of the females is greater than the mean income of the males. The region of rejection in this case is to the right (upper) tail of the curve. An example is shown next: Sampling Distribution for the Statistic Z for a One-Tailed Test, 0.05 Level of Significance 11 Critical value 1.645 = Z Region of rejection 0.95 Probability 0.05 z TWO-TAILED TESTS OF SIGNIFICANCE 12 A test is two-tailed when no direction is specified in the alternate hypothesis H1, such as: H0: The mean income of the females is equal to the mean income of the males. H1: The mean income of the females is not equal to the mean income of the males. The region of rejection in this case is divided equally into the two tails of the curve. An example is shown next: Sampling Distribution for the Statistic Z for a Two-Tailed Test, 0.05 Level of Significance 13 Critical Critical value value Do not -1.96 = Z 1.96 = Z reject H0 Region of Region of rejection rejection 0.95 0.025 0.025 z TESTING FOR THE POPULATION MEAN: LARGE SAMPLE, POPULATION 14 STANDARD DEVIATION KNOWN The test statistic is given by: z X / n EXAMPLE 15 The processors of Mets Catsup indicate on the label that the bottle contains 16 ounces of catsup. Mets’ Quality Control Department is responsible for monitoring the amount included in the bottle. A sample of 36 bottles is selected hourly and the contents weighed. Last hour a sample of 36 bottles had a mean weight of 16.12 ounces with a standard deviation of 0.5 ounces. At the 0.05 significance level can we conclude that the process is out of control? (i.e. not meeting weight goals) EXAMPLE (continued) 16 Step 1: State the null and the alternative hypotheses. H0: = 16 H1: 16 Step 2: State the decision rule. H0 is rejected if z < -1.96 or z > 1.96.(2-side) Step 3: Compute the value of the test statistic. z= [16.12 - 16]/[0.05/36] = 1.44. Step 4: What is the decision on H0? H0 is not rejected, because 1.44 is less than the critical value of 1.96. Sampling Distribution for the Statistic Z for a Two-Tailed Test, 0.05 Level of Significance 17 Critical Critical value value Do not -1.96 = Z 1.96 = Z reject H0 Region of Region of Test Stat. rejection rejection 1.44 0.95 0.025 0.025 z TESTING FOR THE POPULATION MEAN: LARGE SAMPLE, POPULATION 18 STANDARD DEVIATION UNKNOWN Here is unknown, so we estimate it with the sample standard deviation s. As long as the sample size n 30, z can be approximated with z X s/ n EXAMPLE 19 The Thompson’s Discount Store chain issues its own credit card. The credit manager wants to find out if the mean monthly unpaid balance is more than $400. The level of significance is set at 0.05. A random check of 172 unpaid balances revealed the sample mean to be $407 and the sample standard deviation to be $38. Should the credit manager conclude that the population mean is greater than $400, or is it reasonable to assume that the difference of $7 ($407 - $400) is due to chance? EXAMPLE (continued) 20 Step 1: State the null and the alternative hypotheses. H0: 400 H1: > 400 Step 2: State the decision rule. H0 is rejected if z > 1.645. (one-sided) Step 3: Compute the value of the test statistic. z = [407 - 400]/[38/172] = 2.42. Step 4: What is the decision on H0? H0 is rejected. The manager can conclude that mean unpaid balance is greater than $400. 21 Computed z = 2.42 Rejection region 0 1.645 z HYPOTHESIS TESTING: TWO POPULATION MEANS 22 Assume the parameters for the two populations are 1,2,1,and 2. Case I: When 1,2are known, the test statistic is: X1 X 2 z 1 2 2 2 n1 n2 HYPOTHESIS TESTING: TWO POPULATION MEANS 23 Case II: When 1,2are unknown but the sample sizes n1 and n2 are greater than or equal to 30, the test statistic is : X1 X 2 z 2 2 s1 s2 n1 n2 EXAMPLE 24 A study was conducted to compare the mean years of service for those retiring in 1975 with those retiring last year Kentucky Manufacturing Co. The following sample data was obtained. At the 0.01 significance level can we conclude that the workers retiring last year had more service? EXAMPLE (continued) 25 State the null and the alternative hypotheses: Let population 2 refer to those that retired last year. H0:2 1 H1:2 > 1 State the decision rule. Reject H0 if z > 2.33. (one-sided) Compute the value of the test statistic. z 304 256 680 . . . 362 2.92 . 45 40 EXAMPLE (continued) 26 What is the decision on the null hypothesis? Interpret the results. Since z = 6.80 > 2.33, H0 is rejected. Those retiring last year had more years of service. TESTS CONCERNING PROPORTION 27 Proportion: A fraction or percentage that indicates the part of the population or sample having a particular trait of interest. Sample proportion is denoted by p where p Number of successes in the sample Number sampled TEST STATISTIC FOR TESTING A 28 SINGLE POPULATION PROPORTION z p p p(1 p) n p population proportion p sample proportion EXAMPLE 29 In the past 15% of the mail order solicitations for a certain charity resulted in a financial contribution. A new solicitation letter has been drafted. Will this new letter increase the solicitation rate? The new letter is sent to a sample of 200 people and 45 responded with a contribution. At the 0.05 significance level can it be concluded that the new letter is more effective? EXAMPLE (continued) 30 State the null and the alternative hypotheses: H0:p 0.15 H1:p > 0.15. State the decision rule. H0 is rejected if z > 1.645. (one-sided) Compute the value of the test statistic. 45 015 . z 200 2.97 . . (015)(085) 200 EXAMPLE (continued) 31 What is the decision on the null hypothesis? Interpret the results. Since z = 2.97 > 1.645, H0 is rejected. The new letter is more effective. A TEST INVOLVING THE DIFFERENCE 32 BETWEEN TWO POPULATION PROPORTIONS The test statistic in this case is p1 p2 z pc (1 pc ) pc (1 pc ) n1 n2 n1 is the sample size from population 1. n2 is the sample size from population 2. TWO PROPORTIONS (continued) 33 pc is the weighted mean of the two sample proportions, computed by: Total number of successes X 1 X 2 pc = Total number in samples n1 n2 X1 is the number of successes in n1. X2 is the number of successes in n2. EXAMPLE 34 Are unmarried workers more likely to be absent from work than married workers? A sample of 250 married workers showed 22 missed more than 5 days last year for any reason. A sample of 300 unmarried workers showed 35 missed more than 5 days. Use the 0.05 significance level. State the null and alternative hypotheses. H0:p2 p1 H1:p2 > p1 where subscript 2 refers to the population of unmarried workers. EXAMPLE (continued) 35 State the decision rule. H0 is rejected if z > 1.645. (one-sided) Compute the value of the test statistic. p 22 35 01036 . 250 300 c Z 01167 00880 . . 110 . 01036(1 01036) 01036(1 01036) . . . . 300 250 EXAMPLE (continued) 36 What is the decision on the null hypothesis? H0 is not rejected. There is no difference in the proportion of married and unmarried workers missing more than 5 days of work. 37 For Final Exam Chapters 12 and 13: Regression, Inference, and Model Building/ ANOVA James S. Hawkes

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Prof Rushen's Notes for MBA/ BBA students

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