Chapter Study Guide. Chapter 11 Confidence Intervals and Hypothesis Testing for Means


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1 OPRE504 Chapter Study Guide Chapter 11 Confidence Intervals and Hypothesis Testing for Means I. Calculate Probability for A Sample Mean When Population σ Is Known 1. First of all, we need to find out the standard deviation of a sampling distribution. When the population s standard deviation, σ, is known to us, use SD ( ), where n is the sample size. 2. Calculate zstatistic for a value of sample mean z = where is the sample mean and is the population mean 3. Find out probability using Ztable (A32/33 in Appendix C) Q11.1 The weight of potato chips in a mediumsize bag is stated to be 10 ounces (label). The amount that the packaging machine puts in these bags is believed to have a Normal distribution N (10.2, 0.12). Answer the following questions [Question 7 for Chapter 11, p.346]: a) What fraction of all bags sold are underweight? P (w<10) = P ( z< = P (z< ) P (z< 1.67) = (by checking Ztable) More precisely, P (z< ) = P (z<1.6667) = (using excel NORMDIST function). b) Some of the chips are sold in bargain packs of 3 bags. What s the probability that none of the 3 is underweight? The probability of none of 3 bags is underweight = p (bag 1 NOT underweight) x p (bag 2 NOT underweight) x p (bag 3 NOT underweight) = [1 p (bag 1 underweight)] x [1 p (bag 2 underweight)] x [1 p (bag 3 underweight)] = ( ) x ( )x ( ) = c) What s the probability that the mean weight of the 3 bags is below the stated weight? SD ( ) = z = p (Z<2.899) = = Chaodong Han OPRE504 Page 1 of 6
2 d) What s the probability that the mean weight per bag of a 24bag case of potato chips is below 10 ounces? SD ( ) = z = = The probability to see zscore of is almost impossible. P =0.000 Q11.2 Refer to the example in the Textbook (Sharpe 2011, p.324) II. Hypothesis Test for Means without Knowing Population s σ However, since the population parameters are not always known, we can only use SE ( ), to approximate the standard deviation of the sampling distribution, where s is the standard deviation of the sample we happen to use and n is the sample size and the degree of freedom (df) is n State hypotheses (H 0 and H a ) and determine whether a twotailed or onetailed test 2. Based on level and degree of freedom (df= n1), find out the critical value and determine the rejection region(s) using Student s TTable (A34 in Appendix C) 3. Calculate standard error of the sampling distribution SE ( ), s = standard deviation of the sample, n = sample size 4. Calculate Student s tstatistic for the sample t = 5. Compare t and : if t, reject H0; if t <, fail to reject H0. Q11.3 Suppose we would like to know whether the mean GMAT score for all MBA students is 600. Now you take a random sample of 36 MBA students and find that the average GMAT score for this sample is 640 and the standard deviation for this sample 120. State your hypothesis and conduct the test. If an alpha level α of 5% is used, what s your conclusion? If an alpha level α of 10% is used, what is your conclusion? Step 1: State Hypotheses H 0 : μ = 600 Chaodong Han OPRE504 Page 2 of 6
3 H a : μ 600 This is a twotailed test. Step 2: Find Critical Value Using TwoTailed T Table Based on α=5%, df= 361=35: = 2.03 Rejection regions are 2.03 and due to a twotailed test Step 3: Calculate the Standard Error for the Sampling Distribution SE ( ) = 20 Step 4: Calculate Student s tstatistic for the Sample t 35 = = = 2 Step 5: Compare the calculated tstatistic with critical value and make your judgment Since t 35 < and falls outside the rejection regions, we fail to reject H 0 that the average GMAT score is 640 and conclude that the average GMAT score for all MBA students may not differ from 640 at the 5% significance level. REPEAT STEP 2 AND STEP 5 FOR THE NEW α: Step 2: Find out the Critical Value associated with α=10% using twotailed T table = 1.69 Step 5: Compare the calculated tstatistic with critical value and make your judgment Since t 35 > and falls in the rejection region, we reject H 0 that the average GMAT score is 640 and conclude that the average GMAT score for all MBA students may differ from 640 at 10% significance level. Q11.4 Suppose we would like to know whether the mean GMAT score for all MBA students is greater than 600. Now you take a random sample of 36 MBA students and find that the average GMAT score for this sample is 640 and the standard deviation for this sample 120. State your hypothesis and conduct the test. If an alpha level α = 5% is used, what s your conclusion? If an alpha level α = 1% is used, what is your conclusion? Step 1: State Hypothesis Chaodong Han OPRE504 Page 3 of 6
4 H 0 : μ = 600 H a : μ > 600 This is a onetailed test (upper tail). Step 2: Find Critical Value Using OneTailed TTable Based on α=5%, df= 361=35: = 1.69 Rejection regions are 1.69 due to a onetailed test Using Excel formula: TINV (0.10, 35) = 1.69 because TINV only works with 2tailed T. From the T table, we note that twotailed 10% is equivalent to onetailed 5%. Therefore, we need to double alpha levels to use TINV function for a onetailed test. Step 3: Calculate the Standard Error for the Sampling Distribution SE ( ) = 20 Step 4: Calculate Student s tstatistic for the Sample t 35 = = = 2 Step 5: Compare the calculated tstatistic with critical value and make your judgment Since t 35 > and falls within the rejection region, we reject H 0 that the average GMAT score is 640 and conclude that the average GMAT score for all MBA students may differ from 640 at the 5% significance level. REPEAT STEP 2 AND STEP 5 FOR THE NEW α: Step 2: Find out the Critical Value associated with α=1% using onetailed Ttable = Step 5: Compare the calculated tstatistic with the Critical Value and make your judgment Since t 35 < and falls outside the rejection region, we fail to reject H 0 that the average GMAT score is 600 and conclude that the average GMAT score for all MBA students may not differ from 600 at 1% significance level. Textbook Examples of Hypothesis Tests for Means Guided Example Insurance Profits Revisited (pp ); This is also onetailed test with lower tail. Chaodong Han OPRE504 Page 4 of 6
5 III Confidence Intervals for Means Model: Onesample tinterval = Estimate Marginal Error (ME) ME = X SE (, CI = X SE (, SE ( =, where n= sample size, df = n1 1. Determine level for confidence interval (CI): = 1CI 2. Calculate Standard Error of the Sampling Distribution: SE ( 3. Find out the critical value using TwoTailed TTable (, df= n1) 4. CI = X SE (, SE ( Q11.5 The average gasoline price per gallon for a random sample of 30 stations in a region is $4.49 with a standard deviation of $0.29. [Textbook, Q13, p.346]. a) Find a 95% confidence interval for the mean price of gasoline for all stations in that region Given n = 30, = 4.49, s= 0.29, and 95% confidence interval 1. = = SE ( = = = CI = X SE ( = X = CI = (4.382, 4.598) if rounded to tenth cents b) Find a 90% CI for mean price of gasoline for all stations in that region Given n = 30, = 4.49, s= 0.29, and 90% confidence interval 1. = = SE ( = = = CI = X SE ( = X = CI = (4.400, 4.580) c) If we had the sample statistics from a sample of 60 stations, what would be the 95% confidence interval? Given n = 60, = 4.49, s= 0.29, and 95% confidence interval 1. = = SE ( = = = CI = X SE ( = X = CI = (4.415, 4.565) Chaodong Han OPRE504 Page 5 of 6
6 IV Determine Sample Size CI upper = Mean Estimate + Marginal Error (ME) CI lower = Mean Estimate  ME ME = CI upper Mean Estimate SE (, ME = x ( = 1 confidence interval) [ Z is used because we don t know n and need to find it] n = Q11.6 Police departments often try to control traffic speed by placing speedmeasuring machines on roads that tell motorists how fast they are driving. Traffic safety experts must determine where machines should be placed. In one recent test, police recorded the average speed clocked by cars driving on one busy street close to an elementary school. For a sample of 25 clocked speeds, it was determined that the average amount over the speed limit for the 25 clocked speeds was 11.6 mph with a standard deviation of 8 mph. The 95% confidence interval estimate for this sample is 8.30 mph to mph [Sharpe 2011, Q5758, p.353] a) What is the margin of error for this problem? Given CI = (8.30, 14.90) and the mean estimate = 11.6 mph, ME = Upper Limit Mean Estimate = = 3.3 mph b) The researchers commented that the interval was too wide. Explain what should be done to reduce the margin of error to no more than ±2 mph. Given n is unknown, s = 8, = = 0.05, Z * = 1.96 n= = x 8 2 = clocked speeds c) To ensure that error rates are estimated accurately, the researchers want to take a large enough sample that will ensure that usable and accurate interval estimates of how much the machines may be off in measuring actual speeds. Specifically, the researchers want the margin of error for a single speed measurement to be no more than ±1.5 mph at the 95% confidence interval. How may the researchers obtain a reasonable estimate of the standard deviation of error in the measured speeds? Conduct a pilot study or use findings from previous studies d) Suppose the standard deviation of for the error in the measured speeds is believed to be 4 mph from a pilot study, what should be the sample size for next study to ensure that the margin of error is no larger than ±1 mph. 1.0 = 1.96* ( ), n = = x 4 2 = Chaodong Han OPRE504 Page 6 of 6
1. What is the critical value for this 95% confidence interval? CV = z.025 = invnorm(0.025) = 1.96
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