testing m when s is unknown：检测m当s是未知的

Ch 12 實習 2l We shall develop techniques to estimate and test three population parameters.l Population mean m l Population variance s2l Population proportion p Introduction3Recall that when s is known we use the following statistic to estimate and test a population meanWhen s is unknown, we use its point estimator s, and the z-statistic is replaced then by the t-statisticInference About a Population Mean When the Population Standard Deviation Is Unknown4The t - Statistics0The t distribution is mound-shaped, and symmetrical around zero.The “degrees of freedom”,(a function of the sample size)determine how spread thedistribution is (compared to the normal distribution)d.f. = v2d.f. = v1v1 450l The t statisticd.f. = n - 1 = 49Testing m when s is unknown 9l Solution continued (solving by hand)l The rejection region is t ta,n 1ta,n - 1 = t.05,49 t.05,50 = 1.676.Testing m when s is unknown 10l The test statistic isl Since 1.89 1.676 we reject the null hypothesis in favor of the alternative. l There is sufficient evidence to infer that the mean productivity of trainees one week after being hired is greater than 450 packages at .05 significance level.1.676 1.89Rejection regionTesting m when s is unknown 11Estimating m when s is unknown l Confidence interval estimator of m when s is unknown12l Example 2l An investor is trying to estimate the return on investment in companies that won quality awards last year.l A random sample of 83 such companies is selected, and the return on investment is calculated had he invested in them.l Construct a 95% confidence interval for the mean return.Estimating m when s is unknown 13l Solution (solving by hand)l The problem objective is to describe the population of annual returns from buying shares of quality award-winners.l The data are interval.l Solving by handl From the data we determine t.025,82 t.025,80Estimating m when s is unknown 14Checking the required conditionsl We need to check that the population is normally distributed, or at least not extremely nonnormal.l There are statistical methods to test for normality l From the sample histograms we see15A Histogram for Example 1PackagesA Histogram for Example 2Returns16Summary of Test Statistics to be Used in aHypothesis Test about a Population Meann 30 ?s known ? Popul. approx.normal ?s known ?Use s toestimate sUse s toestimate sIncrease nto 30YesYesYesYesNoNoNoNo17Example 1l A federal agency responsible for enforcing laws governing weights and measures routinely inspects packages to determine whether the weight of the contents is at least as great as that advertised on the package. A random sample of 18 containers whose packaging states that the contents weigh 8 ounces was drawn. The contents were weighted and the results follows. Can we concluded at the 1% significance level that on average the containers are mislabeled? (Assume the random variable is normally distributed)l 7.80 7.91 7.93 7.99 7.94 7.757.97 7.95 7.79 8.06 7.82 7.897.92 7.87 7.92 7.98 8.05 7.9118Solutionl H0:=8H1: t0.025,9 = 2.262Test statistic: t = 1.285Conclusion: Dont reject H0. We cant infer at the 5% significance level that the population mean is not equal to 20.l The condition is that ages in the population are normally distributed. A histogram of the data can be used to check if the normality assumption is satisfied.2223Inference About a Population Variancel Sometimes we are interested in making inference about the variability of processes.l Examples:l The consistency of a production process for quality control purposes.l Investors use variance as a measure of risk. l To draw inference about variability, the parameter of interest is s2.24l The sample variance s2 is an unbiased, consistent and efficient point estimator for s2.l The statistic has a distribution called Chi-squared, if the population is normally distributed. d.f. = 5d.f. = 10Inference About a Population Variance25l Example 3 (operation management application)l A container-filling machine is believed to fill 1 liter containers so consistently, that the variance of the filling will be less than 1 cc (.001 liter).l To test this belief a random sample of 25 1-liter fills was taken, and the results recorded l Do these data support the belief that the variance is less than 1cc at 5% significance level?Testing the Population Variance 26l Solutionl The problem objective is to describe the population of 1-liter fills from a filling machine. l The data are interval, and we are interested in the variability of the fills.l The complete test is:H0: s2 = 1H1: s2 1We want to know whether the process is consistentTesting the Population Variance 27There is insufficient evidence to reject the hypothesis thatthe variance is less than 1. Solving by hand Note that (n - 1)s2 = S(xi - x)2 = Sxi2 (Sxi)2/n From the sample, we can calculate Sxi = 24,996.4, and Sxi2 = 24,992,821.3 Then (n - 1)s2 = 24,992,821.3-(24,996.4)2/25 =20.78 Testing the Population Variance 2813.8484 20.8Rejectionregiona = .05 1-a = .95Do not reject the null hypothesisTesting the Population Variance 29Testing and Estimating a Population Variance l From the following probability statementP(c21-a/2 c2 c2a/2) = 1-awe have (by substituting c2 = (n - 1)s2/s2.)30Example 4l With gasoline prices increasing, drivers are becoming more concerned with their cars gasoline consumption. For the past 5 years, a driver has tracked the gas mileage of his car and found that the variance from fill-up to fill-up was 2=23 mpg2. Now that his