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1 Chapter 12 Inference About One Population

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2 12.1 Introduction In this chapter we utilize the approach developed before to describe a population.In this chapter we utilize the approach developed before to describe a population. –Identify the parameter to be estimated or tested. –Specify the parameter’s estimator and its sampling distribution. –Construct a confidence interval estimator or perform a hypothesis test.

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3 We shall develop techniques to estimate and test three population parameters. –Population mean –Population variance 2 –Population proportion p 12.1 Introduction

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4 Recall that when is known we use the following statistic to estimate and test a population mean When is unknown, we use its point estimator s, and the z-statistic is replaced then by the t-statistic 12.2 Inference About a Population Mean When the Population Standard Deviation Is Unknown

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5 The t - Statistic s Z t s s s s When the sampled population is normally distributed, the t statistic is Student t distributed. Z Z Z ZZt t t t t t t t t s s s s s t

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6 The t - Statistic s 0 The t distribution is mound-shaped, and symmetrical around zero. The “degrees of freedom”, (a function of the sample size) determine how spread the distribution is (compared to the normal distribution) d.f. = v 2 d.f. = v 1 v 1 < v 2 t Using the t-table

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7 Testing when is unknown Example 12.1 - Productivity of newly hired Trainees

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8 Example 12.1 –In order to determine the number of workers required to meet demand, the productivity of newly hired trainees is studied. –It is believed that trainees can process and distribute more than 450 packages per hour within one week of hiring. –Can we conclude that this belief is correct, based on productivity observation of 50 trainees (see file Xm12-01).Xm12-01 Testing when is unknown

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9 Example 12.1 – Solution –The problem objective is to describe the population of the number of packages processed in one hour. –The data are interval. H 0 : = 450 H 1 : > 450 –The t statistic d.f. = n - 1 = 49 We want to prove that the trainees reach 90% productivity of experienced workers We want to prove that the trainees reach 90% productivity of experienced workers Testing when is unknown

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10 Solution continued (solving by hand) –The rejection region is t > t ,n – 1 t ,n - 1 = t.05,49 t.05,50 = 1.676. Testing when is unknown

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11 The test statistic is Since 1.89 > 1.676 we reject the null hypothesis in favor of the alternative. 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 1.89 Rejection region Testing when is unknown

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12 Testing when is unknown Since.0323 <.05, we reject the null hypothesis in favor of the alternative. 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..05.0323

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13 Estimating when is unknown Confidence interval estimator of when is unknown

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14 Example 12.2 –An investor is trying to estimate the return on investment in companies that won quality awards last year. –A random sample of 83 such companies is selected, and the return on investment is calculated had he invested in them. –Construct a 95% confidence interval for the mean return. Estimating when is unknown

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15 Solution (solving by hand) –The problem objective is to describe the population of annual returns from buying shares of quality award-winners. –The data are interval. –Solving by hand From the Xm12-02 we determineXm12-02 t.025,82 t.025,80 Estimating when is unknown

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16 Estimating when is unknown

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17 Checking the required conditions We need to check that the population is normally distributed, or at least not extremely nonnormal. There are statistical methods to test for normality (one to be introduced later in the book). From the sample histograms we see…

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18 A Histogram for Xm12- 01 Packages A Histogram for Xm12- 02 Returns

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19 12.3 Inference About a Population Variance Sometimes we are interested in making inference about the variability of processes. Examples: –The consistency of a production process for quality control purposes. –Investors use variance as a measure of risk. To draw inference about variability, the parameter of interest is 2.

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20 The sample variance s 2 is an unbiased, consistent and efficient point estimator for 2. The statistic has a distribution called Chi- squared, if the population is normally distributed. d.f. = 5 d.f. = 10 12.3 Inference About a Population Variance

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21 Testing and Estimating a Population Variance From the following probability statement P( 2 1- /2 < 2 < 2 /2 ) = 1- we have (by substituting 2 = [(n - 1)s 2 ]/ 2.)

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22 Example 12.3 (operation management application) –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). –To test this belief a random sample of 25 1-liter fills was taken, and the results recorded ( Xm12-03) Xm12-03 –Do these data support the belief that the variance is less than 1cc at 5% significance level? Testing the Population Variance

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23 Solution –The problem objective is to describe the population of 1-liter fills from a filling machine. –The data are interval, and we are interested in the variability of the fills. –The complete test is: H 0 : 2 = 1 H 1 : 2 <1 We want to know whether the process is consistent Testing the Population Variance

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24 There is insufficient evidence to reject the hypothesis that the variance is less than 1. There is insufficient evidence to reject the hypothesis that the variance is less than 1. Solving by hand –Note that (n - 1)s 2 = (x i - x) 2 = x i 2 – ( x i ) 2 /n –From the sample (Xm12-03) we can calculate x i = 24,996.4, and x i 2 = 24,992,821.3Xm12-03 –Then (n - 1)s 2 = 24,992,821.3-(24,996.4) 2 /25 =20.78 Testing the Population Variance

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25 13.848420.8 Rejection region =.05 1- =.95 Do not reject the null hypothesis Testing the Population Variance

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26 Estimating the Population Variance Example 12.4 –Estimate the variance of fills in Example 12.3 with 99% confidence.Example 12.3 Solution –We have (n-1)s 2 = 20.78. From the Chi-squared table we have 2 /2,n-1 = 2.005, 24 = 45.5585 2 /2,n-1 2.995, 24 = 9.88623

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27 The confidence interval estimate is Estimating the Population Variance

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28 12.4 Inference About a Population Proportion When the population consists of nominal data, the only inference we can make is about the proportion of occurrence of a certain value. The parameter p was used before to calculate these probabilities under the binomial distribution.

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29 Statistic and sampling distribution –the statistic used when making inference about p is: – Under certain conditions, [np > 5 and n(1-p) > 5], is approximately normally distributed, with = p and 2 = p(1 - p)/n. 12.4 Inference About a Population Proportion

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30 Testing and Estimating the Proportion Test statistic for p Interval estimator for p (1- confidence level)

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31 Example 12.5 (Predicting the winner in election day) –Voters are asked by a certain network to participate in an exit poll in order to predict the winner on election day. –Based on the data presented in Xm12-05 where 1=Democrat, and 2=Republican), can the network conclude that the republican candidate will win the state college vote?Xm12-05 Testing the Proportion Additional example

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32 Solution –The problem objective is to describe the population of votes in the state. –The data are nominal. –The parameter to be tested is ‘p’. –Success is defined as “Vote republican”. –The hypotheses are: H 0 : p =.5 H 1 : p >.5 More than 50% vote Republican Testing the Proportion

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33 –Solving by hand The rejection region is z > z = z.05 = 1.645. From file we count 407 success. Number of voters participating is 765. The sample proportion is The value of the test statistic is The p-value is = P(Z>1.77) =.0382 Testing the Proportion

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34 There is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. At 5% significance level we can conclude that more than 50% voted Republican. Testing the Proportion

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35 Nielsen Ratings –In a survey of 2000 TV viewers at 11.40 p.m. on a certain night, 226 indicated they watched “The Tonight Show”. –Estimate the number of TVs tuned to the Tonight Show in a typical night, if there are 100 million potential television sets. Use a 95% confidence level. –Solution Estimating the Proportion

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36 A confidence interval estimate of the number of viewers who watched the Tonight Show: LCL =.099(100 million)= 9.9 million UCL =.127(100 million)=12.7 million Solution Estimating the Proportion

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37 Selecting the Sample Size to Estimate the Proportion Recall: The confidence interval for the proportion is Thus, to estimate the proportion to within W, we can write

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38 Selecting the Sample Size to Estimate the Proportion The required sample size is

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39 Example –Suppose we want to estimate the proportion of customers who prefer our company’s brand to within.03 with 95% confidence. –Find the sample size. –Solution W =.03; 1 - =.95, therefore /2 =.025, so z.025 = 1.96 Since the sample has not yet been taken, the sample proportion is still unknown. We proceed using either one of the following two methods: Sample Size to Estimate the Proportion

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40 Method 1: –There is no knowledge about the value of Let. This results in the largest possible n needed for a 1- confidence interval of the form. If the sample proportion does not equal.5, the actual W will be narrower than.03 with the n obtained by the formula below. Sample Size to Estimate the Proportion Method 2: –There is some idea about the value of Use the value of to calculate the sample size

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