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Slide1: 

Chapter 12

Introduction: 

Introduction 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.

Slide3: 

Inference About A Population… We will develop techniques to estimate and test three population parameters: Population Mean μ Population Variance σ2 Population Proportion p

Inference About a Population Mean When the Population Standard Deviation Is Unknown: 

Recall that when s is known we use the following statistic to estimate and test a population mean When s is unknown, we use its point estimator s, and the z-statistic is replaced then by the t-statistic Inference About a Population Mean When the Population Standard Deviation Is Unknown x - μ Z = σ / √ n

Slide5: 

Inference With Variance Unknown… Previously, we looked at estimating and testing the population mean when the population standard deviation ( σ ) was known or given: But how often do we know the actual population variance? Instead, we use the Student t-statistic, given by: x - μ Z = σ/√ n x - μ t = S/√ n

Test Statistic for m When s is Unknown : 

Test Statistic for m When s is Unknown Confidence Interval Estmator of m When s is Unknown x - μ t = S/√ n s x ± tα/2 √ n ν = n -1

Slide7: 

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). Testing m when s is unknown

Slide8: 

505 400 499 415 418 467 551 444 481 429 480 466 477 445 413 537 484 418 465 496 487 373 416 424 471 427 509 410 515 435 482 442 465 449 523 488 508 432 405 440 409 501 440 444 485 475 470 485 469 450 Data Xm 12-01: Example 12.1 Number of Packages

Example 12.1…: 

Example 12.1… Our objective is to describe the population of the numbers of packages processed in 1 hour by new workers, that is we want to know whether the new workers’ productivity is more than 90% of that of experienced workers. Thus we have: H1: μ > 450 Therefore we set our usual null hypothesis to: H0: μ = 450 IDENTIFY

Example 12.1…: 

Example 12.1… Our test statistic is: With n=50 data points, we have n–1=49 degrees of freedom. Our hypothesis under question is: H1: μ > 450 Our rejection region becomes: t > tα,ν = t.05,49 ≈ t.05,50 = 1.676 Thus we will reject the null hypothesis in favor of the alternative if our calculated test static falls in this region. COMPUTE x - μ t = S / √ n ν = n - 1

Example 12.1…: 

Example 12.1… From the data, we calculate x = 460.38, s =38.83 and thus: Since we reject H0 in favor of H1, that is, there is sufficient evidence to conclude that the new workers are producing at more than 90% of the average of experienced workers. COMPUTE

INTERPRET: 

INTERPRET Example 12.1… 0 1.676 1.89 t Since 1.89 > 1.676 we reject the null hypothesis. This means that there is enough evidence to infer that the mean number of parcels processed by trainees is more than 90% of that of the experienced workers

Slide13: 

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. ( Data Xm 12-02 ) Estimating m when s is unknown

Return on Investment for Quality Award Winners Example 12.2 - DATA Xm 12-02: 

Return on Investment for Quality Award Winners Example 12.2 - DATA Xm 12-02 16.2 15.34 19.55 13.53 6.76 13.9 11.72 17.63 10.97 22.1 0.58 16.65 24.6 -2.27 -0.67 12.8 26.01 15.46 -1.16 7.41 25.47 27.25 25.46 18.07 2.62 13.45 9.45 12.74 13.6 -3.6 11.42 28.26 21.45 21.36 23.8 9.01 26.58 6.51 18.63 24.69 16.34 22.69 -0.21 7.76 2.28 6.89 23.94 9.62 13.72 21.39 -1.06 22.45 7.23 26.2 21.93 21.02 17.85 7.9 12.54 20.95 15.52 16.82 17.52 29.68 18.02 22.37 25.32 22.78 14.65 21.15 5.14 14.3 24.77 23.33 9.88 18.41 2.26 16.52 5.66 6.39 15.45 21.58 14.15

Slide15: 

Example 12.2… Can we estimate the return on investment for companies that won quality awards? We have are given a random sample of n = 83 such companies. We want to construct a 95% confidence interval for the mean return, i.e. what is: IDENTIFY ?? s x ± tα/2 √ n

Slide16: 

Example 12.2… From the data, we calculate: For this term and so: COMPUTE LCL = 13.20 UCL = 16.84

Slide17: 

Example 12.2… We are 95% confident that the population mean, μ , i.e. the mean return of all publicly traded companies that win quality awards, lies between 13.20% and 16.84% INTERPRET s x ± tα/2 = 15.02±1.990 = 15.02 ±1.82 √ n √83 8.31

Slide18: 

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…

Slide19: 

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

Estimating Totals of Finite Populations…: 

Estimating Totals of Finite Populations… Large populations are defined as “populations that are at least 20 times the sample size” We can use the confidence interval estimator of a mean to produce a confidence interval estimator of the population total: Where N is the size of the finite population. s x ± tα/2 √ n N

Estimating Totals of Finite Populations…: 

Estimating Totals of Finite Populations… For example, a sample of 500 households (in a city of 1 million households) reveals a 95% confidence interval estimate that the household mean spent on Halloween candy lies between $20 & $30. We can estimate the total amount spent in the city by multiplying these lower and upper confidence limits by the total population: Thus we estimate that the total amount spent on Halloween in the city lies between $20 million and $30 million. s x ± tα/2 √ n N = 1,000,000($20…….$30)

Identifying Factors…: 

Identifying Factors… Factors that identify the t-test and estimator of μ:

Inference About Population Variance…: 

Inference About Population Variance… If we are interested in drawing inferences about a population’s variability, the parameter we need to investigate is the population variance: The sample variance (s2) is an unbiased, consistent and efficient point estimator for σ2 . Moreover, the statistic, , has a chi-squared distribution, with n–1 degrees of freedom. (n – 1)s2 χ2 = σ2

Testing & Estimating Population Variance: 

Testing & Estimating Population Variance The test statistic used to test hypotheses about σ2 is: (which is chi-squared with ν = n–1 degrees of freedom). (n – 1)s2 χ2 = σ2

Testing & Estimating Population Variance: 

Testing & Estimating Population Variance Combining this statistic: With the probability statement: Yields the confidence interval estimator for σ2 : lower confidence limit upper confidence limit (n – 1)s2 χ2 = σ2 P ( χ2 < χ2 < χ2 ) = 1 - α α/2 1-α/2 (n – 1)s2 LCL = χ2 α/2 (n – 1)s2 UCL = χ2 1-α/2

Example 12.3…: 

Example 12.3… Consider a container filling machine. Management wants a machine to fill 1 liter (1,000 cc’s) so that that variance of the fills is less than 1 cc2. A random sample of n=25 1 liter fills were taken. Does the machine perform as it should at the 5% significance level? IDENTIFY 1000.3 1001.3 999.5 999.7 999.3 999.8 998.3 1000.6 999.7 999.8 1001 999.4 999.5 998.5 1000.7 999.6 999.8 1000 998.2 1000.1 998.1 1000.7 999.8 1001.3 1000.7 Data Xm12-03

Example 12.3: 

Example 12.3 Variance is less than 1 cc2 (n – 1)s2 χ2 = σ2 We want to show that: H1: σ2 < 1 (so our null hypothesis becomes: H0: σ2 = 1). We will use this test statistic: IDENTIFY

Example 12.3…: 

Example 12.3… Since our alternative hypothesis is phrased as: H1: σ2 < 1 We will reject H0 in favor of H1 if our test statistic falls into this rejection region: We computer the sample variance to be: s2=.8088 And thus our test statistic takes on this value… COMPUTE compare χ2 < χ2 = χ21-.05, 25-1 = χ2.95,24= 13.8484 1-α,n-1 (n – 1)s2 χ2 = σ2 (25 – 1)(.8088) = = 19.41 1

Slide29: 

13.8484 19.41 Rejection region a = .05 1-a = .95 Do not reject the null hypothesis Example 12.3… Since: There is not enough evidence to infer that the claim is true. INTERPRET

Example 12.4…: 

Example 12.4… As we saw, we cannot reject the null hypothesis in favor of the alternative. That is, there is not enough evidence to infer that the claim is true. Note: the result does not say that the variance is greater than 1, rather it merely states that we are unable to show that the variance is less than 1. We could estimate (at 99% confidence say) the variance of the fills…

Example 12.4…: 

Example 12.4… In order to create a confidence interval estimate of the variance, we need these formulae: we know (n–1)s2 = 19.41 from our previous calculation, and we have from Table 5 in Appendix B: COMPUTE lower confidence limit upper confidence limit χ2α/2,n-1 = χ2.005,24 = 45.5585 χ21-α/2,n-1 = χ2.995,24 =9.88623

Example 12.4…: 

Example 12.4… Thus the 99% confidence interval estimate is: That is, the variance of fills lies between .426 and 1.963 cc2. COMPUTE

Identifying Factors…: 

Identifying Factors… Factors that identify the chi-squared test and estimator of σ2 :

Inference: Population Proportion…: 

Inference: Population Proportion… When data are nominal, we count the number of occurrences of each value and calculate proportions. Thus, the parameter of interest in describing a population of nominal data is the population proportion p. This parameter was based on the binomial experiment. Recall the use of this statistic: where p-hat ( p ) is the sample proportion: x successes in a sample size of n items. ^ x P = n ^

Inference: Population Proportion…: 

Inference: Population Proportion… When np and n(1–p) are both greater than 5, the sampling distribution of p is approximately normal with mean: standard deviation: Hence: ^

Inference: Population Proportion…: 

Inference: Population Proportion… Test statistic for p : The confidence interval estimator for p is given by: (both of which require that np>5 and n(1–p)>5)

Example 12.5…: 

Example 12.5… At an exit poll, voters are asked by a certain network if they voted Democrat (code=1) or Republican (code=2). Based on their small sample, can the network conclude that the Republican candidate will win the vote? That is: H1: p > .50 And hence our null hypothesis becomes: H0: p = .50 IDENTIFY

Example 12.5…: 

Example 12.5… Since our research hypothesis is: H1: p > .50 our rejection region becomes: Looking at the data, we count 407 (of 765) votes for code=2. Hence, we calculate our test statistic as follows… COMPUTE

Example 12.5…: 

Example 12.5… Since: …we reject H0 in favor of H1, that is, there is enough evidence to believe that the Republicans win the vote. INTERPRET

Selecting the Sample Size…: 

Selecting the Sample Size… The confidence interval estimator for a population proportion is: Thus the (half) width of the interval is: Solving for n, we have:

Selecting the Sample Size…: 

Selecting the Sample Size… For example, we want to know how many customers to survey in order to estimate the proportion of customers who prefer our brand to within .03 (with 95% confidence). I.e. our confidence interval after surveying will be p ± .03, that means W=.03 Substituting into the equation… Uh Oh. Since we haven’t taken a sample yet, we don’t have this sample proportion… ^

Selecting the Sample Size…: 

Selecting the Sample Size… Two methods – in each case we choose a value for then solve the equation for n. Method 1 : no knowledge of even a rough value of . This is a ‘worst case scenario’ so we substitute = .50 Method 2 : we have some idea about the value of . This is a better scenario and we substitute in our estimated value. p ^ p ^ p p p ^ ^ ^

Selecting the Sample Size…: 

Selecting the Sample Size… Method 1 : no knowledge of value of , use 50%: Method 2 : some idea about a possible value, say 20%: Thus, we can sample fewer people if we already have a reasonable estimate of the population proportion before starting.

Estimating Totals for Large Populations…: 

Estimating Totals for Large Populations… In much the same way as we saw earlier, when a population is large and finite we can estimate the total number of successes in the population by taking the product of the size of the population (N) and the confidence interval estimator: The Nielsen Ratings (used to measure TV audiences) uses this technique. Results from a small sample audience (say 2,000 viewers) is extrapolated to the total number of TV sets (say 100 million)…

Nielsen Ratings Example…: 

Nielsen Ratings Example… Problem: describe the population of television shows watched by viewers across the country (population), by examining the results from 2,000 viewers (sample). We take these values and multiply them by N=100 million to estimate that between 9.9 million and 12.7 million viewers are watching the “Tonight Show”. COMPUTE

Identifying Factors…: 

Identifying Factors… Factors that identify the z-test and interval estimator of p :

Flowchart of Techniques…: 

Flowchart of Techniques… Describe a Population Data Type? Nominal Interval z test & estimator of p Type of descriptive measurement? Central Location Variability t test & estimator of μ u. X2 test & estimator of χ2 σ2

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