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UNC-Chapel Hill STOR 151 - Homework 8

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HW8, due 04/02/09:Chapter 7, questions 7.20, 7.28, 7.66, 7.76.Comments on 7.66: The question asks you to look up some dataon the Web, but you should not have any trouble finding it if youfollow the web link given and search for the variable FEHELPin the year 1998. However, interpreting it may be a little moredifficult — I believe they want you to combine together thecategories labelled ”Strongly agree” and ”Agree” and omit thecases for which no data are available. Even then, I think there isone tiny error in the question, though one which has no materialimpact on the final answer.1Confidence intervals for a population meanSo far we have talked about confidence intervals only in the caseof a sample proportion. What about the other case covered inChapter 6, where we are interested in a population mean µ, andwe use a sample mean ¯x based on a sample of n observations.2Example: Annual average rainfalls over Australia (mm)Year Rainfall Year Rainfall Year Rainfall Year Rainfall1970 384 1979 456 1988 460 1997 5271971 494 1980 433 1989 484 1998 5651972 365 1981 535 1990 418 1999 5841973 661 1982 421 1991 469 2000 7271974 784 1983 499 1992 452 2001 5591975 603 1984 555 1993 499 2002 3411976 528 1985 399 1994 341 2003 4871977 472 1986 392 1995 5231978 526 1987 453 1996 470Mean 496.1; standard deviation 98.88; min 341; max 784.Data plots show no overall trend up or down, though there seemto be high periods around 1975 and 2000. Histogram is approx-imately normal though somewhat right-skewed.3Histogram of AustralianRainfallAustralianRainfallFrequency300 400 500 600 700 8000 2 4 6 8 10 121970 1975 1980 1985 1990 1995 2000400 500 600 700 800Scatterplot of Australian Rainfall with Linear TrendlineYearAustralianRainfall4Problem: calculate a confidence interval for the mean precipita-tion per year in Australia. Use sample means 1970–2003 as arandom sample from the “population” of all annual precipitationmeans in Australia.Step 1: Calculate the sample mean — in this case, ¯x = 496.1.Step 2: Calculate the standard error,σ√n.However since we don’t know the true σ, we substitute the sam-ple standard deviation, which is s = 98.88. Therefore, the stan-dard error is98.88√34= 16.96.Step 3: For a 95% confidence interval, we estimate the marginof error as 1.96 × 16.96 = 33.2 to 1 decimal place.5Therefore, our tentative confidence interval is 496.1 ± 33.2 =(462.9, 529.3).However, although this is close to the right answer, it turns outit isn’t quite the correct calculation.6The t distributionThe problem is that the standardized statistict =Observed value − Population meanStandard error(∗∗)doesn’t have exactly the normal distribution. It would have anormal distribution if in calculating the standard error we usedthe true σ. But we didn’t use σ — instead, we substituted thesample standard deviation s. It turns out that this changes thenormal distribution into another distribution called the t distribu-tion (or Student’s t distribution) — that’s why we use the lettert in place of z in (**).7The t distribution is quite close to a normal distribution, but alittle bit fatter-tailed (see figure, next slide).Also, the t distribution depends on another quantity that we callthe degrees of freedom, abbreviated df. For the set up we havehere, the df is just n −1 where n is the sample size. For Tables,see Table B of the Appendix (page A3).8−4 −2 0 2 40.0 0.1 0.2 0.3 0.4VariableDensityNormalt (df=5)9Confidence leveldf 80% 90% 95% 98% 99% 99.8%1 3.078 6.314 12.706 31.821 63.657 318.3092 1.886 2.920 4.303 6.965 9.925 22.3273 1.638 2.353 3.182 4.541 5.841 10.2154 1.533 2.132 2.776 3.747 4.604 7.1735 1.476 2.015 2.571 3.365 4.032 5.8936 1.440 1.943 2.447 3.143 3.707 5.2087 1.415 1.895 2.365 2.998 3.499 4.7858 1.397 1.860 2.306 2.896 3.355 4.5019 1.383 1.833 2.262 2.821 3.250 4.29710 1.372 1.812 2.228 2.764 3.169 4.14411 1.363 1.796 2.201 2.718 3.106 4.02512 1.356 1.782 2.179 2.681 3.055 3.93013 1.350 1.771 2.160 2.650 3.012 3.85214 1.345 1.761 2.145 2.624 2.977 3.78715 1.341 1.753 2.131 2.602 2.947 3.73316 1.337 1.746 2.120 2.583 2.921 3.68617 1.333 1.740 2.110 2.567 2.898 3.64618 1.330 1.734 2.101 2.552 2.878 3.61019 1.328 1.729 2.093 2.539 2.861 3.57920 1.325 1.725 2.086 2.528 2.845 3.55221 1.323 1.721 2.080 2.518 2.831 3.52722 1.321 1.717 2.074 2.508 2.819 3.50523 1.319 1.714 2.069 2.500 2.807 3.48524 1.318 1.711 2.064 2.492 2.797 3.46725 1.316 1.708 2.060 2.485 2.787 3.45026 1.315 1.706 2.056 2.479 2.779 3.43527 1.314 1.703 2.052 2.473 2.771 3.42128 1.313 1.701 2.048 2.467 2.763 3.40829 1.311 1.699 2.045 2.462 2.756 3.39630 1.310 1.697 2.042 2.457 2.750 3.38540 1.303 1.684 2.021 2.423 2.704 3.30750 1.299 1.676 2.009 2.403 2.678 3.26160 1.296 1.671 2.000 2.390 2.660 3.23280 1.292 1.664 1.990 2.374 2.639 3.195100 1.290 1.660 1.984 2.364 2.626 3.174Inf 1.282 1.645 1.960 2.327 2.576 3.09110What we actually do to calculate a 95% confidence interval:Step 3 (in place of the above): Calculate df. In this case n −1 =33.Step 4: For the desired confidence coefficient and the appropri-ate df, look up the critical value in Table B. In this case, we areagain assuming 95% confidence, so we look in the fourth columnfrom the left (headed “Confidence level 95%”).There isn’t an entry for df=33, so we take the nearest df in thetable, which is 30. Thus the critical value is 2.042.Step 5: Calculate the margin of error using this critical value.So we calculate 2.042 ×16.96 = 34.6 to one decimal place. Thisis in place of the 33.2 calculated if we ignored the distinctionbetween a normal and t distribution.11The final interval is 496.1 ± 34.6 = (461.5, 530.7). This is notvery different from what we calculated earlier, but it’s a littlemore precise because we are allowing for the fact that we hadto estimate σ in calculating the standard error.12Assumptions behind this method1. The data are a random sample from a distribution whose truemean is µ and true standard deviation σ (both unknown).2. The distribution of the individual observations is normal.Assumption 1 is the one that really matters. None of thesestatistical methods works if random sampling assumptions arenot satisfied.Assumption 2 is less critical because the method still works rea-sonably if the distribution is a little skewed. But we need to becareful about extreme outliers.13History of the methodThe t distribution was discovered by William S. Gosset, whowas a brewer for Guinness in Ireland, and developed the


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UNC-Chapel Hill STOR 151 - Homework 8

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