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Introduction to Inference Chapter 6 Statistical Inference Want to draw conclusions based on sample data i.e.-- say something about an entire population based on information in a sample. Conclusions are subject to sampling error We want to quantify the margin of error we are likely to encounter. Examples 1. I’m interested in estimating the mean income of loggers in the interior of BC. 2. Gallup poll. Confidence intervals Our goal is to obtain an estimate of some parameter of a population. We want our estimate to be of the general form Best guess +/- error of estimation To get our estimate, we take a random sample from the population and proceed on the basis of the information we obtain from the sample. What is our “best guess”?? How do we obtain our “error of estimation” ? Confidence intervals for population mean Best Guess Our “best guess” is xbar, the sample mean. How to determine the estimation error? Determining the estimation error To determine estimation error, we’ll use our knowledge of sampling distribution of xbar. We know that xbar has mean µ, with standard deviation, /sqrt(n), where n is our sample size. If our original population is normal, we also know that xbar is normally distributed. Even if the original population isn’t normal, xbar approximately follows a normal distribution if the sample size, n , is large (by CLT). Determining the error of estimation -- continued For example, we are about 95% sure that xbar lies in the range µ +/- 2 /sqrt(n). (*) E.g., if = 45 and n = 100, we are 95% sure that xbar lies in the range µ +/- 9. (verify). How to use (*) to get our confidence interval?? A confidence interval for population mean It turns out that our 95% c.i. for µ is just xbar +/- 2 /sqrt(n). Why does this work? From our picture ( to be added in class), we see that our interval will “trap” µ approximately 95% percent of the time. A numerical example I collect SRS of n = 100 loggers. I find average income in sample is $17,000. Obtain a 95% c.i. for the mean income of all loggers. Assume that std. dev. of loggers incomes is known to be = 2500. (In Ch 7 we will drop the assumption that is known -- will be estimated from the sample data) Level C confidence intervals Tradeoffs For a given sample size, higher level of confidence leads to wider confidence interval. Designing a confidence interval Suppose I want a level C confidence interval of the form Xbar +/- m, where m is a desired margin of error that I supply in advance. I also specify C in advance. Required sample size is n = (z*/m)2 where z* is the z value required for level C confidence Where does this formula come from? Example--Designing a confidence interval Recall the logger example. I have = $2500. Suppose I want a 99% c.i. of the form xbar +/- 300. How big a sample do I need? What if I want a 99% confidence interval of form xbar +/- 150? Example 6.12 (Text) A study of career paths of hotel general managers sent questionnaires to a SRS of 160 hotels belonging to major US hotel chains. There were 114 responses. The average time these 114 general managers had spent with their current company was 11.78 years. Give a 99% c.i. for the mean number of years general managers of major-chain hotels have spent with their current company. (Take it as known that the std dev of time with the company for general managers is 3.2 years). A margin of error of +/- 1 year is considered acceptable. What is the minimum sample size that would be required to achieve this level of accuracy with a confidence level of 99% Confidence interval summary Assume I have a random sample and that the population std dev, , is known. A level C c.i. for the population mean is xbar +/- z*/sqrt(n). (value of z* will depend on C) The c.i, is exact when the underlying population is normal. If the pop. is not normal, the c.i. is approximately correct for large samples (by CLT). We can find the sample size, n, required to obtain a c.i. with a specified margin error, m, by using the formula n = (z*/m)2 Tests of Significance Confidence interval Goal is to estimate some parameter of a population. Test of Significance Goal is to assess the evidence provided by the data in favor of some claim about the population. Motivational Example Four randomly selected students do a 20 hour SAT prep course at a special school. After the course, they write the SAT. Their scores turn out to be 560, 600, 590, 490. We find xbar = 560. Scores of students who take the course are known to be normally distributed with a standard deviation of 50. National SAT test scores are normally distibuted with a mean of 500 and a standard deviation of 50. Does the sample data provide strong support for the school’s claim that the prep course is effective in increasing SAT scores? Motivational Example (cont.) Let’s use what we know about the behavior of xbar to assess this claim. In particular, we ask What is the probability of observing a sample mean of 560 or larger if the population mean score for those who took the course is 500? (i.e., course doesn’t help on average) Motivational Example (cont.) What if the sample mean had been xbar = 700? What if the sample mean were xbar = 520? Formalization of our example: First, state hypotheses: H0: µ = 500 (course has no effect). Ha: µ > 500 (course increases mean score). Terminology H0 is the null hypothesis. Ha is the alternative hypothesis. Usually, Ha is the claim we hope to establish. (Equivalently, H0 is the claim we want to falsify). In our example, the more the sample mean exceeds 500, the more evidence we have in favor of Ha ( i.e. against H0 ). Formalization of our example: (cont.) Second: We compute the test statistic z = (xbar – mu0)/(sigma/sqrt(n)) = (560 -500)/(50/sqrt(4)) = 2.4 Third: Assess the strength of the evidence against the null hypothesis by computing a pvalue. p-value = Prob (z >=2.4) = 0.0082 (The p-value is obtained from the z value that results from the specific form of our hypotheses.) Fourth: state a conclusion. “strong” evidence in favor of company’s claim. Other possible forms of H0 and Ha in our example Suppose that I suspect that the school has a “reverse” effect and actually decreases average SAT performance. How would I set up hypotheses in such a way that a low average test score will support my suspicions? What if I suspect that the course has some effect on SAT scores, but I’m not sure whether it increases or decreases the average score. How would I set up appropriate “two sided” hypotheses for this situation? Either a low or a high average test score will support my suspicions. General form of the z test Example SSHA is a psychological test that measures motivation, attitude towards school, and study habits of students. Scores range from 0 to 200. The mean score for US college students is about 115 with a std. dev of about 30. A teacher who suspects that older students have better attitudes towards school gives the SSHA to 20 students who are at least 30 years of age. Their mean score is 135.2 (a) (b) (c) State appropriate null and alternative hypotheses. Report the p-value of your test and state your conclusion clearly. Your test required 2 important assumptions in addition to the assumption that sigma = 30. What are they? Which is more important? Another Example A study of the pay of corporate CEOs examined the increase in cash compensation for the CEOs of 104 companies, adjusted for inflation, in a recent year. The average inflation adjusted increase in the sample was xbar = 6.9% with a sample standard deviation of s = 55%. Is this good evidence that the mean real compensation increased in the past year? Because the sample size is large, s is close to the population sigma, so it is reasonable to assume that σ =55%. P-values and statistical significance Example –statistical significance The mean yield of corn in the US is about 120 bushels per acre. A survey of 40 farmers this year gives a sample mean of xbar = 123.8 bushels per acre. We want to know if this provides good evidence that the national mean this year is not 120 bushels per acre. Assume that the farmers surveyed constitute a SRS from the population of all commercial corn growers and that the population has a std dev of sigma = 10 bushels per acre. (a) Set up the appropriate hypotheses. Give the p-value for the test. Is the result significant at the 5% level? (b) Are you convinced that the population mean is not 120 bushels per acre? (c) Is your conclusion correct if the distribution of corn yields is somewhat non-normal? Why? Another Example A computer has random number generator that generates random numbers uniformly distributed between 0 and 1. If this is true, the numbers generated come from a population with µ = 0.5 and = .2887. A command to generate 100 random numbers gives an average of 0.4365. Is the generator working properly? Another Example A union leader claims that the average school teacher makes less than $40,000 per year. A random sample of 400 school teachers finds a sample mean of xbar = $39,650. The standard deviation in school teachers incomes is known to be $5,000. Assess the union leader’s claim. Relationship between confidence intervals and two sided tests A level 2-sided significance test rejects the hypothesis H0: µ = µ0 and accepts the alternative Ha: µ ≠ µ0 precisely when the value µ0 lies outside a level 1- confidence interval for µ. Example (confidence interval and 2 sided hypothesis test) Diameters of a certain machine part are normally distributed with a standard deviation of 0.1 mm. A random sample of 25 parts yields an average diameter of 11.9 mm. Find a 99% c.i. for the true mean diameter. Based on your sample, can you conclude, at the 1% level of significance, that the true mean diameter differs from 12 mm? Comments about hypothesis testing P-values tell us more than setting, in advance, a fixed level of significance. Statistical significance is not necessarily the same as practical significance. Statistical testing is not always valid –e.g. faulty data; bias in questionnaires etc. Two types of error Example – If there are more than 32,000 trees on a plot of land it will be economical to log the plot. I sample the plot and get the following 95% c.i. for the total number of trees 32,500 +/- 3,500 Should I log the lot? Type I and Type II errors