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Chapter 7 Inference for Distributions 1 • The previous chapter emphasized the reasoning of tests and confidence intervals • Now we emphasize statistical practice – we no longer assume that population standard deviations are known 2 Section 7.1 Inference for the Mean of a Population 3 Introduction • So far, in all our inference for the population mean , we have assumed that we know . • The sampling distribution of x-bar depends on – Thus, both CI’s and significance tests depend on x 0 x (z*) , z n / n • But usually we don’t know . Then what? – A sensible idea would be to use s, the sample standard deviation, as an estimate to , the population standard deviation. 4 Introduction • Even though we are primarily interested in , we are now forced to first estimate • We know that s changes from sample to sample. – So we are adding some variability into our equations. 5 Introduction • When is known, n is the standard deviation of the sampling distribution of x-bar. • When is not known, it is estimated with the sample standard deviation s – Then we must estimate the standard deviation of x-bar by: – The standard error of the sample mean: 6 Introduction • Now, we also know that if x-bar is normal then x z ~ N (0,1) n s • But if we use the standard error, , does n x ? ~ N (0,1) s n 7 Introduction • Unfortunately it doesn’t. But it does follow a distribution called the T-distribution. x t ~ Tn-1 s/ n • Where n-1 is the degrees of freedom • From now on df = degrees of freedom – Notice then that for each sample size there is a different T-distribution. • The degrees of freedom from this t statistic come from the sample standard deviation s in the denominator of the t statistic 8 History of the T distribution • The T distributions were discovered in 1908 by William S. Gosset, a statistician working for the Guinness brewing company • He published under the pen name “Student” because Guinness didn’t want competitors to know that they were gaining an industrial advantage from employing statisticians 9 Brilliant! Properties of the T-distributions • SIMILAR (but not the same) to a Normal dist. – Symmetric – Mean = 0 – Bell shaped • **The spread of the tk distribution is a bit greater than that of the standard Normal distribution – This is due to the extra variability caused by substituting the random variable s for the fixed parameter • As the degrees of freedom k increase, the tk density curve approaches the N(0,1) curve – This reflects the fact that s approaches as the sample size increases (Law of Large Numbers!!) 10 Properties of the T-distributions • The t distribution has more probability in the tails and less in the center than does the standard Normal distribution • Table D in the back of the book gives critical values for the t distributions – For convenience, the table entries have also been labeled by the confidence level C (in percent) required for confidence intervals 11 D Tea-table, anyone? – Notice that it is not nearly as comprehensive as the standard normal table – This makes getting p-values a little more difficult, but not much. 12 The One-Sample t Confidence Interval • How does using s affect confidence intervals for the mean ? • The one-sample t confidence interval is similar in both reasoning and computational detail to the z confidence interval of Chapter 6 – Note: “One-sample” does NOT mean that we have a sample size of n=1 13 The One-Sample t Confidence Interval • Suppose that an SRS of size n is drawn from a population with an unknown mean and standard deviation . A level C confidence interval for is: – Where t* is the value for the Tn-1 density curve with area C between -t* and t*. – The margin of error is 14 One-sample t confidence interval The area between the critical values –t* and t* under the Tn-1 curve is C. Tn-1 Curve Prob=(1-C)/2 Prob=(1-C)/2 15 -t* 0 +t* Example 1 • The following data are the amounts of vitamin C, measured in milligrams per 100 grams of blend, for a random sample of size 8 from a production run: 26 31 23 22 11 22 14 31 • We want to find a 95% confidence interval for , the mean vitamin C content. 16 n=8 x-bar = 22.5 s = 7.19 s 7.19 SE x 2.54 n 8 From Table D we find that for 95% CI t*7 = 2.365 s 7.19 x t * 22.5 2.365 (16.5, 28.5) n 8 We are 95% confident that the mean vitamin C content for this production run is between 16.5 and 28.5 mg/100g. 17 • In this example we have given the actual interval (16.5, 28.5) as our answer • Sometimes, we prefer to report the mean and margin of error: The mean vitamin C content is 22.5 mg/100g with a margin of error of 6.0 mg/100g. 18 Testing • In tests of significance, as in confidence intervals, we allow for unknown by using s and replacing z by t • Let n be the sample size x – If σ is known and n is large then z ~ N (0, 1) n – If σ is NOT known then x t ~ Tn 1 s n 19 The one-sample t test • Suppose that a SRS of size n is drawn from a population having unknown mean • To test the hypothesis H0 : = 0 based on a SRS of size n, compute the one-sample t statistic: x 0 t s n 20 The one-sample t test • The P-value for a test of H0 against – Ha : > 0 is P(Tn-1 t) – Ha : < 0 is P(Tn-1 t) – Ha : 0 is 2P(Tn-1 |t|) • These P-values are exact if the population distribution (X) is Normal and are approximately correct for large n in other cases 21 …Example 1 • Recall that n = 8, x-bar = 22.50, and s = 7.19 • Suppose we want to test whether the mean vitamin C content in the final product is 40 • Hypotheses: H0: µ = 40 Ha: µ ≠ 40 22 …Example 1 • Test statistic: x 0 22 .5 40 t 6.88 s n 7.2 8 • P-value: 2P T7 6.88 2P T7 5.408 2P T7 5.408 2(.0005) 0.001 P value 0.001 • Decision and Conclusion: – We reject H0 – There is significant evidence to conclude that the mean vitamin C content for this run is not equal to 40. 23 Two-sided alternative and CI’s • We can use (1-)% Confidence intervals to test a two-sided alternative alternative hypothesis at the significance level – If the confidence contains the null mean (the mean that is assumed to be true) then this is a plausible value and fail to reject H0. 24

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posted: | 1/17/2013 |

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