Hypothesis Testing and CI using t distribution Lecture 20 6 8 11 1 Recall Section 6 2 Tests for population mean Confidence Intervals Connection between the 2 procedures 382 390 Today Section 7 1 Tests for population mean when the population SD is unknown 418 428 Lecture 20 6 8 11 2 In a discussion of SAT scores someone comments Because only some students take the test the scores overestimate the ability of typical seniors The mean SAT M score is about 475 but I think if all seniors took the test the mean would be 450 You gave the test to an SRS of 500 seniors from California They had an average score of 461 The SAT M scores follow a normal distribution with a standard deviation of 100 Is there sufficient evidence against the claim that the mean for all California seniors is 450 with a significance level of 0 05 Give a 95 CI for the mean score of all seniors Lecture 20 6 8 11 3 A level 2 sided test Accepts H0 0 exactly when the value 0 falls inside a level 1 confidence interval for rejects H0 when the value 0 falls outside the CI CI can be used to test hypotheses Calculate the 1 level confidence interval then if 0 falls within the interval accept the null hypothesis Otherwise reject the null hypothesis Lecture 20 6 8 11 4 The hypotheses are H0 450 Ha 450 A 95 confidence interval for is x z n 461 1 96 100 500 452 2 469 8 The value 450 does not lie in the above interval At 5 significance level we reject the null hypothesis Lecture 20 6 8 11 5 Problem of interest Population mean of a normal distribution known Z confidence interval x z x Z test x 0 z n Ha 0 0 0 n z 2 n p value 2 P Z z P Z z P Z z What if is unknown Lecture 20 6 8 11 6 An investor is estimating the return on investment in companies that won quality awards last year A random sample of 50 companies is selected and the return on investment is calculated had he invested in them The data is summarized as follows x 14 75 s 8 18 Construct a 95 CI for the mean return Lecture 20 6 8 11 7 For CI and hypothesis testing about a normal mean when is not known the sample standard deviation s is used to estimate X z n Recall where is assumed to be known When X is the mean of a random sample of size n from a normal distribution with mean then X t s n has a Student t distribution with n 1 degrees of freedom df Lecture 20 6 8 11 8 Student t distribution The t distribution is also bell shaped and symmetric about zero d f n2 d f n1 n1 n2 0 The degrees of freedom determines how spread the distribution is Note all t curves are more spread out compared to the standard normal distribution which is equivalent to having d f Lecture 20 6 8 11 9 A 05 tA n 1 t 100 t 05 t 025 t 01 t 005 3 078 1 886 1 299 6 314 2 92 1 676 12 706 4 303 2 009 31 821 6 965 2 403 200 1 286 1 282 1 653 1 645 1 972 1 96 2 345 2 326 63 657 9 925 2 678 2 601 2 576 Degrees of Freedom 1 2 50 Table D t distribution critical values Lecture 20 6 8 11 10 A 100 1 confidence interval for is s s x t 2 n 1 x t 2 n 1 n n or more compactly x t 2 n 1 s n Lecture 20 6 8 11 11 An investor is estimating the return on investment in companies that won quality awards last year A random sample of 50 companies is selected and the return on investment is calculated had he invested in them The data is summarized as follows x 14 75 s 8 18 n 50 Construct a 95 CI for the mean return x t 2 n 1 s 8 18 14 75 2 009 12 43 17 07 n 50 Lecture 20 6 8 11 12 t distribution Table D d f Comparison to N 0 1 curve similarity difference connection Application when to use N 0 1 when to use tdistribution SD known or unknown Lecture 22 more about t tests Exercises don t turn in 7 15 7 21 7 24 b c Lecture 20 6 8 11 13
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