Hypothesis Testing and CI using t-distribution 6/8/11 Lecture 20 1Recall (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) 6/8/11 Lecture 20 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. 6/8/11 Lecture 20 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. 6/8/11 Lecture 20 4].8.469 ,2.452[500/10096.1461/* nzx • The hypotheses are H0 : = 450. Ha : 450. • A 95% confidence interval for is • The value 450 does not lie in the above interval • At 5% significance level we reject the null hypothesis. 6/8/11 Lecture 20 5 Problem of interest: ◦ Population mean of a normal distribution ◦ known Z-confidence interval: Z-test: Ha p-value 0 2 P(Z |z| ) > 0 P(Z z) < 0 P(Z z) nzxnzx2/*nxz/0What if is unknown? 6/8/11 Lecture 20 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: Construct a 95% CI for the mean return. .18.8 ,75.14 sx6/8/11 Lecture 20 7 For CI and hypothesis testing about a normal mean , when is not known, the sample standard deviation s is used to estimate . Recall: where is assumed to be known. When is the mean of a random sample of size n from a normal distribution with mean , then has a Student t distribution with n - 1 degrees of freedom (df). XnsXt/nXz/6/8/11 Lecture 20 8• The t distribution is also bell-shaped, and symmetric about zero. • 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. = ). 0 d.f. = n2 d.f. = n1 n1 < n2 Student t distribution 6/8/11 Lecture 20 9 Degrees of Freedom1 3.078 6.314 12.706 31.821 63.6572 1.886 2.92 4.303 6.965 9.925. . . . . .. . . . . .50 1.299 1.676 2.009 2.403 2.678. . . . ... . . . ..200 1.286 1.653 1.972 2.345 2.6011.282 1.645 1.96 2.326 2.576tA, n-1 t.100 t.05 t.025 t.01 t.005 A Table D: t distribution critical values = .05 6/8/11 Lecture 20 10 A 100(1 - )% confidence interval for is or, more compactly, ] ,[1,2/1,2/nstxnstxnn./1,2/nstxn6/8/11 Lecture 20 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: n=50. Construct a 95% CI for the mean return: ,18.8 ,75.14 sx 07.17,43.125018.8009.275.141,2nstxn6/8/11 Lecture 20 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 t-distribution ? SD known or unknown ? Lecture 22: more about t-tests … Exercises: (don’t turn in) 7.15, 7.21, 7.24 (b) (c) 6/8/11 Lecture 20
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