Download Comparison of Means and Proportions: One-Way Analysis of Variance and Hypothesis Testing and more Exams Biostatistics in PDF only on Docsity! STAT 541 DISCUSSION 9 TA: Lane Burgette Office: 1245F MSC, 1300 Universtiy Avenue E-mail: burgette@stat.wisc.edu URL: www.stat.wisc.edu/˜burgette/541.html or naviagate from stat.wisc.edu Office Hours: 9:30-10:30 T, R 1 Comparison of means Suppose we take ni samples from k different normal distributions with mean µ1, µ2,...,µk. (Assume that these k distributions have the same variance). 1. One-way Analysis of Variance • H0 : µ1 = µ2 = ... = µk HA : at least two of the means differ • Test statistic: F = S2B S2W here, s2B = n1(x̄1 − x̄) 2 + n2(x̄2 − x̄) 2 + ... + nk(x̄k − x̄) 2 k − 1 s2W = (n1 − 1)s 2 1 + (n2 − 1)s 2 2 + ... + (nk − 1)s 2 k n − k Under the null, F has F distribution with degrees of freedom (k − 1, n − k). k − 1 is the degree of freedom for the numerator and n − k is the degree of freedom for the denominator. • critical value: Fk−1,n−k,α (Table A.5 gives the critical values with different degrees of freedom and α). If F > Fk−1,n−k,α, then reject the null. 2. Multiple Comparisons Prodedures • H0 : µ1 = µ2 = ... = µk HA : at least two of the means differ • Here we do the (n 2 ) t-tests, which are, for different i and j, H ij0 : µi = µj H ij A : µi 6= µj • Test statistic: Tij = X̄i − X̄j √ s2W (1/ni + 1/nj) Under H ij0 , Tij has the t-distribution with n − k degrees of freedom. • level of significance for each test: α∗ = α (k 2 ) • This is called the Bonferroni correction, and it tends to be rather conservative, especially when k is large. 1 2 Hypothesis Testing on Proportions 1. One-sample H0 : p = p0 HA : p 6= p0 Test statistic: Z = p̂ − p0 √ p0(1−p0) n Under the null, Z is approximately distributed as a standard normal. (p̂ is just what you would guess it should be: the number of successes divided by the number of trials.) An approximate 100 × (1 − α)% confidence interval for p is ( p̂ − zα/2 √ p̂(1 − p̂) n , p̂ + zα/2 √ p̂(1 − p̂) n ) 2. Two-sample H0 : p1 = p2 HA : p1 6= p2 The test statistic is: Z = p̂1 − p̂2 √ p̂(1 − p̂)[1/n1 + 1/n2] here p̂ = n1p̂1+n2p̂2n1+n2 . Under the null Z has an approximate standard normal distribution. 3 Examples 1. A local government agency wishes to investigate the prevailing rate of unemployment. It reasons correctly that this assessment could be made accurately and efficiently by sampling a small fraction of the labor force. Among 500 randomly selected persons interviewed, 41 are found to be unemployed. Compute a 95% confidence interval for the rate of unemployment. 2. Soccer has become a popular sport, especially among grade school children. Eighty second- grade boys and seventy-five second-grade girls were asked whether they play on a youth soccer team. Among the boys 34 play in some youth team while 41 don’t not. And 59 girls play in soccer team while 21 do not. Determine whether there are different participation rates for second grade boys and girls. (a) Formulate the null and alternative hypotheses. (b) Test the null hypothesis at level α = 0.05 2
