HW 10 Solution

HW 10 Solution - 5. use t test as above 6. state rejection...

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HW 10 Solution 1. load datahw10.dat wc = datahw10; n = length(wc); % Part I ybar = sum(wc)/n; sy = 0; for i = 1:n sy = sy + (wc(i)-ybar)^2/(n-1); end sy = sqrt(sy); 2. % Part II: confidence intervals a = [0.1 0.05 0.01]; d = sy/sqrt(n)*tinv(1-a/2,n-1); L = ybar - d; U = ybar + d; 3. % Part III: test of hypothesis mu0 = 69; a = 0.05; % 95% confidence level d = sy/sqrt(n)*tinv(1-a/2,n-1); % same as part II L = mu0 - d; U = mu0 + d; % Rejection region % ybar does not fall within this region, so H0 is not supported 1. length of data set, mean and standard deviation computed in part one 2. null hypothesis: mu0 = 69% 3. alternate hypothesis: mu0 ~= 69% 4. 95% confidence interval computed as above

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Unformatted text preview: 5. use t test as above 6. state rejection region for the statistic as above 7. does ybar fall within L and U? 8. no, so reject statistic 4. % Part IV: Type I error sig = 2; % Assume true variance is 2% zL = (L-mu0)/(sig/sqrt(n)); zU = (U-mu0)/(sig/sqrt(n)); % This one matches z about right on the one-sided Matlab t-table x = 0.9862; tinv(x,Inf) 2*(1-x) % So probability of a Type I error is 2.76% 5. % Part V: Type II error mu = 70.2; % Assume this is actual mean and keep sig = 2 zL = (L-mu)/(sig/sqrt(n)); zU = (U-mu)/(sig/sqrt(n)); % This one matches zU pretty well tinv(0.316,Inf) % P(z&lt;zU) = 31.6% tinv(1e-6,Inf) % P(z &lt; zL) ~ 0...
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HW 10 Solution - 5. use t test as above 6. state rejection...

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