Researcher at the University of South Florida conducted a study of drug usage of

Question

Researcher at the University of South Florida conducted a study of drug usage of usage of U.S. physicians. The anonymous survey of 5, 430 randomly selected physicians revealed that experienced substance abuse or drug dependency in their lifetime. Test the hypothesis the more than 5% of U.S. physicians have used or depended on drugs in their lifetime at = 0.55. Test statistic z = p - p_0/squareroot p_0 middot (1 - p_0)/n and Confidence interval p cap plusminus z_ /z squareroot pq/n As a professional courtesy, physicians have traditionally provided health care free of charge or at a reduced rate to other physicians and their families, in 1986, 94% of a sample of 1, 000 physicians offered this professional courtesy. To assess the extent to which this practice has changed over the years, The New England Journal of Medicine conducted a survey of 2, 224 physicians of which 1, 957 currently offer free or reduced-rate health care to fellow physicians. Test statistic z = p_1 - p_2/squareroot p cap Q (1/n_1 + 1/n_2) with p cap = x_1 + x_2/n_1 + n_2 In a study of drug usage, researchers surveyed the type of drug list injected by 102 subjects with results as follows: Heroin 42 Speed 36 Other 24 At alpha = 05, test that the probabilities for the three groups were equal. (1/3, 1/3, 1/3). Chi Squared Formula: X^2 = Sigma^n_i=1 (_i - E_i)^z/E_i

Explanation / Answer

3) Ho : p = 0.05

Ha: p > 0.05

b) p^ = X/n = 425/5430 =0.07826887

TS = (p^ - p)/sqrt(pq/n) = (0.07826887 - 0.05)/sqrt(0.05*0.95/5430) = 9.55787967

z-critical for 5% level of significance is 1.645

since

TS > critical value

we reject the null hypothesis

c) 95 % confidence interval

z-critical for two-sided 95% confidence is 1.96

(0.07826887-1.96*sqrt(0.07826887*(1-0.07826887)/5430) , (0.07826887+1.96*sqrt(0.07826887*(1-0.07826887)/5430)

=(0.0711246,0.085413057)

d) we are 95 % confident are true proportion lie in this confidence interval

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