version 2 Questions 16 18 It is desired to estimate the mean length of stay in t
ID: 1126456 • Letter: V
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version 2 Questions 16 18 It is desired to estimate the mean length of stay in the intensive care unit (ICU) after a surgery measured as the number of nights. Following is a summary of data collected from 100 patients Number of nights in ICU Frequency 50 30 Usethe eapproximations: 103500.59, v044~0.66, v032~0.72, v ~0.78, and vmeo Q16 Find the minimum variance unbiased estimator for the mean length of stay in ICU. these approximations Va35 0.59, 0.440.66, V0.52 0.72, 0.6i 0.78, and v0.72 0.85. (e) 2 Q17 Find the confidence interval with 95% confidence level. (a) 11.37295, 1.92843 (b) [1.47064, 1.72936 de 11.54712, 1.852881 (d) 11.25272, 1.98351] (e) [1.15491, 2.01745 Q18 For a more accurate analysis, the margin of error needs to be smaller than 0.1 with 95% confidence level. What is the required sample size? (a) 234 (c) 168 151Explanation / Answer
16. Option d 1.7
Mean = sum of observations/total number of observations= (1*50 + 2*30 + 3*20)/100 = 1.70
17. Option c [correctly marked]
Variance, S2 = [50*(1-1.7)^2 + 30*(2-1.7)^2 + 20*(3-1.7)^2]/100 = 0.61
S = 0.61^0.5 = 0.7810
Standard error, SE = S/n^0.5 = 0.7810/10 = 0.0781
t_critical = 1.984( see t-table at 99 degree of freedom)
Confidence interval = (1.7 - 1.96*0.0781, 1.7 + 1.96*0.0781) = (1.54,1.85)
18. Option a 234
ME = t*s/n^0.50
0.1 = 1.96*0.7810/n^0.5
solving for n will give n= 234
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