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The American Medical Association conducted a study of how long patients who are admitted to hospitals from emergency rooms tend to stay in the hospital.

The American Medical Association conducted a study of how long patients who are admitted to hospitals from emergency rooms tend to stay in the hospital. The dependent variable in this analysis is length of stay (in days). The independent variables are patient age (in years), patient gender, (where males are coded as 0 and females are coded as 1), and number of medicines a patient is currently prescribed to take. The relevant multiple regression output is as follows: Regression Statistics Multiple R 0.9659 R Square 0.9331 Adjusted R Square 0.9294 Standard Error Observations 0.0327 58 ANOVA SS MS F Significance F Df Regression Residual 0.8094 0.2698 0.0580 0.0010 251.1995 3 54 0.8675 Total 57 Standard Error Coefficients t Stat P-value 0.0000 Intercept Age -0.0045 Gender -0.2392 prescriptions 1. a. b. c. d. e. 0.0136 37.4675 0.0000 0.0034 -1.3276 0.1898 0.0123 -19.3942 0.0000 0.0123 -0.0214 0. 9829 0.5118 -0.0002 Is this model statistically significant? How do we know? Yes, the ANOVA-F-test value is significant at p<.01 Yes, the degrees of freedom is >50 No, the Multiple R value is less than .99 No, the ANOVA-F-test value is zero No, the sample size is less than 120 2. What is the value of the appropriate coefficient of determination for this model? Interpret the meaning of this value in terms of this problem. a. 96.59% of the variation in length of stay in days is explained by the dependent variables in the model. b. 93.31% of the variation in length of stay in days is explained by the dependent variables in the model. c. 92.94% of the variation in length of stay in days is explained by the dependent variables in the model. d. 3.27% of the variation in length of stay in days is explained by the dependent variables in the model. 3. In terms of this problem, what does the constant 0.5118 tell you? a. b. c. d. That female adult patients currently on 2 prescription drugs will stay on average for half a day (0.5118 days). That male adult patients currently on 2 prescription drugs will stay on average for fifty-one days (0.5118 *100 days). That female babies below 1 year of age not currently on any prescription drugs will stay on average for fifty-one days (0.5118 *100 days). That male babies below 1 year of age not currently on any prescription drugs will stay on average for half a day (0.5118 days) 4. In terms of this problem, interpret the meaning of the coefficient -0.0045. a) b) c) d) For each additional prescription drug the patient the patient will stay on average .0045 days less. For each additional prescription drug the patient the patient will stay on average .0045 days longer. For each additional year in age the patient the patient will stay on average .0045 days less. For each additional year in age the patient the patient will stay on average .0045 days longer. 5. In terms of this problem, interpret the meaning of the coefficient -0.2392. a) b) c) d) For each additional prescription drug the patient the patient will stay on average .2392 days less. For each additional prescription drug the patient the patient will stay on average .2392 days longer. Women tend to stay on average .2392 days longer than men. Men tend to stay on average .2392 days longer than women. 6. Is the effect of age on length of stay significant at the 0.05 level? How do we know? No, because the p-value of 0.1898 is greater than .05 No, because the p-value of the ANOVA F-Test is less than .05 Yes, because the p-value of 0.1898 is greater than .05 Yes, because the p-value of the ANOVA F-Test is less than .05 7. At what level is the effect of gender on length of stay significant? a) p>.05 b) p>.1 c) p<.01 d) it is not significant (p=0.0000) 8. Predict the length of stay for a 72 old male patient who is on 6 prescriptions. a) = 0.5118-0.0045(72)-0.2392(1)-0.0002(6)=0.0526 days b) = 0.5118-0.0045(72)-0.2392(0)-0.0002(6)=0.1866 days c) = 0.5118-0.0045(6)-0.2392*(0)-0.0002(72)=0.4704 days d) = 0.5118-0.0045(6)-0.2392(1)-0.0002(72)= 0.2312 days

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