Question

I need help with the proportion. My teacher wants us to test it and I'm not sure I'm doing it right or if she meant something different. I got 13/30(100) to get 43% for the female proportion but i don't know what she means by test???

recent extensive study of the hospital indicated that thell avelage standard deviation of $6500. The data was obtained from public medical records. You are to prepare an addiuisa l of the report that will be part of a presentation to the hospital board. 254 DIED (1-died, O-lived) 26 63) Determine the 99% confidence interval for the population proportion of individuals that died during their hospital stay. Would you expect 5 patients out of 50 that are admitted to WeCare Hospital with a primary diagnosis of acute myocardial infarction to die? . にPI 3o GENDER Studies indicate the proportion of males and the proportion of females that seek treatment for heart attacks are not the same. The hospital is interested in testing to see if the proportion of females and males admitted is the same. Test to see if the proportion of females is different than 50%. ° State whether or not the results of this study support the national studies indicating the proportions are not the ° same. GENDER Vs DIED · Determine the 90% confidence interval for the population proportion of female patients that died during their . Determine the 90% confidence interval for the population proportion of males that died during their hospital . Compare the estimated proportions of females to males that died during their hospital stay when the primary hospital stay. stay. diagnosis for admittance in the hospital is acute myocardial infarction. GE Determine the 95% confidence interval for the population age of patients that were admitted to the hospita with a primary diagnosis of acute myocardial infarction. Determine whether or not a 60 year old patient would be expected to be admitted to WeCare with a prima diagnosis of acute myocardial infarction. . 1 000-00 |23233222223 1222123212123221111 2222222222222|222222222222 366726431 38 38 ao o 5680 92440702 4 757678777667759376786667 674585 13649 4167 15 5 11 16 8②9-8816 5 9-3 8 7 9プ61 11 78-150 | 1 6 9 7 1 0 2 5 4 2 5 2 8 09|1|5 48.6-s-2-85 3790,6 55-739 E 536 -3 9 | 2 3 0 4 9004|168 49375 77180 140-6327119 8-5-9-1-5 71383 2 100 | | | 2 8011 Ricard, Kristen A Math 203 Data Quiz 3 Data Set 5 DIED CHARGES AGE LOS 121 46 M 4916.5 122 76 M 8646.97 122 121 122 123 13830.1 8200 61 M 7491.5 8790.14 10130.92 18745.5 10519.2 14698 74 M 60 M 122 2 10188.29 58 M 10003.05 17844 2 72 M 64 F 74 F 80 M 67 F 60 M 62 M 123 8191 23505 10883.2 7968 1851.5 4 7 丨1a.ie)

Answer 1

You have to do a t-test to see if the observed proportion of 43% (which you have also calculated) is statistically any different than 50%. So your Null hypothesis will be Proportion(p) = 0.5 and alternative will be Proportion(p) $\neq$ 0.5.