Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
HSA 523 Homework #5 Answer Key Dr. Robert Jantzen Economics Department 1. A. The occupancy rate that places a hospital in the top 10% (at the 90th percentile) is 77.84% (see Excel calculation below). B. If St.Elsewhere is at 50%, it is .215 standard deviations below average (see Z score of -.215 from Excel calculation below). 41.5% of hospitals have rates below it. Common Data Mean Standard Deviation 54 18.6 Find X and Z Given Cum. Pctage. Cumulative Percentage 90.00% Z Value 1.281552 X Value 77.83687 Probability for X <= X Value 50 Z Value -0.215054 P(X<=50) 0.4148627 2. Given mean of 5 and standard deviation of .05 mg., probability that the content will be less than 4.9 is .02275, so 2.28% of the capsules will contain < 4.9 mgs. The probability that the content will be > 5.2 is 0 so none of the capsules will be > 5.2 mgs. Common Data Mean Standard Deviation 5 0.05 Probability for X <= X Value 4.9 Z Value -2 P(X<=4.9) 0.0227501 Probability for X > X Value 5.2 Z Value 4 P(X>5.2) 0.0000 3. a. The probability that the duration is between 250 and 300 days is .8246, so 82.46% of the pregnancies will be there. b. The top 5% duration (95th percentile) is 292.32 days and the top 1% duration th (99 percentile) is 303.22 days. c. The probability that the duration will be > 275 days is .2869 so only 28.69% of pregnancies will be covered. Common Data Mean Standard Deviation 266 16 Probability for X > X Value 275 Z Value 0.5625 P(X>275) 0.2869 Probability for a Range From X Value 250 To X Value 300 Z Value for 250 -1 Z Value for 300 2.125 P(X<=250) 0.1587 P(X<=300) 0.9832 P(250<=X<=300) 0.8246 Find X and Z Given Cum. Pctage. Cumulative Percentage 95.00% Z Value 1.644853 X Value 292.3177 Find X and Z Given Cum. Pctage. Cumulative Percentage 99.00% Z Value 2.326347 X Value 303.2216 4.a. To test whether a set of numbers are normally distributed you can assess whether the #s conform to the theoretical properties of a normal distribution which state that: Bell shaped symmetrical distribution Mean = median Interquartile Range = 1.33 times the Standard Deviation (SD) #s follow the 68 – 95 – 99.7 rule, i.e., 68% are within 1 SD of the mean, 95% are within 2 SDs, and 99.7% are within 3 SDs. An alternative test is to generate a normal probability plot. If the plot is a straight line, the #s are normally distributed. 4.b. (generate the statistics w/ SPSS by using <Analyze><DescriptiveStats> <Frequencies> or <Analyze><DescriptiveStats><<Explore> Statistics # of staffed services N Valid Missing Descriptives Statistic 23.3400 50 0 Mean 23.3400 Median 23.0000 Std. Deviation 6.53596 # of staffed services Mean 95% Confidence Interval for Mean Lower Bound Upper Bound 21.4825 25.1975 Minimum 11.00 5% Trimmed Mean Maximum 23.3556 36.00 Median 23.0000 Percentiles 25 19.0000 Variance 50 23.0000 Std. Deviation 75 28.0000 Minimum 11.00 Maximum 36.00 Range 25.00 Interquartile Range Std. Error .92432 42.719 6.53596 9.00 A boxplot shows the #s clustered around the middle and symmetrically distributed. The mean is about equal to the median The pearson skewness measure is (23.34-23)/6.54 = .052 which is indicative of a symmetrical distribution. The IQR is 9 while 1.33 * the SD is 8.7 The mean +- 1 SD = 16.8 thru 29.9 captures 62% (31 of 50) of the #s (versus 68%). The mean +- 2 SDs = 10.3 thru 36.4 captures 94% (49 of 50) of the #s (versus 95%). The mean +- 3 SDs = 3.75 thru 42.9 captures 100% of the #s (versus 99.7%) Hence it appears the #s have close to a normal distribution. The normal probability plot plots the actual z values for each # against the expected z values. It’s a pretty straight line indicating a normal distribution. # of staffed services 35.00 30.00 25.00 20.00 15.00 Normal Probability Plot of # of staffed services 4 Expected Normal Value 2 0 -2 -4 -4 -2 0 Standardized Observed Value 2 4 5. Given a failure rate of .2 for the hip replacement parts and a sample size of 10. Binomial Probabilities: This spreadsheet (adapted from PHSTAT2) makes some common calculations for outcomes that can be modeled with the binomial distribution. NOTES: Edit the values in BLUE to reflect the sample size and probability of success in your sample. (Note: the sample size must be <= 20). The spreadsheet will then calculate the probability of observing exactly X successes in the sample, as well as <= X successes, > X successes, etc. Data Sample size Probability of success 10 0.2 The probability that 0 of 10 fail is .107374 The probability that >2 fail is .3222 The probability that all 10 fail is 1.02 E-07 = .000000102 Statistics Mean Variance Standard deviation 2 1.6 1.264911 Binomial Probabilities Table X 0 1 2 3 4 5 6 7 8 9 10 P(X) 0.107374 0.268435 0.30199 0.201327 0.08808 0.026424 0.005505 0.000786 7.37E-05 4.1E-06 1.02E-07 P(<=X) 0.107374 0.37581 0.6778 0.879126 0.967207 0.993631 0.999136 0.999922 0.999996 1 1 P(<X) 0 0.107374 0.37581 0.6778 0.879126 0.967207 0.993631 0.999136 0.999922 0.999996 1 P(>X) 0.892626 0.62419 0.3222 0.120874 0.032793 0.006369 0.000864 7.79E-05 4.2E-06 1.02E-07 0 P(>=X) 1 0.892626 0.62419 0.3222 0.120874 0.032793 0.006369 0.000864 7.79E-05 4.2E-06 1.02E-07 Note: binomial model can be used when: Drawing a random sample for a binary outcome (yes/no, works/doesn’t, lives/dies, etc.) The probability of “success” remains the same across sampled observations. This generally holds true if the sample size is <= .05 of the population size, or if sampling w/ replacement. The binomial model calculates the probability of observing any number (X) of “successes” within the drawn sample.