Question

The critical value of z for 95% level of confidence is +/- 1.96 the picture of a spreadsheet contains the proportions of Medicare cases in 29 southwest Pennsylvania hospitals.

A. How would you characterize children’s hospital?

B. What should we do with Children’s If we are interested in calculating the mean?

C. The mean proportion of Medicare cases is including children’s is 0.449. Without children’s it would be .465. The standard deviation is 0.131. Use a z test (two tails) to determine if children’s hospitals Medicare case los is significantly different from the regions mean at the 95% level of confidence. Decal that observed z is (observation mean) /standard deviation/square root of n. Why should we have anticipated this?

D. Use a z test to determine if Jefferson regional hospitals Medicare case load is significantly different from the regions mean at the 95% level of confidence.

Calibri / Format Pa K21 ital Medicare 2 Children'sHospPgh/UPM 0.031 0.449(0.42)0.18 3 MageeWomens/UPMC 0.153 0.449(0.30) 0.09 4 AdvancedSurgical 0.223 0.449(0.23) 0.05 5 WesternPennsylvania0.293 0.449 (0.16) 0.02 6 UPMCPresbyShadyside 0.377 0.449 (0.07) 0.01 7 AlleghenyGeneral 8 UPMCPassavant 9 HeritageValleySewickle 0.419 0.449 (0.03) 0.00 10 Washington 11 UPMCMercy 12 LatrobeArea 13 Highlands 14 ButlerMemorial 15 UPMCStMargaret 16 ACMH 17 Frick 18 ExcelaHithWestmorelan 0.496 0.449 0.05 0.00 19 HeritageValleyBeaver 0.508 0.4490.060.00 20 StClairMemorial 21 WesternPAHosp/Forbes 0.528 0.4490.08 0.01 22 SouthwestRegionalMC 0.547 0.449 0.10 0.01 23 Uniontown 24 CanonsburgGeneral 25 MonongahelaValley 26 OhioValleyGeneral 27 UPMCMcKeesport 28 Alle-Kiski 29 JeffersonRegional 30 Mean 0.403 0.449 (0.05) 0.00 0.415 0.449 (0.03) 0.00 0.433 0.449 (0.02) 0.00 0.4430.449(0.01)0.00 0.452 0.4490.00 0.00 0.456 0.449 0.01 456 0449 0.4590.4490.01 0.476 0.4490.03 0.00 0.479 0.449 0.03 0.486 0.4490.04 0.00 0.00 0.00 0.524 0.4490.070.01 0.5500.4490.10 0.01 0.5520.449 0.10 0.01 0.566 0.4490.12 0.01 0.575 0.449 0.130.02 0.576 0.4490.13 0.02 0.5800.4490.130.02 0.5810.4490.13 0.02 0.449 0.01 0.48 0.02 variance 32 0.130784 std dev

Answer 1

A. How would you characterize children’s hospital?
It is an outlier

B. What should we do with Children’s If we are interested in calculating the mean?
We must drop the Children hospital while calcuating the mean of the hospitals in the region.

c. C. The mean proportion of Medicare cases is including children’s is 0.449. Without children’s it would be .465. The standard deviation is 0.131. Use a z test (two tails) to determine if children’s hospitals Medicare case los is significantly different from the regions mean at the 95% level of confidence. Decal that observed z is (observation mean) /standard deviation/square root of n. Why should we have anticipated this?

Hypothesis

$\\H_0 : \mu = 0.464\\ H_1 : \mu \ne 0.464$

In this case we use a two tail test.

Hence $\alpha/2 = 0.05/2 = 0.025$

$Z = \frac{X- 0.464}{\frac{0.131}{\sqrt{28}}}= \frac{0.031- 0.464}{\frac{0.131}{\sqrt{28}}}$
Z = -17.49
Using a z table, we get the pvalue = 0

Hence since the pvalue is less than 0.025, we reject the null hypothesis and conclude that the medicare of children hospital is significantly different from the regions mean

(Decal that observed z is (observation mean) /standard deviation/square root of n. Why should we have anticipated this?) Unable to understand this.

D. Use a z test to determine if Jefferson regional hospitals Medicare case load is significantly different from the regions mean at the 95% level of confidence.

Hypothesis

$H_0 : \mu = 0.464\\ H_1 : \mu \ne 0.464$

In this case we use a two tail test.

Hence $\alpha/2 = 0.05/2 = 0.025$

$Z = \frac{X- 0.464}{\frac{0.131}{\sqrt{28}}}= \frac{0.581- 0.464}{\frac{0.131}{\sqrt{28}}}$
Z = 4.7259
Using a z table, we get the pvalue = 0

Hence since the pvalue is less than 0.025, we reject the null hypothesis and conclude that the medicare of JeffersonRegional is significantly different from the regions mean