Question

A hospital director is told that 43% of the emergency room visitors are uninsured. The director wants to test the claim that the percentage of uninsured patients is over the expected percentage. A sample of 310 patients found that 155 were uninsured. At the 0.05 level, is there enough evidence to support the director's claim? . Step 4 of 7: Determine the P-value of the test statistic. Round your answer to four decimal places.

Answer 1

A hospital hospital director is told that 43.6 are Of visitors the Yoom emergency uninsured p=0-43 given claim: Percentage of uninsured patients over the expected expected percentage is As mull and alternative hypothesis each other- States always opposite to and either null or alternative hypothesis represents the claim. Ho: P = 0.43 so Null hypothesis Alternative hypothesis Ha? P>0-43 test statistic Z- 7 -P Pli n X=155, 7-310 P Х 155 0.5 310 0.5-0-43 a 0-4 (1-0-43) CS Scanned with CamScanner 310

score of 2.49 Z-2.49 P-value; be calculated by using can alternative hypot thesis a Z with distribution or claim condition P-value Plz > 2 2.49) 1- P(Z <2.49 Z =1 left tail area of from E-tables with Z with Z row starting 2.4 and the column of o.og Value P(Z 72,47 ife 1-0-99361 1 Z=2.49 P value = 0.00639 ď then decision If p-value is less than H. Otheywise not null hypothesis reject the P-value -0.00634 22 (0.05) null So There is Reject the hypothesis (1) Conclusion a sufficient evidence to support them claime, that percentage of unínsured patients is over Scanned with CS expected percentage