Question

A study investigated survival rates for in-hospital patients who suffered cardiac arrest. Among 58,593 patients who had a cardiac arrest during the day, 11,604 survived. Among 28.155 patients who suffered cardiac arrest at night, 4139 survived. Use a 0.01 significance level to test the claim that the survival rates are the same for day and night.

1. State the degrees of freedom (i.e., DF = __) or write “not applicable” if not relevant.
2. State all the requirements with respect to the problem.
3. Verify any requirement that should require a calculation. If none, write “not applicable”.
4. Identify which test statistic to use and write the associated formula using symbols.
5. Compute the value of the test statistic. If using StatCrunch, circle it.
6. State the P-value. If using StatCrunch, circle it.
7. Show and apply the P-value decision rules (with values substituted appropriately) that lead you to “reject H₀”or “fail to reject H₀”.
8. State the critical value(s). If using StatCrunch, circle it.
9. Show and apply the critical value decision rules (with values substituted appropriately) that lead you “reject H₀”or “fail to reject H₀”.
10. Write the conclusion utilizing the “correct” words using Table 8-3, p. 366. Make sure you insert the relevant portions of the claim into the conclusion.
11. Construct the associated confidence interval (CI).
12. State the CI using the correct notation.
13. State the margin of error (E = __).
14. State the point estimate using the correction notation.
15. Does the CI support the hypothesis test conclusion? Explain/interpret. This may require a sentence or two, not just a single word.
16. Extra Credit: Express the Type I error in the context of the problem using the words from the “helper” document. Make sure the conclusion is worded such that it addresses the claim (p. 368).
17. Extra Credit: Express the Type II error in the context of the problem using the words from the “helper” document. Make sure the conclusion is worded such that it addresses the claim (p. 368).
    

Answer 1

Ho:   p1 - p2 =   0      
Ha:   p1 - p2 ╪   0      

two tail
              
sample #1   ----->  day  
first sample size,     n1=   58593      
number of successes, sample 1 =     x1=   11604      
proportion success of sample 1 , p̂1=   x1/n1=   0.1980      
              
sample #2   ----->   night
second sample size,     n2 =    28155      
number of successes, sample 2 =     x2 =    4139      
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.147      
              
difference in sample proportions, p̂1 - p̂2 =     0.1980   -   0.1470   =0.0510
              
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.1815      
              
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0028      
Z-statistic = (p̂1 - p̂2)/SE = (   0.051   /   0.0028   ) =18.2609
              
z-critical value , Z* =        2.5758   [excel formula =NORMSINV(α/2)]  
p-value =        0.0000   [excel formula =2*NORMSDIST(z)]  
decision :    p-value<α,Reject null hypothesis           
................

level of significance, α =   0.01              
Z critical value =   Z α/2 =    2.576   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0027          
margin of error , E = Z*SE =    2.576   *   0.0027   =   0.0069
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    0.051   -   0.0069   =   0.0441
upper limit = (p̂1 - p̂2) + E =    0.051   +   0.0069   =   0.0579
                  
so, confidence interval is (   0.0441   < p1 - p2 <   0.0579   )  
...........................

THANKS

revert back for doubt

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