Question

Homework: Review Questions III - 6, 7, 8, 9, 10 ITUT ZU 1:51 PI Sas Score: 0 of 1 pt 8 of 42 (37 complete) HW Score: 86.9%, 36.5 of 42 7.2.18 Question Help In the week before and the week after a holiday, there were 10,000 total deaths, and 4988 of them occurred in the week before the holiday a. Construct a 90% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday b. Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday? a.<<(Round to three decimal places as needed)

Answer 1

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   4988          
Sample Size,   n =    10000          
                  
Sample Proportion ,    p̂ = x/n =    0.4988          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.005000          
margin of error , E = Z*SE =    1.645   *   0.00500   =   0.0082
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.49880   -   0.00822   =   0.4906
Interval Upper Limit = p̂ + E =   0.49880   +   0.00822   =   0.5070
                  
90%   confidence interval is (   0.491   < p <    0.507   )

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