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6.3.6. Among the early attempts to revisit the death postponement theory introduced in Case Study 6.3.2 was an examination of the birth dates and death dates of three hundred forty-eight U.S. celebrities (144). It was found that sixteen of those individuals had died in the month preceding their birth month. Set up and test the appropriate H0 against a one-sided H1. Use the 0.05 level of significance.

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Answer:

So the Null hypothesis is rejected in this case

Step-by-step explanation:

  The number of celebrities is  n = 348

   

So to solve this we would assume that p is the percentage of people that died on the month preceding their birth month  

     

  Generally if there is no death postponement then p will be mathematically evaluated as

            [tex]p = \frac{1}{12}[/tex]  

This implies the probability of date in one month out of the 12 months

Now from the question we can deduce that the hypothesis we are going to be testing is  

  [tex]Null Hypothesis \ \ H_0 : p = 0.083[/tex]

This is a hypothesis is stating that a celebrity  dies in the month preceding their birth

  [tex]Alternative \ Hypothesis H_1 : p < 0.083[/tex]

   This is a hypothesis is stating that a celebrity does not die in the month preceding their birth

       is c is the represent probability for each celebrity which either c = 0 or c = 1

Where c = 0 is that the probability  that the celebrity does not die on the month preceding his/ her birth month

     and  c =  1  is that the probability  that the  celebrity dies on the month preceding his/ her birth month

  Then it implies that

   for  

       n= 1 + 2 + 3 + .... + 348  celebrities

Then the sum of c for each celebrity would be  [tex]c_s = 16[/tex]

i.e The number of celebrities that died in the month preceding their birth month

We are told that the significance level is  [tex]\alpha = 0.05[/tex], the the z value of [tex]\alpha[/tex] is

              [tex]z_{\alpha } = 1.65[/tex]

This is obtained from the z-table

Since this test is carried out on the left side of the area under the normal curve then the critical value will be

                 [tex]z_{\alpha } = - 1.65[/tex]

So what this implies is that  [tex]H_o[/tex] will be rejected if

                [tex]z \le -1.65[/tex]

Here z is the test statistics

Now z is mathematically evaluated as follows

                  [tex]z = \frac{c - np}{\sqrt{np_o(1- p_o)} }[/tex]

                [tex]z = \frac{16 - (348 *0.083)}{\sqrt{348*0.083 (1- 0.083)} }[/tex]

                [tex]z =-2.50[/tex]

From our calculation we see that the value of z is less than [tex]-1.65[/tex] so the Null hypothesis will be rejected

   Hence this tell us that the  evidence provided is not enough to conclude  that 16 celebrities died a month to their birth month

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