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TEA [102]
3 years ago
7

6.3.6. Among the early attempts to revisit the death postponement theory introduced in Case Study 6.3.2 was an examination of th

e 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.
Mathematics
1 answer:
kakasveta [241]3 years ago
8 0

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

            p = \frac{1}{12}  

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  

  Null Hypothesis  \  \  H_0 : p = 0.083

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

  Alternative \ Hypothesis  H_1 : p < 0.083

   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  c_s = 16

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

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

              z_{\alpha } =  1.65

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

                 z_{\alpha } = - 1.65

So what this implies is that  H_o will be rejected if

                z \le -1.65

Here z is the test statistics

Now z is mathematically evaluated as follows

                  z = \frac{c - np}{\sqrt{np_o(1- p_o)} }

                z = \frac{16 - (348 *0.083)}{\sqrt{348*0.083 (1- 0.083)} }

                z =-2.50

From our calculation we see that the value of z is less than -1.65 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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Abe has 150 t-shirts to sell before the playoffs begin. The table shows the linear relationship between the number of t-shirts r
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Answer:

Y= -12x + 150

Step-by-step explanation:

1. You do the change in y over the change in x to get the slope.

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3 years ago
The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales
andrezito [222]

Answer:

t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657    

p_v =P(t_{(39)}>6.657)=3.17x10^{-8}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=44 represent the sample mean

s=1.9 represent the sample standard deviation

n=40 sample size  

\mu_o =42 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 42, the system of hypothesis would be:  

Null hypothesis:\mu \leq 42  

Alternative hypothesis:\mu > 42  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=40-1=39  

Since is a one side test the p value would be:  

p_v =P(t_{(39)}>6.657)=3.17x10^{-8}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

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3 years ago
I need help on this problem
Helen [10]

Answer:

if you solve it for the t your answer would be: pt+30pte−0.2t=600

but if you solved it for the p your answer would be: p=600t(1+30e−0.2t)

hope it works, sweete

8 0
3 years ago
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