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joja [24]
3 years ago
5

Six months ago you won $1,000,000 on a scratch-off lottery ticket and invested your winnings with a financial advisor at the inv

estment firm Dewey, Lie, and Howe. The earnings on the investment are compounded monthly.You complain to the financial advisor that your returns after 6 months are inadequate and do not even cover the advisor's fees. The 6 monthly interest rates (in decimal form) have been R1 = -0.4, R2= 0.67, R3 = 1.0, R4 = -0.5, R5 = 0.2, R6 = -0.165.1. What is the total worth of your investment after 6 months?$The financial advisor responds that you shouldn't complain because the average return over the 6 months has been an impressive 13.4% (0.134 in decimal form). You angrily respond that the average return is not representative of the 6-month investment performance, and that the geometric mean should be used instead.2. What is the geometric mean of the above monthly returns? (express as a decimal; round off to 3 decimal places)geometric mean
Mathematics
1 answer:
statuscvo [17]3 years ago
5 0

Answer:

The total worth of the investment after 6 months is T =  $ 1004004

The geometric mean of the above monthly returns is   \= G  =  0.001

Step-by-step explanation:

From the question we are told that

  The growth for each month are

             R1 = -0.4, R2= 0.67, R3 = 1.0, R4 = -0.5, R5 = 0.2, R6 = -0.165

   The amount invested is  A = \$ 1,000,000

   The number period of the investment is  6 months

Generally the worth of the investment after each month is

              G_i =  G_p * (1 + R_i)

Here G_p is the worth of the investment the previous year

          R_i is the growth for that month

So considering the first month

          G_1 =  G_p (1 + R_1)

Here G_p =  A

So

             G_1 = 1000000 (1 -0.4)

              G_1 = 600000

Considering the second month

  Here G_p =  600000

So

    G_2 =  600000 (1 + 0.67)

=>  G_2 =  1002000

Considering the third month

    Here G_p =  1002000

So    

    G_3 =  1002000 (1 + 1)

    G_3 =  2004000

Considering the fourth month

    Here G_p =  2004000

So  

    G_4= 2004000 (1 + -0.5)

    G_4= 1002000

Considering the fifth  month

    Here G_p =   1002000

So  

    G_5= 1002000 (1 + 0.2)

    G_5= 1202400

Considering the six month

       Here G_p =  1202400

So  

    G_6= 1202400 (1 -0.165)

    G_6= 1004004

Generally the total  worth of the investment after 6 months is T =  $ 1004004

Generally the geometric mean of the monthly returns is  

           \= G  =  \sqrt[n]{ [(1 + R_1 ) *  \cdots (1 + R_n)} ]-1

Here  n represents the number of months which has a value  n =  6

So

  \= G = \sqrt[6]{[(1+ (-0.4 )) * (1 + 0.67) * \cdots * (1 + (-0.165))]} - 1

     

    \= G  =  0.001

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Answer:

From the question we are told that

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    The sample size is  n  =  30  

     The  sample mean is  \= x =  16.32

     The  population standard deviation is  \sigma  =  0.8

      The  level of significance is  \alpha  = 0.10

Step 1: State hypotheses:

The  null hypothesis is  H_o :  \mu = 16

The alternative hypothesis is  H_a :  \mu \ne  16

Step 2: State the test statistic. Since we know the population standard deviation and the sample is large our test statistics is

          t = \frac{ \= x  -\mu }{ \frac{\sigma}{ \sqrt{n} } }

=>       t = \frac{ 16.32  -16  }{ \frac{0.8 }{ \sqrt{30} } }

=>       t =2.191

Generally the degree of freedom is mathematically represented as

          df  =  n - 1

=>      df  =  30  - 1

=>      df  =29

Step 3: State the critical region(s):

From the student t-distribution table the critical value corresponding to  \alpha  = 0.10  is

         t = 1.311

Generally the critical regions is mathematically represented as  

         - 1.311 < T <  1.311

Step 4: Conduct the experiment/study:

Generally the from the value obtained we see that the t value is outside the critical region so  the decision is [Reject the null hypothesis ]  

Step 5: Reach conclusions and state in English:

  There is  sufficient evidence to show that the filling weight has to be adjusted

Step 6: Calculate the p-value associated with this test. How does this the p-value support your conclusions in Step 5?

From the student t-distribution table the probability value to the right corresponding to t =2.191 at  a degree of freedom of  df  =29  is

        P( t > 2.191) =  0.0183

Generally the p-value is mathematically represented as

       p-value  = 2 * P( t >  2.191 )

=>    p-value  = 2 * 0.0183

=>    p-value  =  0.0366

Generally  looking at the value obtained we see that p- value <  \alpha hence

The decision rule is

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Step-by-step explanation:

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