An investor with a stock portfolio worth several hundred thousand dollars sued his broker and brokerage firm because he felt that lack of diversification in his portfolio led to poor performance for many years in a row. in an effort to avoid close public scrutiny, the firm agreed to settle the conflict by an arbitration panel. the arbitration panel compared a sample of 39 months of the investor's returns with the average of the standard & poor's 500-stock index for the same period in order to determine whether there was a substantial decrease. historically, the s& p has a mean return of 0.95. suppose that you are a member of this arbitration panel. conduct a hypothesis test to determine if the investor's portfolio performed significantly worse than the performance of the s& p 500. use a level of significance of alpha=0.05 . open the data file rates of return. use this information to answer questions 1 through 5.

1. give a 95% confidence interval for the true mean return of the investor's portfolio. round your answer accurate to three decimal places in interval notation. be sure to put the lower bound in the first box and the upper bound in the second. [example: (42.335, 54.859)]
( , )

2. is the alternative hypothesis for this test one-tailed or two-tailed?

a. one-tailed
b. two-tailed
3. what is the t-score for this test? give your answer accurate to three decimal places. (example: -3.234)
4. what is the p-value for this test? give your answer accurate to three decimal places. (example: 0.034)
5. based on the results of this test, is there enough evidence to say that the investor's portfolio performed significantly worse than the s& p 500?

a. yes, because we rejected the null.
b. yes, because we failed to reject the null.
c. no, because we rejected the null.
d. no, because we failed to reject the null.

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1