Consider the following time series data: Month 1 2 3 4 5 6 7 Value 23 14 20 11 19 24 15 (a) Compute MSE using the most recent value as the forecast for the next period. If required, round your answer to one decimal place. What is the forecast for month 8? If required, round your answer to one decimal place. Do not round intermediate calculation. (b) Compute MSE using the average of all the data available as the forecast for the next period. If required, round your answer to one decimal place. Do not round intermediate calculation. What is the forecast for month 8? If required, round your answer to one decimal place. (c) Which method appears to provide the better forecast?

a) MSE = 61.33

Forecast for 8th month = 15

b) MSE = 34.51

Forecast for 8th month = 18

c) The average method provides a better forecast because of its low MSE

Explanation:

Check the attached file for the solvings.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

honestly bro aint no one got time to answer these dumb question just do it p***y

step-by-step explanation: