Question

Upload speeds for data are increasing with technological advances. The speeds are recorded in the table for each year.

a-Make a graph of the data large enough to include 2019.

b-Find the slope between years 2008 and 2011, accurate to 3 decimal places.

c- Show your calculation for the mean point; plot and label the mean point.

d- Add a trend line to the graph.

e-Use your trend line to make a prediction for the upload speed in 2018.

f-Find the equation of the line passing trough 2008&2011.

g-Find the equation of the line passing trough 2009&2010

h-Find the equation of your trend line using the mean point and one other point that is on your trend line. Identify the point you are using on your trend line . rounded to 3 decimal places.

i-Find the equation for the best fit line using a linear regression tool(Desmos,graphing calculator,other). rounded to one decimal place.

j-Interpret the slope of the best fit line in a complete sentence.

k-use your regression equation to predict the upload speed in 2019. rounded to the nearest Kb/s. Plot and label this point on your graph. Explain the meaning of this point an a complete sentence.

l-Use your regression equation to predict the year the upload speed will reach 6500 Kb/s. Plot and label this point on your graph. Explain the meaning of this point in a complete sentence.

Year Speed(Kb/s(

2008 1028

2009 1727

2010 2208

2011 2704

Answer 1

a.)

b.) Slope between year 2008 and 2011.

Slope = ( Change in Y coordinates ) / (Change in X coordinates)

initial value, X₁ = 2008 , Y₁ = 1028

final value , X₂ = 2011, Y₂ = 2704

Slope = ( Y₂- Y₁ ) / ( X₂ - X₁ )

Slope = ( 2704 - 1028) / (2011 - 2008)

= 1676 / 3

= 558.667

Therefore, slope between 2008 and 2011 is 558.667 kbps/year.

c.) Mean line will give the mean value of kbps speed between the years 2008 and 2011. It is simply sum of individual values divided by total number of observations i.e., 4 in this case.

mean = [ (1028+ 1727 + 2208 + 2704) / 4 ]

= 7667 / 4

= 1916.75 kbps

Mean line will be a horizontal line with value of Y = 1916.75

d. ) Using basicTrendline Package in R:

model used for trend line is linear i.e., y = ax + b

e. ) For making prediction of speed in year 2018 using the above generated trend line,

y = 550.9 * x - 1105117

here, using x = 2018, we will get the predicted value of speed in year 2018.

=> y = 550.9 * 2018 - 1105117

=> y = 6599.2 kbps.

Therefore, from our trend line, we predict the speed to be 6599.2 kbps in year 2018.