Let the average snowfall for the years 1953 to 1960 be 412 4

Let the average snowfall for the years 1953 to 1960 be 41.2, 41.9, 42.6, 43.4, 44.4, 45.5, 46.6, 47.5, where t=3 represents 1953.

a. Graph the data,

b. Curve fit the data with a least squares regression line.

c. Graph the regression line.

d. What is the curve fitted snowfall for March of 1958?

System:: {

Y = ____________________________

Y(March 1958) = _______________________

a) Graph :
Just plot the points
(3 , 41.2)
(4 , 41.9)
(5 , 42.6)
(6 , 43.4)
(7 , 44.4)
(8 , 45.5)
(9 , 46.6)
(10 , 47.5)

on the graph

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b) We get he line to be :
y = 9.202380952·10-1 x + 38.15595238

y = 0.920238x + 38.155952 ---> ANSWER

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c)
Simply take that curve y = 0.920238x + 38.155952
and plot on the graphing calculator

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d)
1958 ---> t = 8

Plug in t = 8, we get :

y = 0.920238(8) + 38.155952

y = 45.517856

So, approx 45.518

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ANSWERS :

y = 0.920238x + 38.155952

Y(March 1958) = 45.518