Question

isl - u1, wkst 2 - graphing practice 2
name:
the data at right shows information collected in a lab investigation. graph your data in sheets. in the first column below, write the line equation for the trend line in the form of y=mx + b, show the r² value, and show your 5% calculation. in the second column, explain your r² value, and evaluate the 5% rule (do you throw the y - intercept or keep it? rewrite the line equation appropriately). in the third column write a \for every\ statement to explain what the graph means.
equation: (y=mx + b format)
y =

x +

r² value: (always a decimal point)
5% calculation: (show your work)
there is a

% relationship between (iv) and (dv)
r² value explanation: (is there over a 90% relationship?)
5% rule evaluation. (keep or drop b, write the inequality and write/circle final line equation)
keep/drop b because
for every statement
for every 1

of

the

by

average monthly
month
temperature
august
september
october
november
december
january
temp (f)
72
66
56
47
35
29

Explanation:

Step1: Calculate the slope \(m\)

We use the formula \(m=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}\) for a set of \(n\) data - points \((x_i,y_i)\). First, we need to set up a table with columns for \(x\) (months, we can assign August = 1, September = 2, etc.), \(y\) (temperature), \(x^{2}\), \(xy\), and calculate the sums \(\sum x\), \(\sum y\), \(\sum x^{2}\), \(\sum xy\). Let \(x\) be the month number (\(x = 1,2,\cdots,6\)) and \(y\) be the temperature.

Month	\(x\)	\(y\)	\(x^{2}\)	\(xy\)
September	2	66	4	132
October	3	56	9	168
November	4	47	16	188
December	5	35	25	175
January	6	29	36	174

\(\sum x=1 + 2+3 + 4+5 + 6=\frac{6\times(6 + 1)}{2}=21\), \(\sum y=72 + 66+56 + 47+35 + 29 = 305\), \(\sum x^{2}=1 + 4+9 + 16+25 + 36 = 91\), \(\sum xy=72+132 + 168+188+175+174 = 909\).
\(n = 6\).
\(m=\frac{6\times909-21\times305}{6\times91 - 21^{2}}=\frac{5454-6405}{546 - 441}=\frac{-951}{105}\approx - 9.057\)

Step2: Calculate the y - intercept \(b\)

We use the formula \(b=\frac{\sum y-m(\sum x)}{n}\).
\(b=\frac{305-(-9.057)\times21}{6}=\frac{305 + 190.2}{6}=\frac{495.2}{6}\approx82.53\)

Step3: Calculate the \(R^{2}\) value

First, we calculate the mean of \(y\), \(\bar{y}=\frac{\sum y}{n}=\frac{305}{6}\approx50.83\).
The total sum of squares \(SST=\sum(y - \bar{y})^{2}\), and the residual sum of squares \(SSE=\sum(y-(mx + b))^{2}\).
\(SST=\sum(y - 50.83)^{2}=(72 - 50.83)^{2}+(66 - 50.83)^{2}+(56 - 50.83)^{2}+(47 - 50.83)^{2}+(35 - 50.83)^{2}+(29 - 50.83)^{2}\)
\(SST=(21.17)^{2}+(15.17)^{2}+(5.17)^{2}+(-3.83)^{2}+(-15.83)^{2}+(-21.83)^{2}\)
\(SST = 448.1 + 230.1 + 26.7+14.6+250.6+476.6 = 1446.7\)
We calculate the predicted \(y\) - values \(\hat{y}_i=mx_i + b\) for each \(x_i\).
\(\hat{y}_1=-9.057\times1 + 82.53=73.473\), \(\hat{y}_2=-9.057\times2 + 82.53=64.416\), \(\hat{y}_3=-9.057\times3 + 82.53=55.359\), \(\hat{y}_4=-9.057\times4 + 82.53=46.302\), \(\hat{y}_5=-9.057\times5 + 82.53=37.245\), \(\hat{y}_6=-9.057\times6 + 82.53=28.188\)
\(SSE=\sum(y-\hat{y})^{2}=(72 - 73.473)^{2}+(66 - 64.416)^{2}+(56 - 55.359)^{2}+(47 - 46.302)^{2}+(35 - 37.245)^{2}+(29 - 28.188)^{2}\)
\(SSE=(-1.473)^{2}+(1.584)^{2}+(0.641)^{2}+(0.698)^{2}+(-2.245)^{2}+(0.812)^{2}\)
\(SSE = 2.17+2.51+0.41+0.49+5.04+0.66 = 11.28\)
\(R^{2}=1-\frac{SSE}{SST}=1-\frac{11.28}{1446.7}\approx0.992\)

Answer:

Equation: \(y=-9.06x + 82.53\) (rounded to two - decimal places)
\(m=-9.06\), \(b = 82.53\), \(R^{2}=0.992\)
Since \(R^{2}\approx0.992>0.9\), there is a very strong linear relationship between the month and the temperature. The \(y\) - intercept represents the estimated temperature when \(x = 0\) (a non - existent month in our context, but in the linear model). The slope indicates that the temperature decreases by approximately \(9.06^{\circ}F\) per month.