Question

2. Given the data in the table below, answer questions a) 40 e) rainfall (mm) (runoff (mm) 0 0 0 0 O 0 0 0 0 150 200 a) create a scatter plot using these data b) determine the linear regression equation and correlation coeficient include details from the method you used to find the equation: Excel Geodebra, or open office file image of graphing calculator or manual calculations () is this a positive or negative correlation? is ita weak or strong correlation d) a particular storm hod 300mm of rain. what would be a good prediction of the runoff? e) another storm had 180mm of rain with 6.0mm of runoff. Would you consider this dota on Outlier ? Suggest a reason that might explain this data point

Answer 1

(a) The scatter plot is:

80 70 60 50 > 40 30 20 10 100 50 150 200 250 х

(b) The calculations are:

x	y	(x-xbar)	(y-ybar)	(x-xbar)*(y-ybar)	(x-xbar)²	(y-ybar)²
20	5	-83.57	-30.14	2519.08	6984.18	908.59
50	12	-53.57	-23.14	1239.80	2869.90	535.59
80	21	-23.57	-14.14	333.37	555.61	200.02
100	35	-3.57	-0.14	0.51	12.76	0.02
125	41	21.43	5.86	125.51	459.18	34.31
150	58	46.43	22.86	1061.22	2155.61	522.45
200	74	96.43	38.86	3746.94	9298.47	1509.88
xbar	ybar	Σ (x-xbar)	Σ (y-ybar)	Σ (x-xbar)*(y-ybar)	Σ (x-xbar)²	Σ (y-ybar)²
103.57	35.14	0	0	9026.43	22335.71	3710.86
r	0.991
b1	0.404
bo	-6.713

The correlation coefficient is 0.991.

The regression equation is:

y = -6.713 + 0.404*x

(c) This is a strong positive correlation.

(d) The regression equation is:

y = -6.713 + 0.404*x

Put x = 300

y = -6.713 + 0.404*300

y = 114.52

(e) The runoff point is an outlier. So, this point can be considered as an outlier.

Please give me a thumbs-up if this helps you out. Thank you!