a) If the relationship between Y and X is believed to be linear, then the equation for a line may be appropriate:
b)
"mean=\\bar{y}={\\sum y_i \\over n}={8+12+6+14+8 \\over 5}=9.6"
"S_{xx}=\\sum (x_i-\\bar{x})^2=\\sum x_i^2-n\\bar{x}^2=22.8"
"S_{yy}=\\sum (y_i-\\bar{y})^2=\\sum y_i^2-n\\bar{y}^2=43.2"
"S_{xy}=\\sum (x_i-\\bar{x})(y_i-\\bar{y})=\\sum x_iy_i-n\\bar{x}\\bar{y}=30.4"
"B={S_{xy} \\over S_{xx}}={30.4 \\over 22.8}={4 \\over 3}\\approx1.333333"
"A=\\bar{y}-B\\bar{x}=9.6-{4 \\over 3}\\cdot5.2={8 \\over 3}\\approx2.666667"
"y={8 \\over3}+{4 \\over 3}x"
The intercept is the expected mean value of Y when all X=0.
When all inputs = 0, then the expected mean value of output will be "y={8 \\over3}."
c)
The slope is the rate of change, the mean amount of change in y when x increases by 1.
The output y increases by 4/3, when x increases by 1.
d)
The r2 value tells us that 93.8% of the variation in the output is explained by the input.
e)
f)
The "r^2" value tells us that 93.8% of the variation in the output is explained by the input.
6.2 % of the changes in output is due to other raw materials apart from X.
g)
"H_0:" The slope of the regression line is equal to zero.
"H_1:" The slope of the regression line is not equal to zero.
"b_1={4 \\over 3}\\approx1.3333"
"df=n-2=5-2=3"
"t={b_1 \\over SE}\\approx{1.3333 \\over 0.7947}\\approx1.6778"
Determine the p-value (two-tailed test)
Since the P-value (0.1920) is greater than the significance level (0.01), then the null hypothesis is not rejected.
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