Solve using Excel functions:
a) Correlation coefficient: r = CORREL(x,y) = 0.926
b) n = 9, Degrees of freedom: df = n-2 = 7, Level of significance: α = 0.05
H0: ρ = 0, There is not sufficient evidence that index of exposure and cancer mortality rate are linearly correlated
H1: ρ ≠ 0, There is sufficient evidence that index of exposure and cancer mortality rate are linearly correlated
Test statistic: t = r*((1-r*r)/(n-2))^0.5 = 0.926*((1-0.926*0.926)/(9-2))^0.5 = 0.132
Critical value (Using Excel function T.INV.2T(probability,df)) = T.INV.2T(0.05,7) = 2.365
Since test statistic is less than critical value, we fail to reject H0.
So, there is not sufficient evidence that index of exposure and cancer mortality rate are linearly correlated.
c) y = a+bx
a = INTERCEPT(y,x) = 114.72
b = SLOPE(y,x) = 9.23
y = 114.72 + 9.23x
Index exposure:x = 5.6
Mortality rate: y = 114.72 + 9.23*5.6 = 114.72+51.69 = 166.41
The sample data below are the index of exposure (x) to radioactive waste for nine different...
The sample data below are the index of exposure (x) to radioactive waste for nine different Oregon counties and cancer mortality rate (y) (deaths per 100,000). X 2.49 2.57 3.41 1.25 1.62 3.83 11.64 6.41 8.34 y 147.1 130.1 129.9 113.5 137.5 162.3 207.5 177.9 210.3 a) Find the linear correlation coefficient r. (2 points) b) Do the data provide sufficient evidence to conclude that index of exposure and cancer mortality rate are linearly correlated? (3 points) c) Find the...
The index of exposure to radioactive waste, x, and the cancer mortality rates, y, (deaths per 100,000) were recorded for nine different Oregon counties. Use the regression analysis provided below to perform the hypothesis test to determine if the index of exposure is useful as a predictor of cancer mortality rate. The regression equation is = 114.7156 + 9.231456x R-sq = 85.8% 9- 2 = 7 degrees of freedom Predictor Coef Constant 114.7156 Index of Exp 9.231456 SE(Coef) I 8.045663...