Question

Consider the following data on x = rainfall volume (m3) and y = runoff volume (m) for a particular location. x 5 12 14 17 23 30 40 48 55 67 72 83 96 112 127 y | 4 10 13 14 15 25 27 47 38 46 53 75 82 99 103 Use the accompanying Minitab output to decide whether there is a useful linear relationship between rainfall and runoff. The regression equation is runoff = -1.72 + 0.845 rainfall Predictor Coef Stdev t-ratio Constant -1.720 2.421 -0.71 0.490 rainfall 0.84495 0.03722 22.700.000 s = 5.354 R-sq = 97.5% R-sq(adj) = 97.4% State the appropriate null and alternative hypotheses. Ho: B1 = 0 Ha: B1 * 0 Ho: B1 = 0 Ha: B1 > 0 Ho: B1 = 0 Ha: B1 < 0 O Ho: B4 + 0 Ha: B1 = 0 Compute the test statistic value and find the P-value. (Round your test statistic to two decimal places and your p-value to three decimal places.) t= P-value = State the conclusion in the problem context. (Use a = 0.05.) Reject Ho. There is a useful linear relationship between runoff and rainfall at the 0.05 level. O Reject Ho. There is not a useful linear relationship between runoff and rainfall at the 0.05 level. O Fail to reject Ho. There is not a useful linear relationship between runoff and rainfall at the 0.05 level. O Fail to reject Ho. There is a useful linear relationship between runoff and rainfall at the 0.05 level. increase in rainfall volume. (Round your answers to three decimal places.) Calculate a 95% confidence interval for the true average change in runoff volume associated with a 1 m m3 You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

t = rainfall (coef) sainfall (sider) 0.01722 =22.70 0084495 0.037 22 22. 70 P-value =10.0001 p-value of scinfall. 95% CI of slope fainfall (coup) E temos (SPD) kainfall Ta Zn-2 n=15 03722 5 0184495 I to.os, is (0.03722) = 0.84495 ITINV (0.05,13) (0.03722) = 0.884495 + 2.16 + 0.03722 = 0.84495 + 0.08040 = 0.76455, 0.92535 = (10.765 , 10.925 ] ) ms.