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Call: lm(formula = A ~ x1) Residuals: Min 1Q Median 3Q Max -1.03922 -0.07843 0.03922 0.11765 1.13725 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept)
Call: lm(formula = A ~ x1) Residuals: Min 1Q Median 3Q Max -1.03922 -0.07843 0.03922 0.11765 1.13725 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.11765 0.28650 3.901 0.00142 ** x1 -0.01961 0.02796 -0.701 0.49386 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.5648 on 15 degrees of freedom Multiple R-squared: 0.03175, Adjusted R-squared: -0.0328 F-statistic: 0.4918 on 1 and 15 DF, p-value: 0.4939 -----------------Mann-Kendall Trend Test and Sen's Regression------------------------- Corrected Zc new P-value N/N* Original Z old P.value Tau -0.5548762 0.5789793 1.4755418 -0.6740186 0.5002995 -0.1029412 Sen's slope old.variance new.variance 0.0000000 372.0000000 548.9015480 Test for a trend in each phase. Choose an appropriate effect size test. Do the test and report your findings. What do you think this means about the intervention put into place for Jenny
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