1.5/2.5 polints 24 suemisalons Used The ability of ecologists to identi y re ons

Question

1.5/2.5 polints 24 suemisalons Used The ability of ecologists to identi y re ons richness uld have an impact on the p ee vation o genetic diversity, à ma or objective of the World Conservation Strategy. A study used a sar ple or where y-species richness, x,-watershed area, shore width, poor drainage %), xa water color total color units x, sand %), and XG-alkali ity, The SSR and SSE have been calculated to be: SS What percent of the variation in y can be explained by the model? R2 087 Carry out the model utility F test. of greatest species act on the preservation of 37 akes to obtain the estimated egression equation y-3.89 + 0.033x1 + 0.024x2 + 0.023x3-0.0080x4-0.13xs-0.72xG 76S and SSE 3072. X %" (Give your value in percent and round to two decimal places.) State the appropriate typotheses Hai at least one among A, , p6 is not zero Hai at least one among »-·-F6 is zero Compute the test statistic value and obtain a p-value. Round your test statistic to two decimal places and your p-value to rour decimal places.(Hint: You can use R to cetain the p-value, pvalpri test stat,umerator dr, denominator ar) test statistic X p-valve = 0000 context. use a-0.05. 8 Reject Ho. There is significant evidence at least one explanatory variable is a signicant predictor of species richness. O Reject Ho- There is no evidence at least one explanatory variable is a species richness. significant predictor of Fail to reject Ho. There is significant evidence at least one explanatory variable is a significant predictor of species richness O Fail to reject Ho. There is no evidence at least one explanatory variable is a significant predictor of species richness.

Explanation / Answer

1) here % of variation R2 =SSR/(SSR+SSE )=768/(768+307.2)=71.43%

test statistic f =(SSR/6)/(SSE/(37-7))=(768/6)/(307.2/30)=12.5

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