You have the following R output, which describes how stream flow (flow) changes

Question

You have the following R output, which describes how stream flow (flow) changes with the depth of the stream (depth), and the depth of the stream, squared (depthsq).

> stream2<-lm(flow~depth+depthsq)

> summary(stream2)

Call:

lm(formula = flow ~ depth + depthsq)

Residuals:

Min 1Q Median 3Q Max

-0.406145 -0.163666 -0.002649 0.198973 0.327658

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 1.683      1.059 1.589 0.1561

depth      -10.861    4.517     -2.404   0.0472 *

depthsq    23.535    4.274      5.506   0.0009 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.2794 on 7 degrees of freedom

Multiple R-Squared: 0.99, Adjusted R-squared: 0.9871

F-statistic: 346.5 on 2 and 7 DF, p-value: 1e-07

a)When performing an ANOVA on this analysis, the null hypothesis has been defined as H0: b0 = 0. Write out the regression line for your sub-model and your full model.

b) Under the null hypothesis in (a), what is the F-statistic and it’s corresponding p-value?

c) Based on your answer to (b), what decision do you make regarding model selection?

Explanation / Answer

a)When performing an ANOVA on this analysis, the null hypothesis has been defined as H0: b0 = 0. Write out the regression line for your sub-model and your full model.

The regression model line is formed using the coefficients as

Flow = 1.683 -10.861*depth +23.535*depthsq

b) Under the null hypothesis in (a), what is the F-statistic and it’s corresponding p-value?

Again from the regression output , the values are

F-statistic: 346.5 and the p value is 1e-07

c) Based on your answer to (b), what decision do you make regarding model selection?

p-value: 1e-07 as the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis

H0 : Model is not signficant

H1 : Model is statistically signficant

also , the rsquare value is 0.99 , this means that the model is able to explain 99% of the variation of the data

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