The index of biotic integrity (IB) is a measure of the water quality in streams.

Question

The index of biotic integrity (IB) is a measure of the water quality in streams. IBI and land-use measures for a collection of streams in the ozark Highland ecoregion of Arkansas were collected as part of a study. The data data226.dat gives the data for IBI and the area of the watershed in square kilometers for streams in the original sample with area less than or equal to 70km (a) Use numerical and graphical methods to describe the variable IBI. Do the same for area. (Round your answers for to two decimal places and your answers for s to three decimal places.) (b) Run the simple linear regression and summarize the results. Let x area and y BI. Round your slope, intercept, and r to three decimal places, Round Fto two decimal places and your P-value to four decimal places.) ls area of watershed a good predictor for 1BI at the 5% significance level? Yes, the area of watershed is a good predictor for IBI. No, there is insufficient evidence to say the area of watershed is a good predictor for IB interpret the intercept, The area of watershed when IBI value is 0 The increase in the watershed area when IBI is increased by one unit The value of IBI when watershed area is 0 The increase in IBI when the watershed area is increased by one unit

Explanation / Answer

Descriptive Statistics: x, y

Variable   N N*   Mean StDev
x         48   0 28.33 17.90
y         48   0 63.35 18.30

Regression Analysis: y versus x

Analysis of Variance

Source         DF Adj SS Adj MS F-Value P-Value
Regression      1    3051 3050.6    11.05    0.002
x             1    3051 3050.6    11.05    0.002
Error          46   12696   276.0
Lack-of-Fit 28    9179   327.8     1.68    0.127
Pure Error   18    3518   195.4
Total          47   15747

Model Summary

      S    R-sq R-sq(adj) R-sq(pred)
16.6135 19.37%     17.62%      12.06%

Coefficients

Term       Coef SE Coef T-Value P-Value   VIF
Constant 50.60     4.52    11.19    0.000
x         0.450    0.135     3.32    0.002 1.00

Regression Equation

y = 50.60 + 0.450 x

P-value for regression <0.05 so we reject null hypothesisso answer is can conclude thet there sufficient evidenc.

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