Question

Coarse sediments in a stream that are carried by intermittent contact with the streambed by rolling, sliding, and bouncing are called bedload. The rate at which the bedload is transported in a stream can vary dramatically, depending on flow conditions, stream slope, water depth, weight of the sediment, and so on. In a 2007 technical report for the U.S. Department of Agriculture, Forest Service, Rocky Mountain Research Station, regression was used to model bedload transport rate (kilograms per second) as a function of stream discharge rate (cubic meters per second). The data for the study, collected from Hayden Creek in Colorado during snowmelt runoff in the following table.

Answer the following for the variables bedload (dependent variable) and Discharge (independent variable) by using SAS or R (your answer includes the SAS or R code and output).

Obs	Discharge	BedLoad
1	1.03635	0.00373
2	1.31878	0.0118
3	1.12	0.01507
4	1.28595	0.0149
5	1.57864	0.02055
6	1.43772	0.02041
7	1.29477	0.00777
8	1.16407	0.00386
9	1.33357	0.01009
10	1.35399	0.01663
11	1.28252	0.01596
12	1.38463	0.01492
13	1.32336	0.01984
14	1.32336	0.01046
15	1.39484	0.02087
16	0.75154	0.0036
17	0.79239	0.00471
18	0.7107	0.00317
19	0.81281	0.00499
20	0.7107	0.0053
21	0.87408	0.00308
22	0.69028	0.00556
23	0.85365	0.00426
24	0.7107	0.00564
25	0.56775	0.0008
26	0.60859	0.00094
27	0.6939	0.00092
28	0.5612	0.00074
29	0.6478	0.00185
30	0.91584	0.00575
31	0.99158	0.00882
32	1.62547	0.01937
33	1.24061	0.00634
34	1.33107	0.00919
35	1.21859	0.01631
36	1.21859	0.01356
37	1.33107	0.01554
38	1.4495	0.01138
39	1.62547	0.01332
40	1.17524	0.00636
41	1.15391	0.00982
42	1.54858	0.01107
43	0.38707	0.0012
44	1.01107	0.01085

Answer 1

In order to solve this question I used R software.

R codes and output:

Que.a

d=read.table('bedload.csv',header=TRUE,sep=',')
head(d)
Obs Discharge BedLoad
1 1 1.03635 0.00373
2 2 1.31878 0.01180
3 3 1.12000 0.01507
4 4 1.28595 0.01490
5 5 1.57864 0.02055
6 6 1.43772 0.02041
attach(d)
The following objects are masked from d (pos = 3):

BedLoad, Discharge, Obs

plot(Discharge,BedLoad)

Scatter plot :

> fit=lm(BedLoad~Discharge)
> summary(fit)

Call:
lm(formula = BedLoad ~ Discharge)

Residuals:
Min 1Q Median 3Q Max
-0.0066801 -0.0026549 -0.0003797 0.0024205 0.0068250

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.007546 0.001830 -4.124 0.000172 ***
Discharge 0.015537 0.001612 9.640 3.32e-12 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.003495 on 42 degrees of freedom
Multiple R-squared: 0.6887, Adjusted R-squared: 0.6813
F-statistic: 92.93 on 1 and 42 DF, p-value: 3.315e-12

> res=fit$residuals
> plot(Discharge,res)

Que.b

Residual plot:

Que.c

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.007546 0.001830 -4.124 0.000172 ***
Discharge 0.015537 0.001612 9.640 3.32e-12 ***

We use t test statistic for testing

t = 9.64 and p-value = 0.000

Since p-value is less than 0.05, we reject null hypothesis and conclude that

Que.d

Intercept = -0.007546

Slope = 0.015537

Que.e

Fitted regression equation is

Bedload = -0.007546 + 0.015537 * Discharge

Scatter plot with regression line is,

> plot(Discharge,BedLoad)
> abline(fit)