This function runs the quadrants analysis suggested by Cate and Nelson (1971)
Usage
cate_nelson_1971(data = NULL, stv, ry, tidy = TRUE, plot = FALSE)
boot_cn_1971(data, ry, stv, n = 5, ...)Arguments
- data
argument to call a data.frame or data.table containing the data
- stv
argument to call the vector or column containing the soil test value (stv) data
- ry
argument to call the vector or column containing the relative yield (ry) data
- tidy
logical operator (TRUE/FALSE) to decide the type of return. TRUE returns a data.frame, FALSE returns a list. Default: TRUE.
- plot
logical operator (TRUE/FALSE) to decide the type of return. TRUE returns a ggplot, FALSE returns either a list (tidy == FALSE) or a data.frame (tidy == TRUE).
- n
sample size for the bootstrapping Default: 500
- ...
when running bootstrapped samples, the
...(open arguments) allows to add grouping variable/s (factor or character) Default: NULL
Value
returns an object of type ggplot if plot = TRUE.
returns an object of class data.frame if tidy = TRUE,
returns an object of class list if tidy = FALSE.
boot_cn_1971: bootstrapping function
Details
See online-documentation for additional details.
Note
This code was adapted from Mangiafico, S. S. (2013). Cate-Nelson Analysis for Bivariate Data Using R-project. The Journal of Extension, 51(5), Article 33. https://tigerprints.clemson.edu/joe/vol51/iss5/33/
References
Cate & Nelson (1971). A simple statistical procedure for partitioning soil test correlation data into two classes. Soil Sci. Soc. Am. Proc. 35:658-660. doi:10.2136/sssaj1971.03615995003500040048x
Examples
# \donttest{
# Example 1 dataset
dat <- data.frame("ry" = c(65,80,85,88,90,94,93,96,97,95,98,100,99,99,100),
"stv" = c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15))
# Run
fit_example_cn_1971 <- cate_nelson_1971(data = dat,
ry = ry, stv = stv, tidy=FALSE, plot=FALSE)
#> Warning: Chi-squared approximation may be incorrect
fit_example_cn_1971
#> $n
#> [1] 15
#>
#> $CRYV
#> [1] 86.5
#>
#> $CSTV
#> [1] 3.5
#>
#> $R2
#> [1] 0.6942517
#>
#> $AIC
#> [1] 97.24281
#>
#> $BIC
#> [1] 99.36696
#>
#> $RMSE
#> [1] 5.06568
#>
#> $quadrants
#> q.I q.II q.III q.IV positive negative
#> 1 0 12 0 3 15 0
#>
#> $X2
#>
#> Pearson's Chi-squared test with Yates' continuity correction
#>
#> data: data.frame(row.1, row.2)
#> X-squared = 9.401, df = 1, p-value = 0.002169
#>
#>
#> $anova
#> Analysis of Variance Table
#>
#> Response: y
#> Df Sum Sq Mean Sq F value Pr(>F)
#> xgroup 1 874.02 874.02 29.519 0.0001145 ***
#> Residuals 13 384.92 29.61
#> ---
#> Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>
# }