ESCI 340 computer lab Friday 15 January 2016

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# Douglas fir dbh data, from 6 January 2016



dbh.edge <- c(93.3, 107.4, 95, 

106.2, 97.2, 94.3, 116.1, 115.3, 

87.6, 80)



> mean(dbh.edge)

[1] 99.24

> sd(dbh.edge)

[1] 11.76012



dbh.int <- c(61.5, 63.4, 191, 58.7, 54.5, 

146, 71.6, 37.2, 48.1, 75.1)



> mean(dbh.int)

[1] 80.71

> sd(dbh.int)

[1] 48.69147



# One sample t-test using dbh edge data; Ho: mu=110cm

> t.edge = (mean(dbh.edge) - 110) / (sd(dbh.edge) / sqrt(10))

> t.edge

[1] -2.893347



# Determine p-value for this t-value:

# Multiply by 2 to get 2-tailed probability

> 2*pt(t.edge, df=9)

[1] 0.0177870





#  t-test function in R produces identical result:

# documentation:

## Default S3 method:

t.test(x, y = NULL,

       alternative = c("two.sided", "less", "greater"),

       mu = 0, paired = FALSE, var.equal = FALSE,

       conf.level = 0.95, ...)





> t.test(dbh.edge, mu=110)



        One Sample t-test



data:  dbh.edge

t = -2.8933, df = 9, p-value = 0.01779

alternative hypothesis: true mean is not equal to 110

95 percent confidence interval:

  90.82732 107.65268

sample estimates:

mean of x 

    99.24 



# ********************************************************************



# Two sample t-test using edge and interior dbh data, Ho: mu_edge = mu_int

# t = [mean(edge) - mean(int)] / se

# First, calculate se, from pooled variance:

> sp.dbh <- ((9*var(dbh.edge) + 9*var(dbh.int)) / (10+10))

> sp.dbh

[1] 1129.122

> se.dbh <- sqrt(sp.dbh/10 + sp.dbh/10)

> se.dbh

[1] 15.02745

# Insert se value into formula for t:

> t.dbh <- (mean(dbh.edge) - mean(dbh.int)) / se.dbh  # d.f. = 10 + 10 - 2 = 18

> t.dbh

[1] 1.233077

> # determine p-value, 2-tailed:

> 2*(1-pt(t.dbh, df=18))

[1] 0.2334088



#  2-sample t-test function in R produces similar result:

> t.test(dbh.edge, dbh.int)



        Welch Two Sample t-test



data:  dbh.edge and dbh.int

t = 1.1698, df = 10.046, p-value = 0.2691

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

 -16.74234  53.80234

sample estimates:

mean of x mean of y 

    99.24     80.71 



# slight difference in t-value and p-value due to inequality in variances;

#    t.test function applies Welch's approximation



# ************************************************************************

# One-tailed t-test:   Ha: mu_edge > mu_int;  Ho: mu_edge <= mu_int



> t.dbh <- (mean(dbh.edge) - mean(dbh.int)) / se.dbh  # d.f. = 10 + 10 - 2 = 18

> # determine p-value, 1-tailed

> 1-pt(t.dbh, df=18)

[1] 0.1167044





# 2-sample, one-tailed t-test function in R:

> t.test(dbh.edge, dbh.int, alt="greater")



        Welch Two Sample t-test



data:  dbh.edge and dbh.int

t = 1.1698, df = 10.046, p-value = 0.1345

alternative hypothesis: true difference in means is greater than 0

95 percent confidence interval:

 -10.16651       Inf

sample estimates:

mean of x mean of y 

    99.24     80.71 



# ******************************************************************************

#

# Plot maple seed "dispersal" data from a prior year

# NOTE: for ESCI 340 winter 2016 assignment 2, must use data listed on that assignment.

#

# Maple seed dispersal data:

# dist.cm <- c(335, 323, 400, 314, 48, 461, 132, 125, 146, 202,

              160, 156, 177, 118, 151, 170, 268, 233, 202, 220,

              238, 75, 76, 66, 122, 71, 112, 99, 111, 110,

              100, 55, 116, 160, 90, 13, 161, 168, 141, 129,

              160, 178, 150, 272, 324, 156, 287, 235, 212, 155,

              114, 187, 380, 250, 352, 279, 290, 82, 117, 200,

              239, 290, 253, 208)



par(mfrow = c(1,1))

hist(dist.cm, xlim=c(0,1000), xlab="Distance (cm)")  # Plot histogram



# Superimpose normal curve on histogram, fit to the data

x.cm <- seq(0, 1000, by=10)  # Create seqence of distance values



# Use dnorm function to plot normal probabilities

# syntax: dnorm(x-values, mean, standard deviation)

# multiply dnorm probabilities by sample size, n=64, and histogram interval width = 50cm

# Note: sample size in winter 2016, n=60

lines(x.cm, 64*50*dnorm(x.cm, 186.3, 93.9))



# alternatively, could use: 

#   lines(x.cm, 64*50*dnorm(x.cm, mu=mean(dist.cm), sd=sd(dist.cm)))



# Superimpose Uniform distribution on histogram, over range [0, 600]

# Use dunif function to plot uniform distribution probabilities

# syntax: dunif(x-values, lower limit, upper limit)

# As above, multiply dunif probabilities by sample size, n=64, and histogram interval width = 50cm

# Plot as dashed line, "lty=2")

lines(x.cm, 64*50*dunif(x.cm, 0, 600), lty=2)