# NOT RUN {
# Show how potential confidence intervals for the difference between two means
# will look assuming standard deviations of 1, 2, or 3, differences between
# the two means of 2, 1, or 0, and a sample size of 10 in each group.
ciTableMean()
# Diff=2 Diff=1 Diff=0
#SD=1 [ 1.1, 2.9] [ 0.1, 1.9] [-0.9, 0.9]
#SD=2 [ 0.1, 3.9] [-0.9, 2.9] [-1.9, 1.9]
#SD=3 [-0.8, 4.8] [-1.8, 3.8] [-2.8, 2.8]
#==========
# Show how a potential confidence interval for a mean will look assuming
# standard deviations of 1, 2, or 5, a sample mean of 5, 3, or 1, and
# a sample size of 15.
ciTableMean(n1 = 15, diff.or.mean = c(5, 3, 1), SD = c(1, 2, 5), sample.type = "one")
# Mean=5 Mean=3 Mean=1
#SD=1 [ 4.4, 5.6] [ 2.4, 3.6] [ 0.4, 1.6]
#SD=2 [ 3.9, 6.1] [ 1.9, 4.1] [-0.1, 2.1]
#SD=5 [ 2.2, 7.8] [ 0.2, 5.8] [-1.8, 3.8]
#==========
# The data frame EPA.09.Ex.16.1.sulfate.df contains sulfate concentrations
# (ppm) at one background and one downgradient well. The estimated
# mean and standard deviation for the background well are 536 and 27 ppm,
# respectively, based on a sample size of n = 8 quarterly samples taken over
# 2 years. A two-sided 95% confidence interval for this mean is [514, 559],
# which has a half-width of 23 ppm.
#
# The estimated mean and standard deviation for the downgradient well are
# 608 and 18 ppm, respectively, based on a sample size of n = 6 quarterly
# samples. A two-sided 95% confidence interval for the difference between
# this mean and the background mean is [44, 100] ppm.
#
# Suppose we want to design a future sampling program and are interested in
# the size of the confidence interval for the difference between the two means.
# We will use ciTableMean to generate a table of possible confidence intervals
# by varying the assumed standard deviation and assumed differences between
# the means.
# Look at the data
#-----------------
EPA.09.Ex.16.1.sulfate.df
# Month Year Well.type Sulfate.ppm
#1 Jan 1995 Background 560
#2 Apr 1995 Background 530
#3 Jul 1995 Background 570
#4 Oct 1995 Background 490
#5 Jan 1996 Background 510
#6 Apr 1996 Background 550
#7 Jul 1996 Background 550
#8 Oct 1996 Background 530
#9 Jan 1995 Downgradient NA
#10 Apr 1995 Downgradient NA
#11 Jul 1995 Downgradient 600
#12 Oct 1995 Downgradient 590
#13 Jan 1996 Downgradient 590
#14 Apr 1996 Downgradient 630
#15 Jul 1996 Downgradient 610
#16 Oct 1996 Downgradient 630
# Compute the estimated mean and standard deviation for the
# background well.
#-----------------------------------------------------------
Sulfate.back <- with(EPA.09.Ex.16.1.sulfate.df,
Sulfate.ppm[Well.type == "Background"])
enorm(Sulfate.back, ci = TRUE)
#Results of Distribution Parameter Estimation
#--------------------------------------------
#
#Assumed Distribution: Normal
#
#Estimated Parameter(s): mean = 536.2500
# sd = 26.6927
#
#Estimation Method: mvue
#
#Data: Sulfate.back
#
#Sample Size: 8
#
#Confidence Interval for: mean
#
#Confidence Interval Method: Exact
#
#Confidence Interval Type: two-sided
#
#Confidence Level: 95%
#
#Confidence Interval: LCL = 513.9343
# UCL = 558.5657
# Compute the estimated mean and standard deviation for the
# downgradient well.
#----------------------------------------------------------
Sulfate.down <- with(EPA.09.Ex.16.1.sulfate.df,
Sulfate.ppm[Well.type == "Downgradient"])
enorm(Sulfate.down, ci = TRUE)
#Results of Distribution Parameter Estimation
#--------------------------------------------
#
#Assumed Distribution: Normal
#
#Estimated Parameter(s): mean = 608.33333
# sd = 18.34848
#
#Estimation Method: mvue
#
#Data: Sulfate.down
#
#Sample Size: 6
#
#Number NA/NaN/Inf's: 2
#
#Confidence Interval for: mean
#
#Confidence Interval Method: Exact
#
#Confidence Interval Type: two-sided
#
#Confidence Level: 95%
#
#Confidence Interval: LCL = 589.0778
# UCL = 627.5889
# Compute the estimated difference between the means and the confidence
# interval for the difference:
#----------------------------------------------------------------------
t.test(Sulfate.down, Sulfate.back, var.equal = TRUE)
#Results of Hypothesis Test
#--------------------------
#
#Null Hypothesis: difference in means = 0
#
#Alternative Hypothesis: True difference in means is not equal to 0
#
#Test Name: Two Sample t-test
#
#Estimated Parameter(s): mean of x = 608.3333
# mean of y = 536.2500
#
#Data: Sulfate.down and Sulfate.back
#
#Test Statistic: t = 5.660985
#
#Test Statistic Parameter: df = 12
#
#P-value: 0.0001054306
#
#95% Confidence Interval: LCL = 44.33974
# UCL = 99.82693
# Use ciTableMean to look how the confidence interval for the difference
# between the background and downgradient means in a future study using eight
# quarterly samples at each well varies with assumed value of the pooled standard
# deviation and the observed difference between the sample means.
#--------------------------------------------------------------------------------
# Our current estimate of the pooled standard deviation is 24 ppm:
summary(lm(Sulfate.ppm ~ Well.type, data = EPA.09.Ex.16.1.sulfate.df))$sigma
#[1] 23.57759
# We can see that if this is overly optimistic and in our next study the
# pooled standard deviation is around 50 ppm, then if the observed difference
# between the means is 50 ppm, the lower end of the confidence interval for
# the difference between the two means will include 0, so we may want to
# increase our sample size.
ciTableMean(n1 = 8, n2 = 8, diff = c(100, 50, 0), SD = c(15, 25, 50), digits = 0)
# Diff=100 Diff=50 Diff=0
#SD=15 [ 84, 116] [ 34, 66] [-16, 16]
#SD=25 [ 73, 127] [ 23, 77] [-27, 27]
#SD=50 [ 46, 154] [ -4, 104] [-54, 54]
#==========
# Clean up
#---------
rm(Sulfate.back, Sulfate.down)
# }
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