diff --git a/boxplot.go b/boxplot.go
index d745545..a0332f6 100644
--- a/boxplot.go
+++ b/boxplot.go
@@ -45,6 +45,7 @@ func main() {
median := stat.Quantile(0.50, stat.Empirical, ttls, nil)
q3 := stat.Quantile(0.75, stat.Empirical, ttls, nil)
max := stat.Quantile(1.00, stat.Empirical, ttls, nil)
+ mean := stat.Mean(ttls, nil)
odata = append(odata, []string{
cc,
strconv.Itoa(len(ttls)),
@@ -53,6 +54,7 @@ func main() {
strconv.FormatFloat(median, 'f', -1, 64),
strconv.FormatFloat(q3, 'f', -1, 64),
strconv.FormatFloat(max, 'f', -1, 64),
+ strconv.FormatFloat(mean, 'f', -1, 64),
})
}
diff --git a/poster.png b/poster.png
index 19800eb..592e029 100644
Binary files a/poster.png and b/poster.png differ
diff --git a/poster.svg b/poster.svg
index e54c729..247ef2d 100644
--- a/poster.svg
+++ b/poster.svg
@@ -75916,8 +75916,8 @@
inkscape:pageopacity="0"
inkscape:pageshadow="2"
inkscape:zoom="0.42426407"
- inkscape:cx="2232.7531"
- inkscape:cy="1879.9857"
+ inkscape:cx="2262.3592"
+ inkscape:cy="2018.4275"
inkscape:document-units="mm"
inkscape:current-layer="layer1"
showgrid="false"
@@ -76195,7 +76195,7 @@
style="font-size:65.00000954%;baseline-shift:sub"
id="flowSpan51354">a: μ1 < μ1 > μ2SRS: Let's assume the sample to be a simple random sample. Independence: Independence: Because of randomisation, we are willing to take the two groups as independent samples. We are not sampling without replacement from a population of samples. Normality: Normality: From the box plots, there are no visible outliers and the normality assumption seems reasonable. t = t = 95.840P-value = Using conservative estimate, df = 1037955 - 1 = 1037954P-value = 0 (really close to zero)Since the P-value is _ than 0.05, we _ H0 as there is _sufficient evidence to conclude that the mean latency to the US is lower than the mean latency to the Netherlands. Since the P-value is less than 0.05, we reject H0 and conclude that there is sufficient evidence to suggest that the mean latency to the US is lower than the mean latency to the Netherlands.
+