Remember you should
This practice reviews the Tests for continuous data from one sample lecture.
From lecture! Consider if average height of males training at the Australian Institute of Sport is different than average of human population.
These are all one sample tests, but they differ in what we know. If we know the variance of our population, we use a z test (function in BSDA package).
sport <- read.table("http://www.statsci.org/data/oz/ais.txt", header = T)
library(BSDA)
z.test(sport[sport$Sex == "male", "Ht"], mu = 175.6, sigma.x=7)
One-sample z-Test
data: sport[sport$Sex == "male", "Ht"]
z = 14.292, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 175.6
95 percent confidence interval:
184.1474 186.8643
sample estimates:
mean of x
185.5059 If we don’t, we use a t-test
t.test(sport[sport$Sex == "male", "Ht"], mu = 175.6)
One Sample t-test
data: sport[sport$Sex == "male", "Ht"]
t = 12.658, df = 101, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 175.6
95 percent confidence interval:
183.9535 187.0583
sample estimates:
mean of x
185.5059 These both assume the means of the data are normal! If we want to relax that assumption, we can use the Wilcoxon test (also known as Mann-Whitney test, signed binary transform, or other terms!). This assumes the distribution of means is symmetric.
wilcox.test(sport[sport$Sex == "male", "Ht"], mu = 175.6)
Wilcoxon signed rank test with continuity correction
data: sport[sport$Sex == "male", "Ht"]
V = 5052, p-value = 5.714e-16
alternative hypothesis: true location is not equal to 175.6or the sign-test/media test.
SIGN.test(sport[sport$Sex == "male", "Ht"], md = 175.6)
One-sample Sign-Test
data: sport[sport$Sex == "male", "Ht"]
s = 90, p-value = 8.882e-16
alternative hypothesis: true median is not equal to 175.6
95 percent confidence interval:
183.9000 187.4684
sample estimates:
median of x
185.55
Achieved and Interpolated Confidence Intervals:
Conf.Level L.E.pt U.E.pt
Lower Achieved CI 0.9406 183.9 187.3000
Interpolated CI 0.9500 183.9 187.4684
Upper Achieved CI 0.9629 183.9 187.7000Note this is just transforming data to 1/0 and doing a binomial test!
above_175.6 <- nrow(sport[sport$Sex == "male" & sport$Ht > 175.6,])
binom.test(above_175.6, nrow(sport[sport$Sex == "male",]))
Exact binomial test
data: above_175.6 and nrow(sport[sport$Sex == "male", ])
number of successes = 90, number of trials = 102, p-value = 6.125e-16
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
0.8035103 0.9377091
sample estimates:
probability of success
0.8823529 We can also bootstrap the data.
number_of_simulations <- 1000
library(ggplot2)
boostrap_data<- sport[sport$Sex == "male", "Ht"]
boostrap_outcomes <- data.frame(mean = rep(NA, number_of_simulations), sd = NA)
for (i in 1:number_of_simulations){
iris_bootstrap <-sample(boostrap_data, length(boostrap_data), replace = T)
boostrap_outcomes$mean[i] <- mean(iris_bootstrap)
boostrap_outcomes$sd[i] <- sd(iris_bootstrap)
}
ggplot(boostrap_outcomes, aes(x=mean)) +
geom_histogram(color="black") +
labs(title=expression(paste("Bootstrapped means")),
x= "Mean value",
y= "Frequency")`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
and find associated quantile-based 95% confidence intervals:
quantile(boostrap_outcomes$mean, probs=c(.025, .975) ) 2.5% 97.5%
183.9988 187.1167 or using functions in the boot library
library(boot)
results <- boot(data=boostrap_data, statistic = function(x, inds) mean(x[inds]),
R=number_of_simulations)
ggplot(data.frame(results$t), aes(x=results.t)) +
geom_histogram(color="black") +
labs(title=expression(paste("Bootstrapped means")),
x= "Mean value",
y= "Frequency")`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
quantile( results$t, probs=c(.025, .975) ) 2.5% 97.5%
183.9750 186.9786 boot.ci(results) Warning in boot.ci(results): bootstrap variances needed for studentized
intervalsBOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 1000 bootstrap replicates
CALL :
boot.ci(boot.out = results)
Intervals :
Level Normal Basic
95% (184.0, 187.0 ) (184.0, 187.1 )
Level Percentile BCa
95% (184.0, 187.0 ) (183.9, 187.0 )
Calculations and Intervals on Original ScaleAnother option (recommended) is the boot.t.test function in the MKinfer package.
library(MKinfer)
boot.t.test(sport[sport$Sex == "male", "Ht"], mu = 175.6)
Bootstrap One Sample t-test
data: sport[sport$Sex == "male", "Ht"]
bootstrap p-value < 2.2e-16
bootstrap mean of x (SE) = 185.5055 (0.776676)
95 percent bootstrap percentile confidence interval:
183.9892 187.0285
Results without bootstrap:
t = 12.658, df = 101, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 175.6
95 percent confidence interval:
183.9535 187.0583
sample estimates:
mean of x
185.5059 Using the qqplot_example.R code, examine the following distributions and, for the continuous distributions (marked with a “*”), observe how a normal probability plot (qqplot) can be used to visually test for approximate normality.
For this question, its easiest to just source the main file and see what happens. When you source a script, it is run in R without showing any console output (but graphs and objects are still produced!). Try source(“https://raw.githubusercontent.com/jsgosnell/CUNY-BioStats/master/code_examples/qqplot_example.R”)
source("https://raw.githubusercontent.com/jsgosnell/CUNY-BioStats/master/code_examples/qqplot_example.R")
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Notice the spread of DATA of every distribution tend towards normality as sample size increases
Review the central_limit_theorem.R code if you need to convince/remind yourself how common normality of means is for even non-normal data. You can source the code using the same approach noted in Question 1.
Here we are focused on how the means look as sample size increases
#make sure you have VGAM library installed
source("https://raw.githubusercontent.com/jsgosnell/CUNY-BioStats/master/code_examples/central_limit_theorem.R")
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Make sure you are comfortable with null and alternative hypotheses for all examples. You should also feel comfortable graphing the data.
Seven observers were shown, for a brief period, a grill with 161 flies impaled and were asked to estimate the number. The results are given by Cochran (1954). Based on five estimates, they were 183.2, 149.0, 154.0, 167.2, 187.2, 158.0, and 143.0. Test the null hypothesis that the mean of the estimates is 161 flies.
flies <- c(183.2, 149.0, 154.0, 167.2, 187.2, 158.0, 143.0)
library(BSDA)
z.test(x=flies, mu = 161, sigma.x=sqrt(275))
One-sample z-Test
data: flies
z = 0.33276, p-value = 0.7393
alternative hypothesis: true mean is not equal to 161
95 percent confidence interval:
150.8010 175.3704
sample estimates:
mean of x
163.0857 Using a z-test, I found a test statistics of z~=0.33 .This corresponds to a p-value of 0.73. This p value is >.05, so I fail to reject the null hypothesis that the mean of the estimates is 161 flies.
t.test(x=flies,mu = 161)
One Sample t-test
data: flies
t = 0.32656, df = 6, p-value = 0.7551
alternative hypothesis: true mean is not equal to 161
95 percent confidence interval:
147.4576 178.7138
sample estimates:
mean of x
163.0857 Using a t-test, which is appropriate when the variance must be estimated from the sample and the means of the data may be assumed to follow a normal distribution, I found a test statistics of t6=0.32. This corresponds to a p-value of 0.76. This p-value is >.05, so I fail to reject the null hypothesis that the mean of the estimates is 161 flies.
wilcox.test(flies, mu=161)
Wilcoxon signed rank exact test
data: flies
V = 15, p-value = 0.9375
alternative hypothesis: true location is not equal to 161Using a Wilcoxon signed rank test, which is appropriate when normality assumptions can’t be met and the distribution of the data appears to be symmetric, I found a test statistics of V = 15 .This corresponds to a p-value of 0.94. This p-value is >.05, so I fail to reject the null hypothesis that the mean of the estimates is 161 flies.
SIGN.test(flies, md=161)
One-sample Sign-Test
data: flies
s = 3, p-value = 1
alternative hypothesis: true median is not equal to 161
95 percent confidence interval:
144.8857 185.9429
sample estimates:
median of x
158
Achieved and Interpolated Confidence Intervals:
Conf.Level L.E.pt U.E.pt
Lower Achieved CI 0.8750 149.0000 183.2000
Interpolated CI 0.9500 144.8857 185.9429
Upper Achieved CI 0.9844 143.0000 187.2000Using a sign test, which is appropriate when the data is continuous and other assumptions can’t be met, I found a test statistics of s = 3 .This corresponds to a p-value of 1. This p-value is >.05, so I fail to reject the null hypothesis that the median (Note change here) of the estimates is 161 flies.
Note there are several ways to load the data! You can make a list (since the list is short):
flies <- c(183.2, 149.0, 154.0, 167.2, 187.2, 158.0, 143.0 )or make a dataframe in a spreadsheet software (eg, Excel, Google Sheets) and then upload using a read.csv command. We did this in the introduction to R!
Yields of 10 strawberry plants in a uniformity trial are given by Baker and Baker (1953) as 239, 176, 235, 217, 234, 216, 318, 190, 181, and 225 g. Test the hypothesis that µ = 205 * Assuming variance = 1500
strawberries <- c(239, 176, 235, 217, 234, 216, 318, 190, 181, 225)
z.test(x=strawberries,mu = 205, sigma.x=sqrt(1500))
One-sample z-Test
data: strawberries
z = 1.4779, p-value = 0.1394
alternative hypothesis: true mean is not equal to 205
95 percent confidence interval:
199.0954 247.1046
sample estimates:
mean of x
223.1 Using a z-test, I found a test statistics of z=1.48. This corresponds to a p-value of 0.14. This p-value is >.05, so I fail to reject the null hypothesis that the population mean is equal to 205.
t.test(x=strawberries,mu = 205)
One Sample t-test
data: strawberries
t = 1.4164, df = 9, p-value = 0.1903
alternative hypothesis: true mean is not equal to 205
95 percent confidence interval:
194.1922 252.0078
sample estimates:
mean of x
223.1 Using a t-test, which is appropriate when the variance must be estimated from the sample and the means of the data may be assumed to follow a normal distribution, I found a test statistics of t9=1.42. This corresponds to a p-value of 0.19. This p-value is >.05, so I fail to reject the null hypothesis that the population mean is equal to 205.
wilcox.test(strawberries, mu=205)Warning in wilcox.test.default(strawberries, mu = 205): cannot compute exact
p-value with ties
Wilcoxon signed rank test with continuity correction
data: strawberries
V = 40.5, p-value = 0.2023
alternative hypothesis: true location is not equal to 205Using a Wilcoxon signed rank test, which is appropriate when normality assumptions can’t be met and the distribution of the data appears to be symmetric, I found a test statistics of V=40.5. This corresponds to a p-value of 0.20. This p-value is >.05, so I fail to reject the null hypothesis that the population mean is equal to 205.
SIGN.test(strawberries, md=205)
One-sample Sign-Test
data: strawberries
s = 7, p-value = 0.3437
alternative hypothesis: true median is not equal to 205
95 percent confidence interval:
183.9200 237.7022
sample estimates:
median of x
221
Achieved and Interpolated Confidence Intervals:
Conf.Level L.E.pt U.E.pt
Lower Achieved CI 0.8906 190.00 235.0000
Interpolated CI 0.9500 183.92 237.7022
Upper Achieved CI 0.9785 181.00 239.0000Using a sign test, which is appropriate when the data is continuous and other assumptions can’t be met, I found a test statistics of s= 7. This corresponds to a p-value of 0.34. This p-value is >.05,so I fail to reject the null hypothesis that the population median (Note change here) is equal to 205.
Evolutionary geneticists predicts the family sex ratio will be 80% female in broods of eagles that successfully fledge >3 young. Nests that fledge 3 or more chicks are very rare but a sample of 30 chicks are obtained from such nests and they yield 25 females and 5 males. Test the hypotheses that that: * a) the sex ratio is 50% females
#a
binom.test(25,30, p=.5)
Exact binomial test
data: 25 and 30
number of successes = 25, number of trials = 30, p-value = 0.0003249
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
0.6527883 0.9435783
sample estimates:
probability of success
0.8333333 A binomial test was used as we are comparing an observed proportion against a set value. Given a p-value of <.001, I reject the null hypothesis that the proportion of sons is equal to .5.
binom.test(25,30, .8)
Exact binomial test
data: 25 and 30
number of successes = 25, number of trials = 30, p-value = 0.8205
alternative hypothesis: true probability of success is not equal to 0.8
95 percent confidence interval:
0.6527883 0.9435783
sample estimates:
probability of success
0.8333333 A binomial test was used as we are comparing an observed proportion against a set value. Given a p-value of <.001, I fail to reject the null hypothesis that the proportion of sons is equal to .8.
Studies of flying snakes have led researchers to posit the mean undulation rate is 1.4 Hz. You wish to test this hypothesis using the small sample of undulation rates shown below. Create a small dataset of the paradise tree snake undulation rates and choose and justify a test you can use to assess the data.
Undulation rates (in Hz): 0.9, 1.4, 1.2, 1.2, 1.3, 2.0, 1.4, 1.6
snakes <- c(0.9, 1.4, 1.2, 1.2, 1.3, 2.0, 1.4, 1.6)
t.test(snakes, mu=1.4)
One Sample t-test
data: snakes
t = -0.21822, df = 7, p-value = 0.8335
alternative hypothesis: true mean is not equal to 1.4
95 percent confidence interval:
1.104098 1.645902
sample estimates:
mean of x
1.375 Using a t-test, which is appropriate when the variance must be estimated from the sample and the means of the data may be assumed to follow a normal distribution, I found a test statistics of t7=-.22. This corresponds to a p-value of 0.83. This p-value is >.05, so I fail to reject the null hypothesis that the mean undulation rate is 1.4 Hz.
Using data from Australian athletes (http://www.statsci.org/data/oz/ais.html for details), determine if the average male training at the Australian Institute of Sport differs in weight from the average Australian male (85.9 kg) using bootstrapping techniques. Data at
sport <- read.table("http://www.statsci.org/data/oz/ais.txt", header = T,
stringsAsFactors = T)library(MKinfer)
boot.t.test(sport[sport$Sex == "male", "Wt"], mu= 85.9)
Bootstrap One Sample t-test
data: sport[sport$Sex == "male", "Wt"]
bootstrap p-value = 0.008801
bootstrap mean of x (SE) = 82.53556 (1.22002)
95 percent bootstrap percentile confidence interval:
80.21223 84.95838
Results without bootstrap:
t = -2.7487, df = 101, p-value = 0.007089
alternative hypothesis: true mean is not equal to 85.9
95 percent confidence interval:
80.08671 84.96035
sample estimates:
mean of x
82.52353 Using a bootstrap test wtih 10,000 samples, we found a p-value of .007; we thus reject the null hypothesis that males training at the AIS have the same weight as the average Australian male. Data indicated they weigh less.