University of Sydney
Recap
Example: Soft drink cans
Soft drink contents in a pack of six cans (in milliliters) are:
Is the soft drink content less than the 375 mL claimed on the label?
Example: Soft drink cans
Revision of hypothesis testing
Hypothesis testing
- Hypotheses:   \(H_0: \ \theta=\theta_0\) vs \(H_1: \ \theta>\theta_0\)
- Test statistic:   \(T=f(X_1, X_2,..., X_n)\).
- Assumptions:   Sample \(X_1, X_2, ..., X_n \sim F_{\theta}\)
- Significance:   \(p\mbox{-value} = P(T\geq t_0)\)
- Decision:   If \(p\mbox{-value} < \alpha\), there is evidence against \(H_0\).
Hypothesis
- The statement against which you search for evidence is called the null hypothesis, and is denoted by \(H_0\). It is generally a “no difference” statement.
- The statement you claim is called the alternative hypothesis, and is denoted by \(H_1\).
Typical Hypotheses:
  \(H_0 : \theta = \theta_0\)
    VS
  \(H_1 : \theta > \theta_0\) (upper-side alternative)
  \(H_1 : \theta < \theta_0\) (lower-side alternative)
  \(H_1 : \theta \ne \theta_0\) (two-sided alternative)
Test statistic
- Since observations \(X_i\) vary from sample to sample we can never be sure whether \(H_0\) is true or not.
- We use a test statistic \(T = f(X_1, ..., X_n)\) to test if the data are consistent with \(H_0\) such that…
- the distribution of \(T\) is known assuming \(H_0\) is true;
- the large (positive or negative depending on \(H_1\)) observed value of \(T\) is taken as evidence of poor agreement with \(H_0\).
Assumptions
- Each observation \(X_1, X_2, ..., X_n\) is chosen at random from a population.
- We say that such random variables are iid (independently and identically distributed).
Significance
- The \(p\)-value (observed significance level) is defined as the probability of getting the observed test statistic \(t_0\) and more extreme values assuming that \(H_0\) is true…
  \(p\)-value \(= P(T \ge t_0 )\); for \(H_1: \theta > \theta_0\)
  \(p\)-value \(= P(T \le -t_0 )\); for \(H_1: \theta < \theta_0\)
  \(p\)-value \(= P(T \ge t_0 )\); for \(H_1: \theta \ne \theta_0\)
## TableGrob (1 x 3) "arrange": 3 grobs ## z cells name grob ## 1 1 (1-1,1-1) arrange gtable[layout] ## 2 2 (1-1,2-2) arrange gtable[layout] ## 3 3 (1-1,3-3) arrange gtable[layout]
Decision
An observed large positive or negative value of \(t_0\) and hence small \(p\)-value is taken as evidence of poor agreement with \(H_0\).
If the p-value is small, then either \(H_0\) is true and the poor agreement is due to an unlikely event, or \(H_0\) is false. Therefore..
- the smaller the p-value, the stronger the evidence against \(H_0\) in favour of \(H_1\).
A large \(p\)-value does not mean that there is evidence that \(H_0\) is true
The level of significance, \(\alpha\), is the strength of evidence needed to reject \(H_0\) (often \(\alpha = 0.05\)).
t-test
Suppose we have a sample \(X_1, X_2, ..., X_n\) of the size \(n\) drawn from a normal population with an unknown variance \(\sigma^2\). Let \(x_1, x_2, ..., x_n\) be the observed values. We want to test the population mean \(\mu\).
t-test
Suppose we have a sample \(X_1, X_2, ..., X_n\) of the size \(n\) drawn from a normal population with an unknown variance \(\sigma^2\). Let \(x_1, x_2, ..., x_n\) be the observed values. We want to test the population mean \(\mu\).
- Hypothesis: \(H_0: \mu = \mu_0\) vs \(H_1: \mu > \mu_0, \ \mu < \mu_0\) (one-sided) or \(\mu \ne \mu_0\) (two-sided)
- Test statistic: \(\tau_0 = \frac{\bar{x} - \mu_0}{s /\sqrt{n}}\)
- Assumptions: \(X_i\) are iid rv and follow \(N(\mu, \sigma^2)\).
- \(P\)-value: \(P(t_{n-1} \ge \tau_0)\), \(P(t_{n-1} \le \tau_0)\) or \(P(t_{n-1} \ge |\tau_0|)\)
- Decision: Reject \(H_0\) in favour of \(H_1\) if the p-value is less than \(\alpha\).
Example: Soft drink cans
Soft drink contents in a pack of six cans (in milliliters) are:
Is the soft drink content less than the 375 mL claimed on the label?
x = c(374.8, 375, 375.3, 374.8, 374.4, 374.9) mean(x)
## [1] 374.8667
sd(x)
## [1] 0.294392
In R
t.test(x, mu = 375, alternative = "less")
## ## One Sample t-test ## ## data: x ## t = -1.1094, df = 5, p-value = 0.1589 ## alternative hypothesis: true mean is less than 375 ## 95 percent confidence interval: ## -Inf 375.1088 ## sample estimates: ## mean of x ## 374.8667
What if you have two samples?
There are times that we want to test if the population means of two samples are different.
Here we are left with two possible scenarios
- The two samples are independent
- The two samples are dependent
Discuss and write down a few examples of situations were the two samples would be independent or dependent.
Smokers and blood platelet aggregation
Blood samples are taken from \(11\) smokers and \(11\) non-smokers to measure aggregation of blood platelets.
Non.Smokers = c(25, 25, 27, 44, 30, 67, 53, 53, 52, 60, 28) Smokers = c(27, 29, 37, 36, 46, 82, 57, 80, 61, 59, 43) c(mean(Smokers), mean(Non.Smokers), sd(Smokers), sd(Non.Smokers))
## [1] 50.63636 42.18182 18.89589 15.61293
Is the aggregation affected by smoking?
Smokers and blood platelet aggregation
library(ggplot2)
Blood.platelet.levels = c(Non.Smokers, Smokers)
Group = rep(c("Non-smokers", "Smokers"), c(length(Non.Smokers), length(Smokers)))
qplot(Group, Blood.platelet.levels, geom = "boxplot")
Two-sample t-test
Hypotheses: \(H_0: \mu_x=\mu_y\) vs \(H_1: \mu_x>\mu_y, \ \mu_x<\mu_y\) or \(\mu_x\not =\mu_y\)
Test statistic: \(\tau_0 = \frac{{\bar x} - {\bar y}}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\) where \(s^2_p = \frac{(n_x-1) s_{x}^2 + (n_y-1) s_{y}^2}{n_x+n_y-2}\).
Assumptions: \(X_1,...,X_{n_x}\) are iid \(N(\mu_X,\sigma^2)\), \(Y_1,...,Y_{n_y}\) are iid \(N(\mu_Y,\sigma^2)\) and \(X_i's\) are independent of \(Y_i's\). Hence \(\tau_0 \sim t_{n_x+n_y-2}\)
\(P\)-value: \(P(t_{n_x+n_y-2}\le \tau_{0})\), \(P(t_{n_x+n_y-2}\ge \tau_{0})\) or \(2P(t_{n_x+n_y-2} \ge |\tau_{0}|)\).
Decision: If \(p\mbox{-value}<\alpha\), there is evidence against \(H_0\). If \(p\mbox{-value}> \alpha\), the data are consistent with \(H_0\).
In R
t.test(Smokers, Non.Smokers, alternative = "two.sided")
## ## Welch Two Sample t-test ## ## data: Smokers and Non.Smokers ## t = 1.144, df = 19.313, p-value = 0.2666 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -6.997031 23.906122 ## sample estimates: ## mean of x mean of y ## 50.63636 42.18182
Paired sample t-test
Example: Smoking
Blood samples from \(11\) individuals before and after they smoked a cigarette are used to measure aggregation of blood platelets.
Before = c(25, 25, 27, 44, 30, 67, 53, 53, 52, 60, 28) After = c(27, 29, 37, 36, 46, 82, 57, 80, 61, 59, 43)
Is the aggregation affected by smoking?
Graphical exploration
df = data.frame(Difference = After - Before, group = 1)
p1 = ggplot(df, aes(group, Difference)) + geom_boxplot() + theme_bw() + geom_hline(yintercept = 0,
linetype = "dashed") + ylab("Difference in blood platelet levels") + theme(axis.title.x = element_blank(),
axis.text.x = element_blank(), axis.ticks.x = element_blank())
p1
Paired sample t-test
t.test(Before - After)
## ## One Sample t-test ## ## data: Before - After ## t = -2.9065, df = 10, p-value = 0.01566 ## alternative hypothesis: true mean is not equal to 0 ## 95 percent confidence interval: ## -14.93577 -1.97332 ## sample estimates: ## mean of x ## -8.454545
t.test(Before, After, paired = TRUE)
## ## Paired t-test ## ## data: Before and After ## t = -2.9065, df = 10, p-value = 0.01566 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -14.93577 -1.97332 ## sample estimates: ## mean of the differences ## -8.454545
Paired sample t-test
Examples
- Will a paired-sample t-test always give smaller p-values than a two-sample t-test?
- Discuss and write down a few examples that you think illustrate your answer.