A pharmaceutical company is interested in testing a potential blood pressure lowering medication. Their first examination considers only subjects that received the medication at baseline then two weeks later. The data are as follows (SBP in mmHg):
| Subject | Baseline | Week 2 |
|---|---|---|
| 1 | 140 | 132 |
| 2 | 138 | 135 |
| 3 | 150 | 151 |
| 4 | 148 | 146 |
| 5 | 135 | 130 |
Consider testing the hypothesis that there was a mean reduction in blood pressure? Give the P-value for the associated two sided T test.
The experiment is following to establish the potential benefits of the medication (i.e. the lowering of SBP) by measuring the baseline SBP and the SBP two weeks after the medication administration. The experiment design involves measurements for the same patients, hence the data are paired.
The test that will be used should be a paired \(t\) test, testing for the null hypothesis that the difference between the two groups is zero, i.e. \(H_0: \mu_{b}-\mu_{m} = 0\). The alternative hypothesis is \(H_0: \mu_{b}-\mu_{m} \neq 0\).
The \(t\) test is perfomed in
R via t.test.
For the alternative hypothesis, alternative = two.sided
should be seted.
For paired data, paired = TRUE should be seted.
Since no information is provided about the two groups variances, the
test should be performed under the assumption that the two variances are
not necessarily equal, i.e. var.equal = FALSE (this is the
default value so it doesn’t have to be explicitly stated).
x <- c(140, 138, 150, 148, 135)
y <- c(132, 135, 151, 146, 130)
t.test(y, x, paired = TRUE, alternative = "two.sided", var.equal = TRUE)$p.value## [1] 0.08652278Equivalently
t.test(y-x, alternative = "two.sided", var.equal = TRUE)$p.value## [1] 0.08652278A sample of 9 men yielded a sample average brain volume of 1,100cc and a standard deviation of 30cc. What is the complete set of values of \(\mu_0\) that a test \(H_{0} : \mu = \mu_0\) would fail to reject the null hypothesis in a two sided \(5\%\) Students t-test?
The normalized sample average random variable is following a \(t\) distribution with \(n-1\) degrees of freedom, i.e.
\[ \frac{\bar{X} - \bar{x}}{\text{SE}} \sim t(n-1), \] where \(n\) represents the number of samples, \(SE\) the standard error, \(\text{SE} = \sigma/\sqrt{n}\), defined via the samples standard deviation and \(\bar{x}\) the sample mean.
The \(95\%\) confidence interval corresponding to a significance level \(\alpha = 0.05\) is given by
\[ \mu \pm \text{SE} \times t_{1-\frac{\alpha}{2},n-1}, \]
where \(t_{1-\frac{\alpha}{2}\) represents the quantile function.
In this case, \(n = 9\), \(\bar{x} = 1100\) and \(\sigma = 30\).
n = 9
mu = 1100
sd = 30
SE = sd/sqrt(n)
mu + c(-1, 1) * SE * qt(0.975, n-1)## [1] 1076.94 1123.06Researchers conducted a blind taste test of Coke versus Pepsi. Each of four people was asked which of two blinded drinks given in random order that they preferred. The data was such that 3 of the 4 people chose Coke. Assuming that this sample is representative, report a P-value for a test of the hypothesis that Coke is preferred to Pepsi using a one sided exact test.
The hypothesis can be tested using the Binomial distribution, testing for the success probability \(p\) being greater than \(0.5\).
This boils down to a null hypothesis where we test for the success probability being exactly \(0.5\) since for every success probability greater than \(0.5\) the corresponding \(p\)- value will be greater.
The null hypothesis is \(H_{0}: p = 0.5\). The alternative hypothesis is \(H_{a}: p \geq 0.5\)
The \(p\)-value of observing 3 successes out of 4 trials, or a more extreme value (i.e. 4 successes out of 4) under the null hypothesis is:
\[
P\left( X > 3 \right) = {4 \choose 3} p^3(1-p) + {4 \choose 4} p^4 =
5 \times \frac{1}{2^4} = 0.3125.
\] This can be computed via pbinom, with
q = \(3-1 = 2\),
size $ = 4$ and prob \(=0.5\):
1 - pbinom(2, 4, 0.5)## [1] 0.3125or by simply selecting the upper tail:
pbinom(2, 4, 0.5, lower.tail = FALSE)## [1] 0.3125The exact binomial test is computed via binom.test with
the alternative = "greater"
binom.test(3, 4, p = 0.5, alternative = "less", conf.level = 0.95)##
## Exact binomial test
##
## data: 3 and 4
## number of successes = 3, number of trials = 4, p-value = 0.9375
## alternative hypothesis: true probability of success is less than 0.5
## 95 percent confidence interval:
## 0.0000000 0.9872585
## sample estimates:
## probability of success
## 0.75Infection rates at a hospital above 1 infection per 100 person days at risk are believed to be too high and are used as a benchmark. A hospital that had previously been above the benchmark recently had 10 infections over the last 1,787 person days at risk. About what is the one sided P-value for the relevant test of whether the hospital is below the standard? Given the model, could the observed rate being larger than 0.05 be attributed to chance?
Under the assumption that the count i.e. the number of infection is Poisson distributed,
\[ X \sim \text{Poisson}\left(x \mid t \lambda \right) \]
We test for the null hypothesis corresponding to a \(\lambda\) equal to the benchmark, \[ H_{0} : \lambda = 0.01 \] with the alternative hypothesis corresponding to a rate parameter bellow the standard
\[ H_{0} : \lambda < 0.01 \]
The \(p\)-value corresponds to the probability of observing 10 infections under the null hypothesis, i.e. \(\lambda = 0.01\) over the last 1,787 person days at risk, i.e.
\[ p\text{-value} = \text{Poisson}\left( X \leq 10 \mid 1787\lambda \right) \]
observedRate <- 10/1787
benchmarkRate <- 1/100
lambda = benchmarkRate
ppois(10, 1787*lambda, lower.tail = TRUE)## [1] 0.03237153Suppose that 18 obese subjects were randomized, 9 each, to a new diet pill and a placebo. Subjects’ body mass indices (BMIs) were measured at a baseline and again after having received the treatment or placebo for four weeks. The average difference from follow-up to the baseline (followup - baseline) was −3 kg/m2 for the treated group and 1 kg/m2 for the placebo group. The corresponding standard deviations of the differences was 1.5 kg/m2 for the treatment group and 1.8 kg/m2 for the placebo group. Does the change in BMI appear to differ between the treated and placebo groups? Assuming normality of the underlying data and a common population variance, give a \(p\)-value for a two sided \(t\) test.
mu_t <- -3; sd_t <- 1.5; n_t <- 9
mu_p <- 1; sd_p <- 1.8; n_p <- 9
mu <- mu_t - mu_p
df <- n_p + n_t - 2
pooledSd <- sqrt(((n_t-1)*sd_t**2 + (n_p-1)*sd_p**2)/(n_t + n_p - 2))
sd <- pooledSd * sqrt(1/n_t + 1/n_p)
x <- mu /sd
2*pt(x, df) ## [1] 0.00010251742*pnorm(x)## [1] 3.031546e-07Brain volumes for 9 men yielded a 90% confidence interval of 1,077 cc to 1,123 cc. Would you reject in a two sided 5% hypothesis test of \(H_0 : \mu = 1078\)?
The test wouldn’t be rejected for the \(90\%\) confidence interval, since the value \(\mu = 1078\) belongs to the confidence interval. The \(95\%\) confidence interval is wider than the \(90\%\) confidence interval, hence including \(\mu = 1078\). The hypothesis would not be rejected.
Researchers would like to conduct a study of \(100\) healthy adults to detect a four year mean brain volume loss of \(.01\) mm\(^3\). Assume that the standard deviation of four year volume loss in this population is \(.04\) mm\(^3\). About what would be the power of the study for a 5% one sided test versus a null hypothesis of no volume loss?
The null hypothesis is \[ H_0 : \mu = 0. \] and the alternative hypothesis is \[ H_0 : \mu > 0. \] with the sample mean \(\bar{x} = 0.1\), the population variance \(\sigma = 0.04\) and \(n=100\) the number of samples.
A 5% one sided test versus a null hypothesis of no volume loss means the significance level is \(\alpha = 0.1\).
\[ \alpha = \text{P}\left( \text{reject }H_{0} \mid H_0 \right) \] The test statistic is the normalized average random variable, i.e. $({X} - )/$ which follows a standard normal distribution \[ \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim \text{Normal} \left( x \mid 0, 1\right) . \] The significance level of \(\alpha = 0.1\) means that, under the null hypothesis \(\mu = 0\), “reject \(H_{0}\)” is done for observed sample mean values greater than \(1-\alpha/2\) quantile, i.e. \[ \frac{\bar{X}}{\sigma/\sqrt{n}} > z_{1-\frac{\alpha}{2}} \] The power is then: \[ \begin{align} \text{Power} = \text{P} \left( \text{reject } H_{0} \mid H_{1} \right) = \text{P} \left( \frac{\bar{X}}{\sigma/\sqrt{n}} > z_{1-\frac{\alpha}{2}} \mid \mu = 0.01 \right) = \text{P} \left( \frac{\bar{X}-0.01}{\sigma/\sqrt{n}} > z_{1-\frac{\alpha}{2}} - \frac{0.01}{\sigma/\sqrt{n}} \right) = \text{P} \left( Z > z_{1-\frac{\alpha}{2}} - \frac{0.01}{\sigma/\sqrt{n}} \right) \end{align} \]
alpha = 0.1
mu = 0.01
sigma = 0.04
n = 100
pnorm(qnorm(1-alpha/2)-mu/(sigma/sqrt(n)), lower.tail = FALSE)## [1] 0.8037649A direct way, using the power.t.test, specifing the
number of samples, the difference of means and the population standard
deviation is:
power.t.test(n = 100, delta = 0.01, sd = 0.04,
sig.level = 0.05,
type = 'one.sample',
alt = 'one.sided')$power## [1] 0.7989855Researchers would like to conduct a study of \(n\) healthy adults to detect a four year mean brain volume loss of \(.01\) mm\(^3\). Assume that the standard deviation of four year volume loss in this population is \(.04\) mm\(^3\). About what would be the value of \(n\) needed for \(90\)% power of type one error rate of 5% one sided test versus a null hypothesis of no volume loss?
Via the computations from the before, \[ \begin{align} 0.8 = \text{Power} = \text{P} \left( Z > z_{1-\frac{\alpha}{2}} - \frac{0.01}{\sigma/\sqrt{n}} \right) = \text{P} \left( Z > z_{1-\frac{\alpha}{2}} - \frac{\sqrt{n}}{4} \right) , \end{align} \] hence
\[ \begin{align} z_{1-\frac{\alpha}{2}} - \frac{\sqrt{n}}{4} = z_{\frac{1-0.8}{2}} \Rightarrow \sqrt{n} = 4 \left( z_{0.95}-z_{0.1} \right) \end{align} \]
n = (4 * (qnorm(0.95)-qnorm(0.1)))**2
n## [1] 137.0216Directly, via the power.t.test:
power.t.test(power = 0.9, delta = 0.01, sd = 0.04,
sig.level = 0.05,
type = 'one.sample',
alt = 'one.sided')$n## [1] 138.3856As you increase the type one error rate, \(\alpha\), what happens to power?
Power is decreasing.