In the previous chapter we computed the sampling distribution using resampling techniques to be able to perform hypothesis tests. If the null distribution was already known, or could be computed based on a few assumptions, resampling would not be necessary.
We can follow the same steps as before to perform a hypothesis test:
Define \(H_0\) and \(H_1\)
Select an appropriate significance level, \(\alpha\)
Select appropriate test statistic, \(T\), and compute the observed value, \(t_{obs}\)
Assume that the \(H_0\) and derive the null distribution of the test statistic based on appropriate assumptions.
Compare the observed value, \(t_{obs}\), with the null distribution and compute a p-value. The p-value is the probability of observing a value at least as extreme as the observed value, if \(H_0\) is true.
Based on the p-value either accept or reject \(H_0\).
In this section we will present a few common situation in which the null distribution can be described parametrically.
3.1 One sample, mean
A one sample test of means compares the mean of one sample to a prespecified value.
For example, we might know that the weight of a mouse on normal diet is normally distributed with mean 24.0 g and standard deviation 3 g and want to compare the weight of a sample of 10 mice on high-fat diet to the known mean value for mice on normal diet.
The hypotheses:
\[H_0: \mu = \mu_0\]\[H_1: \mu \neq \mu_0\]
The alternative hypothesis, \(H_1,\) above is for the two-sided hypothesis test. Other options are the one-sided \(H_1\); \(H_1: \mu > \mu_0\) or \(H_1: \mu < \mu_0\).
If \[X \sim N(\mu, \sigma)\] (this could for example be the weight of a mouse on high-fat diet) then the sample mean \[\bar X \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right).\]
If \(\sigma\) is known under the null hypothesis, then the test statistic
\[Z = \frac{\bar X - \mu_0}{\frac{\sigma}{\sqrt{n}}}\] is normally distributed, \(\sim N(0,1)\).
In many situations the the population standard deviation is not known, but can instead be estimated from the sample. For large sample size \(n\), \(\sigma\) can be replaced by the sample standard deviation \(s\) and the test statistic will still be normally distributed according to the central limiit theorem*.
For small \(n\) and unknown \(\sigma\), we can use the one-sample t-test, that uses the test statistic
\[T = \frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}},\]
which is t-distributed with \(df=n-1\) degrees of freedom.
Once we have an appropriate test statistic, \(T\) with known distribution, we can compute the observed value, \(t_{obs}\) and use the null distribution to compute the p-value, \(P(|T| \geq |t_{obs}|)\).
In R, the functions pnorm and pt are useful for computing these probabilities. The one-sample t-test can also be performed using the function t.testlike this;
##The observed 10 mouse weights;x <-c(25, 30, 23, 18, 31, 24, 39, 26, 36, 29, 23, 32)## Under the null hypothesis mu=24##Perform the t-test to investigate if the null hypotheis can be accepted, i.e.##if the sample comes from a normal distribution with expected value 24.t.test(x, mu=24)
One Sample t-test
data: x
t = 2.3153, df = 11, p-value = 0.04092
alternative hypothesis: true mean is not equal to 24
95 percent confidence interval:
24.19742 31.80258
sample estimates:
mean of x
28
3.2 Two samples, mean
A two sample test of means is used to determine if two population means are equal.
Two independent samples are collected (one from each population) and the means are compared. This test can for example be used to determine if a treatment group is different compared to a control group, in terms of the mean of a property of interest.
The null hypothesis;
\[H_0: \mu_2 = \mu_1\] The alternative hypothesis can either be two sided
Assume observations from both populations are normally distributed;
\[
\begin{aligned}
X_1 \sim N(\mu_1, \sigma_1) \\
X_2 \sim N(\mu_2, \sigma_2)
\end{aligned}
\] Then it follows that the sample means will also be normally distributed;
If \(H_0\) is true: \[D = \bar X_2 - \bar X_1 = N\left(0, \sqrt{\frac{\sigma_2^2}{n_2} + \frac{\sigma_1^2}{n_1}}\right)\]
The test statistic: \[Z = \frac{\bar X_2 - \bar X_1}{\sqrt{\frac{\sigma_2^2}{n_2} + \frac{\sigma_1^2}{n_1}}}\] is standard normal, i.e. \(Z \sim N(0,1)\).
However, note that the test statistic require the standard deviations \(\sigma_1\) and \(\sigma_2\) to be known.
What if the population standard deviations are not known?
If the sample sizes are large, we can replace the known standard deviations with our sample standard deviations and according to the central limit theorem assume that
If it can be assumed that \(X_1\) and \(X_2\) both are normally distributed and have equal variances, Studen’t t-test can be used. For equal variances the pooled sample variance can be computed;
is t-distributed with \(n_1+n_2-2\) degrees of freedom.
For unequal variances Welch’s t-test can instead be used.
The t-test is implemented in R, e.g. in the function t.test in the R-package stats, both Student’s t-test for equal variances and Welch’s t-test for unequal variances.
3.3 One sample, proportions
Example 3.1 (Pollen allergy) Let’s get back to the pollen example!
Assume that the proportion of pollen allergy in Sweden is known to be \(0.3\). Observe 100 people from Uppsala, 42 of these were allergic to pollen. Is there a reason to believe that the proportion of pollen allergic in Uppsala \(\pi > 0.3\)?
Null and alternative hypotheses
\[H_0:\, \pi=\pi_0\]
\[H_1:\, \pi>\pi_0,\]
where \(\pi_0=0.30\) in this example. Other potential alternative hypothesis are \(H_1: \pi<\pi_0\) or \(H_1:\pi \neq \pi_0,\) but in this particular example we are only interested in the alternative that \(\pi > \pi_0\).
Significance level
Set the signifgicance level to \(\alpha=0.05\).
Test statistic
Here, we will use \(X\), the number of allergic persons in a random sample of size \(n=100\). The observed value is \(x_{obs} = 42\).
Null distribution
\(X\) is binomially distributed, so there is no need to use resampling here, we can use the binomial distribution to answer the question.
p-value
The probability of \(x_{obs}\) or something higher,
As \(p<0.05\)\(H_0\) is rejected and we conclude that there is reason to believe that the proportion of pollen allergic in Uppsala is higher than 0.3.
This p-value can also be computed using the exact binomial test;
binom.test(42, 100, 0.3, alternative="greater")
Exact binomial test
data: 42 and 100
number of successes = 42, number of trials = 100, p-value = 0.007174
alternative hypothesis: true probability of success is greater than 0.3
95 percent confidence interval:
0.3364797 1.0000000
sample estimates:
probability of success
0.42
An alternative approach is to use the central limit theorem and use the normal approximation.
As a result of the central limit theorem, the distribution of number or proportion of allergic individuals in a sample of size \(n\) is approximately normal. At least if the sample is large enough. A rule of thumb is that the sample size should be \(n>30\).
Here, the sample size is 100!
This test of proportions using the normal approximation is implemented in in the r-function prop.test.
1-sample proportions test with continuity correction
data: 42 out of 100, null probability 0.3
X-squared = 6.2976, df = 1, p-value = 0.006045
alternative hypothesis: true p is greater than 0.3
95 percent confidence interval:
0.3372368 1.0000000
sample estimates:
p
0.42
But can of course also be calculated using only the normal distribution table.
The normal distribution has two parameters, mean and standard deviation.
From the binomial distribution we know that \(E[X] = n\pi\) and \(var(X) = n\pi(1-\pi)\).
With this null distribution and ur observed value \(x_{obs} = 42\), the p-value can be computed.
\(p = P(X \geq 42)\)
As we are now approximating a discrete distribution using a continuous distribution we need to use a trick called continuiuty correction, which simply means that we let each integer be represented by y a region \(\pm 0.5\) from its value. Hence,
where \(P\) is the proportion in the merged sample of size \(n_1 + n_2\). \(Z \in N(0,1)\) and p-value can be computed using the standard normal distribution.
Also the two sample proportions test is implemented in the function prop.test.
3.5 Variance
The test of equal variance in two groups is based on the null hypothesis
\[H_0: \sigma_1^2 = \sigma_2^2\]
If the two samples both come from populations with normal distributions, the sample variances (also random variables) are