Suppose we are interested in testing the population median. The hypotheses are similar to those we have seen before but use the median, \(\eta\), instead of the mean.

If the hypotheses are based on the median, they would look like the following:

For the IRS example, the null and alternative are:

\(H_0\colon \eta=160\) vs \(H_a\colon \eta >160\)

Consider the test statistic, \(S^+\), where

\(S^+=\text{the number of observations greater than 160}\)

Under the null hypothesis, \(S^+\), should be about 50% of the observations. Therefore, \(S^+\) should have a binomial distribution with parameters \(n\) and \(p=0.5\). Let’s review and verify that it is a Binomial random variable.

• The number of trials, \(n\), is fixed and known. Here the number of trials equals the number of observations. Therefore, in this case, \(n\) is fixed and known.
• The outcomes of each trial can be categorized as either a "success" or a "failure", with the probability of success being \(p\). Observations can either be above the median (a "success") or below the median (a "failure") with the probability of being above the median being \(p=½\).
• The probability of "success" remains constant from trial to trial. The probability of being above the median remains the same for each observation.
• The trials are independent. Each of the observations is independent of the next, so we are okay here.

Now, back to our problem. To make a conclusion, we need to find the p-value. It is the probability of seeing what we see or something more extreme given the null hypothesis is true.

In the IRS example, let’s find \(S^+\), or in other words, let's find the number of observations that fall above 160. We find \(S^+=15\).

Using the Binomial distribution function, we can find the p-value as \(P(S^+\ge 15)\):

\begin{align} P(X\ge15)&=\sum_{i=15}^{30} {30\choose i}(0.5)^{30-i}(0.5)^i\\&=\sum_{i=15}^{30} {30\choose i} (0.5)^{30}\\&\approx 0.5722322 \end{align}

If we assume the significance level is 5%, then the p-value\(>0.05\). We would fail to reject the null hypothesis and conclude that there is no evidence in the data to suggest that the median is above 160 minutes.

This test is called the Sign Test and \(S^+\) is called the sign statistic. The Sign Test is also known as the Binomial Test.

Let's recap what we found. The research question was to see if it took longer than 160 minutes to complete the 1040 form. The measurement was the time in minutes to complete the form. Here is a summary:

Here, we have two opposite conclusions from each of the tests. Given the shape of the data, which do you think is the valid conclusion?

As you can see in the Minitab output, you can also find a confidence interval for the population median based on the sign statistic. As you can imagine, finding the confidence interval by hand is a bit tricky. The interpretation of the confidence interval for the median has the same template interpretation as the confidence interval for the population mean.

We present the details of the Sign Test because it can be found based on the material we covered so far in the course. For the next section, we present another test and how to do it in Minitab but leave out the details.

In this section, we briefly present the one-sample Wilcoxon test. This test was developed by Frank Wilcoxon in 1945. It is considered one of the first “nonparametric” tests developed.

The hypotheses are the similar to the ones presented previously for the Sign Test:

The Wilcoxon test needs additional assumptions, however. They are:

1. The random variable of interest is continuous
2. The probability distribution of the population is symmetric.

If we compare the assumptions of the Wilcoxon test to the Sign Test, the Wilcoxon test requires the distribution to be symmetric. For example, we should not be making conclusions for the IRS data using the Wilcoxon test because the data is right-skewed.

The test statistic is typically denoted as \(W\). We will not go into details on how this statistic is found as it involves ranks.

So far we discussed nonparametric tests for only one parameter. There are many tests for two parameters and for more than two parameters. There are also tests like Fisher’s Exact that will test for the association between two categorical variables.

In the table below, we give some examples of nonparametric tests. If you are interested, explore these tests on your own.

Point estimates are helpful to estimate unknown parameters but in order to make inference about an unknown parameter, we need interval estimates. Confidence intervals are based on information from the sampling distribution, including the standard error.

What if the underlying distribution is unknown? What if we are interested in a population parameter that is not the mean, such as the median? How can we construct a confidence interval for the population median?

If we have sample data, then we can use bootstrapping methods to construct a bootstrap sampling distribution to construct a confidence interval.

Bootstrapping is a topic that has been studied extensively for many different population parameters and many different situations. There are parametric bootstrap, nonparametric bootstraps, weighted bootstraps, etc. We merely introduce the very basics of the bootstrap method. To introduce all of the topics would be an entire class in itself.

Let’s show how to create a bootstrap sample for the median. Let the sample median be denoted as \(M\).

If we have the approximate distribution, we can find an estimate of the standard error of the sample median by finding the standard deviation of \(M_1,...,M_B\).

Sampling with replacement is important. If we did not sample with replacement, we would always get the same sample median as the observed value. The sample we get from sampling from the data with replacement is called the bootstrap sample.

Once we find the bootstrap sample, we can create a confidence interval. For a 90% confidence interval, for example, we would find the 5th percentile and the 95th percentile of the bootstrap sample.

You can create a bootstrap sample to find the approximate sampling distribution of any statistic, not just the median. The steps would be the same except you would calculate the appropriate statistic instead of the median.

Video: Bootstrapping

  Sampling R Code from the Bootstrapping Video

sampling.distribution <- function(n = 100, B = 1000, mean = 5, sd = 1, confidence = 0.95) {
  median <- rep(0, B)
  for (i in 1:B) {
    median[i] <- median(rnorm(n, mean = mean, sd = sd))
  }
  med.obs <- median(median)
  c.l <- round((1 - confidence) / 2 * B, 0)
  c.u <- round(B - (1 - confidence) / 2 * B, 0)
  l <- sort(median)[c.l]
  u <- sort(median)[c.u]
  cat(c.l / 1000 * 100, "-percentile:      ", l, "\n")
  cat("Median: ", med.obs, "\n")
  cat(c.u / 1000 * 100, "-percentile:      ", u, "\n")
  return(median)
}

bootstrap.median <- function(data, B = 1000, confidence = 0.95) {
  n <- length(data)
  median <- rep(0, B)
  for (i in 1:B) {
    median[i] <- median(sample(data, size = n, replace = T))
  }
  med.obs <- median(median)
  c.l <- round((1 - confidence) / 2 * B, 0)
  c.u <- round(B - (1 - confidence) / 2 * B, 0)
  l <- sort(median)[c.l]
  u <- sort(median)[c.u]
  cat(c.l / 1000 * 100, "-percentile:      ", l, "\n")
  cat("Median: ", med.obs, "\n")
  cat(c.u / 1000 * 100, "-percentile:      ", u, "\n")
  return(median)
}