Home > Model checking & comparison > Bayesian testing of a point null hypothesis

Bayesian testing of a point null hypothesis

Here is today’s example illustrating a Bayesian test of a point null hypothesis.  In Agresti and Franklin’s intro stats book, there is an example wondering if Americans work, on average, 40 hours per week.   If \mu represents the mean hours worked per week for all Americans, we are interested in testing H: \mu = 40 against the alternative A: \mu \neq 40.

In the example in Agresti and Franklin’s text, a sample of 868 workers were selected and one observes that the workers worked an average of 39.11 hours and one assumes the population standard deviation is equal to \sigma = 14.  The test statistic is

z = \frac{\sqrt{868}(\bar y - 40)}{14} = -1.87

and one computes a p-value of 0.061.  There seems to be moderately significant evidence that the mean working hours for all workers is different from 40.

Here’s an outline of a Bayesian test (I give more details in class):

1.  We assign probabilities to the hypotheses H and A.  Here we assume each hypothesis is equally likely.

2.  Under the hypothesis H, the prior for \mu is a point mass at the hypotheized value of 40.

3.  Under the alternative hypothesis, we’ll place a N(40, \tau) prior on \mu.  This reflects the belief that, under the alternative, values of \mu near 40 are more likely that values far from 40.

Under these assumptions, we give (in class) the posterior probability of the hypothesis H. This probability is programmed in the function mnormt.twosided in the LearnBayes package.  The inputs to this function are (1) the value to be tested, (2) the prior probability of H, (3) a vector of values of \tau and (4) a data vector consisting of \bar y, n, \sigma.  The outputs are the Bayes factor in support of H and the posterior probability of H.

Since it is not clear how to specify the prior standard deviation \tau — we compute the posterior probability for a range of values of \tau.  I write a short function to output the posterior probability as a function of \tau and then I use the curve function to plot this probability for values of \tau from 0.01 to 2.

What we see is that the posterior probability of \mu = 40 exceeds 0.35 for all values of \tau.  This suggests that the p-value overstates the evidence against the null hypothesis.

> data=c(39.11, 868, 14)
> prob.H=function(tau)
+   mnormt.twosided(40,.5,tau,data)$post

> curve(prob.H(x),from=.01,to=2,xlab=”TAU”,ylab=”P(H)”)testing



















  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: