- Construct a chi-square test of the Guinea Pig survival times data (p. 90) to test if the times are normally distributed.
- Continue work on 2.1 or 2.2.
- Histograms of the data, and "best-fit" normals:
- Works with binned data, observed versus expected
- Degrees of freedom: one less than the number of bins
- Would like expected > 5 in each bin.
- Examine Chi-square table, p. 369: interpret alpha.
- chi-square with 4 df
- Probability plots:
- Quantile plots for the two:
- Chi-square test results:
The chi-square test requires counts, so I split the data according to the normal distribution into 5 bins, each supposed to contain 20% of the data: the center 20%, the next "ring" of 20%, and so on. This meant that there was an expected number of counts of almost, but not quite, 5 for each bin (SAS and our authors would complain...!).
Using the sample mean and standard deviation,
- (chi-square-test survival) = 1.42
p=0.84 of a value more extreme
4 degrees of freedom - (chi-square-test recruitment) = 5.04
p=0.28 of a value more extreme
4 degrees of freedom
- (chi-square-test survival) = 1.00
p=0.91 of a value more extreme
4 degrees of freedom - (chi-square-test recruitment) = 4.17
p=0.38 of a value more extreme
4 degrees of freedom
Hence we don't reject a normal distribution on the basis of the chi-square test (although we have few values).
One way to deal with the issue of small numbers is to have a a look at what a normal distribution will do under the same conditions as the data. Here's a look at 1000 simulations of the same calculations:
- (chi-square-test survival) = 1.42
