Information Technology Reference
In-Depth Information
Discussion
I subscribe to a stock market newsletter that is well written for the most part, but
includes a section purporting to identify stocks that are likely to rise. It does this by
looking for a certain pattern in the stock price. It recently reported, for example, that
a certain stock was following the pattern. It also reported that the stock rose six times
after the last nine times that pattern occurred. The writers concluded that the proba-
bility of the stock rising again was therefore 6/9 or 66.7%.
Using prop.test , we can obtain the confidence interval for the true proportion of times
the stock rises after the pattern. Here, the number of observations is n = 9 and the
number of successes is x = 6. The output shows a confidence interval of (0.309, 0.910)
at the 95% confidence level:
> prop.test(6, 9)
1-sample proportions test with continuity correction
data: 6 out of 9, null probability 0.5
X-squared = 0.4444, df = 1, p-value = 0.505
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.3091761 0.9095817
sample estimates:
p
0.6666667
The writers are pretty foolish to say the probability of rising is 66.7%. They could be
leading their readers into a very bad bet.
By default, prop.test calculates a confidence interval at the 95% confidence level. Use
the conf.level argument for other confidence levels, like this:
> prop.test(n, x, p, conf.level=0.99) # 99% confidence level
1.14 Comparing the Means of Two Samples
Problem
You have one sample each from two populations. You want to know if the two popu-
lations could have the same mean.
Solution
Perform a t test by calling the t.test function:
> t.test(x, y)
By default, t.test assumes that your data is not paired. If the observations are paired
(i.e., if each x i is paired with the corresponding y i ), then specify paired=TRUE :
> t.test(x, y, paired=TRUE)
 
Search WWH ::




Custom Search