Re: Basic normal and t table questions
- To: mathgroup at smc.vnet.net
- Subject: [mg110064] Re: Basic normal and t table questions
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Tue, 1 Jun 2010 04:19:39 -0400 (EDT)
On 5/30/10 at 6:48 AM, canopus56 at yahoo.com (Canopus56) wrote: >I would like to use Mathematica to calculate exact values and >inverse values from the standard normal table and Student's t >distribution table. >For example, the following returns the cdf equvialent to standard >normal distribution table for a known mu, sigma (STD) and target >x-bar: >(1 - CDF[NormalDistribution[187000, 2181.9716], 190000]) >The following returns the two-sided t-distribution value for a known >z-statistic: >StudentTPValue[-0.1373, 6, TwoSided -> True] >Q1) How do I get the inverse of these, which the equivalent of >reading a cdf table and a t-distribution table backwards, i.e. - >a) Given a cdf, how do I return the z-statistic from a standard >normal distribution? >b) Given a P-Value and sampling distribution size, how did I return >the z-star critical test value? The inverse function to CDF in Mathematica is Quantile. Using your first example, I can do as follows: In[39]:= p = (1 - CDF[NormalDistribution[187000, 2181.9716], 190000]) Out[39]= 0.0845807 In[40]:= zscore = Quantile[NormalDistribution[], 1 - p] Out[40]= 1.3749 The standard z-score. And to show Quantile is the inverse function to CDF In[41]:= Quantile[NormalDistribution[187000, 2181.9716], 1 - p] Out[41]= 190000. or using the computed zscore In[42]:= zscore 2181.9716 + 187000 Out[42]= 190000. Quantile also solves the second problem as well. But here you need to understand what is being done after loading HypothesisTesting and using StudentTPValue First, doing In[43]:= StudentTPValue[-0.1373, 6, TwoSided -> True] Out[43]= TwoSidedPValue->0.895285 and doing In[45]:= p = 2 CDF[StudentTDistribution[6], -0.1373] Out[45]= 0.895285 results in the same p-Value. So, In[46]:= Quantile[StudentTDistribution[6], p/2] Out[46]= -0.1373 gives the desired result