University:
McMaster University
Course:
STATS 1LL3 | Introduction to Probability and Statistics
Academic year:

2023
Views:

270

Pages:

3
Author:

Sabrina S.

Video 22: Testing Goodness of Fit Is the die producing the right proportion of random numbers for all outcomes? Expected probability for each outcome of a die roll Outcome 1 2 3 4 5 6 Probability 1/6 1/6 1/6 1/6 1/6 1/6 Goodness-of-fit test we compare what we observe in a sample of data with what we would expect under a specified hypothesis We toss a die 60 times: Outcomes 1 2 3 4 5 6 Observed no. of outcomes 7 8 7 11 20 10 Probability of outcome 1/6 1/6 1/6 1/6 1/6 1/6 Expected no. of outcomes = 1/6 * 60 10 10 10 10 10 10 Step 1: Ho: X has the following PMF: p(x) = 1/6 X = 1, 2, 3, 4, 5, 6 0 elsewhere H1: X has some other distribution Step 2: Step 3: α = 0.05 CHI-SQUARED DISTRIBUTION the Chi-squared distribution is similar to the t-distribution in that it has degrees of freedom and its shape changes according to these degrees of freedom. • skewed to the right • is always positive TABLE 3 pg 327 to get the correct degrees of freedom: - the number of categories we are comparing less the number of parameters we estimate in the data less 1. -> 6 - 1 = 5 critical value from the table: 5 df and 0.05 = 11.07 Step 4: we will reject the null hypothesis (Ho) if our test statistic > 11.07 Step 5: Our test statistic is the comparison between what we observe and what we expect if Ho is true If Ho is true, we expect 1/6 of the 60 tosses to produce each outcome D² = Σ(k=1 to k) ((Oi - Ei)² / Ei) k = number of comparisons • if the discrepancy is large, ie large test statistic, this would suggest you are observing something quite different to what you expect if the test statistic is small, it would not constitute much evidence against Ho