April 22 — Hypothesis Testing – M&M’srpruim/courses/m243/stob/2005-04-25-chisq.… · April...

1
April 22 — Hypothesis Testing – M&M’s 1. Advertised distribution of M&M colors; brown 13% yellow 14% red 13% blue 24% orange 20% green 16% 2. The multinomial distribution. k + 1 paramters: n, π 1 ,...,π k such that π 1 + ··· π k = 1. Interpretation: X 1 ,...,X k have a multinomial distribution if there are n independent trials of an experiment which can result in one of k distinct outcomes and X i is the number of trials that result in the i th outcome. 3. Want to test the hypothesis H 0 : π 1 = .13 π 2 = .14 π 3 = .13 π 4 = .24 π 5 = .20 π 6 = .16 or in general the hypothesis H 0 : π 1 = π 1,0 ... π k = π k,0 4. Development of a test statistic. Given the result of the experiment x 1 ,...,x k want a statistic T that measures deviation of the result from what would be expected if H 0 is true. 5. Candidate: 6. Decision: reject H 0 if T is too large. 7. If the null hypothesis is true then the distribution of T is approximately a distribution that is called the distribution. It has one parameter k - 1, called the degrees of freedom. 8. Decision: reject H 0 if Homework, Due Monday, May 2, 1. Read Devore and Farnum 8.3, pages 373-376. 2. Do problems 8.42–44.

Transcript of April 22 — Hypothesis Testing – M&M’srpruim/courses/m243/stob/2005-04-25-chisq.… · April...

Page 1: April 22 — Hypothesis Testing – M&M’srpruim/courses/m243/stob/2005-04-25-chisq.… · April 22 — Hypothesis Testing – M&M’s 1. Advertised distribution of M&M colors; brown

April 22 — Hypothesis Testing – M&M’s

1. Advertised distribution of M&M colors;

brown 13% yellow 14%red 13% blue 24%orange 20% green 16%

2. The multinomial distribution. k + 1 paramters: n, π1, . . . , πk such that π1 + · · ·πk = 1. Interpretation:X1, . . . , Xk have a multinomial distribution if there are n independent trials of an experiment which canresult in one of k distinct outcomes and Xi is the number of trials that result in the ith outcome.

3. Want to test the hypothesis

H0 : π1 = .13 π2 = .14 π3 = .13 π4 = .24 π5 = .20 π6 = .16

or in general the hypothesisH0 : π1 = π1,0 . . . πk = πk,0

4. Development of a test statistic. Given the result of the experiment x1, . . . , xk want a statistic T thatmeasures deviation of the result from what would be expected if H0 is true.

5. Candidate:

6. Decision: reject H0 if T is too large.

7. If the null hypothesis is true then the distribution of T is approximately a distribution that is called thedistribution. It has one parameter k − 1, called the degrees of freedom.

8. Decision: reject H0 if

Homework, Due Monday, May 2,

1. Read Devore and Farnum 8.3, pages 373-376.

2. Do problems 8.42–44.