The GoodnessofFit Test


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1 on the Lecture 49 Section 14.3 HampdenSydney College Tue, Apr 21, 2009
2 Outline 1 on the 2 3 on the 4 5
3 Hypotheses on the (Steps 1 and 2) (1) H 0 : H 1 : H 0 is false. (2) α = p 1 = 0.24 p 2 = 0.20 p 3 = 0.16 p 4 = 0.14 p 5 = 0.13 p 6 = 0.13.
4 Hypotheses on the (Step 3) (3) χ 2 = (O E) 2. E all cells (4) χ 2 =
5 Degrees of Freedom on the Definition (χ 2 degrees of freedom) In a goodnessoffit test, the number of degrees of freedom is one less than the number of cells. χ 2 distribution has an associated degrees of freedom, just like the t distribution. Each χ 2 distribution has a slightly different shape, depending on the number of degrees of freedom. For example, we let χ 2 5 denote the chisquare statistic with 5 degrees of freedom.
6 Degrees of Freedom on the Graph of χ
7 Degrees of Freedom on the Graph of χ
8 Degrees of Freedom on the Graph of χ
9 Degrees of Freedom on the Graph of χ
10 Degrees of Freedom on the Graph of χ
11 Degrees of Freedom on the Graph of χ
12 Degrees of Freedom on the Graph of χ
13 Degrees of Freedom on the Graph of χ
14 Degrees of Freedom on the Graph of χ
15 Degrees of Freedom on the Graph of χ
16 Degrees of Freedom on the Graphs of χ 2 1, χ2 2,..., χ
17 Properties of χ 2 on the chisquare distribution with df degrees of freedom has the following properties. χ 2 0. It is unimodal. It is skewed right (not symmetric!) µ χ 2 = df. σ χ 2 = 2df. If df is large, then χ 2 df is approximately normal with mean df and standard deviation 2df.
18 vs. Normal on the graph of χ 2 8 vs. N(8, 4)
19 vs. Normal on the graph of χ 2 32 vs. N(32, 8)
20 vs. Normal on the graph of χ vs. N(128, 16)
21 vs. Normal on the graph of χ vs. N(512, 32)
22  Probabilities on the Chisquare Probabilities Press 2nd DISTR. Select χ 2 cdf. Enter the lower endpoint, the upper endpoint, and the degrees of freedom. Press ENTER. probability appears in the display.
23  Probabilities on the Practice Find P(χ 2 3 > 6) with df = 3. Find P(20 < χ 2 25 < 30) with df = 25. Find P(χ 2 6 < 10) with df = 6. Find the probability that χ is within one standard deviation of its mean.
24 on the In our example, we found χ 2 = re are 6 categories (colors), so there are 5 degrees of freedom.
25 on the (Steps 5, 6, and 7) (5) pvalue = χ 2 cdf(9.7581,e99,5) = (6) Accept H 0. (7) colors fit the distribution given by the Mars Candy Company.
26 on the ( ) (1) H 0 : p 1 = 0.24, p 2 = 0.20, p 3 = 0.16, p 4 = 0.14, p 5 = 0.13, p 6 = 0.13 H 1 : H 0 is false. (2) α = (O E) 2 E. Color Blue Orange Green Yellow Brown Red (3) χ 2 = all cells (4) Observed (Expected) (27.36) (22.80) (18.24) (15.96) (14.82) (14.82) χ 2 = (5) pvalue = χ 2 cdf(9.7581,e99,5) = (6) Accept H 0. (7) color distribution in plain M&Ms is what the Mars Candy Company advertises it is.
27 on the on the Be careful when using the! re is a function called χ 2 , but it does not perform the goodnessoffit test. Some TI84s have a GOF function. GOF function does perform the goodnessoffit test.
28 on the on the fit test Put the observed counts in list L 1. Put the hypothetical proportions in list L 2. Multiply L 2 by the sample size and store as L 2. se are the expected counts. Calculate (L 1 L 2 ) 2 /L 2. Go to LIST > MATH and select sum (item #5). Enter Ans and press ENTER. value of χ 2 appears. n use χ 2 cdf to find the pvalue.
29  Male vs. Female Births on the ( fit test) Suppose we observe 1000 births and find that 520 are male and 480 are female. Does this indicate that male births and female births are not equally likely?
30  Male vs. Female Births on the ( fit test) (1) Let p 1 = proportion of male births. Let p 2 = proportion of female births. H 0 : p 1 = 0.50, p 2 = 0.50 H 1 : H 0 is not true. (2) α = (3) test statistic is χ 2 = all cells (O E) 2. E
31  Male vs. Female Births on the ( fit test) (4) We have the table Calculate Male Female Observed (Expected) (500) (500) χ 2 = ( ) = = 1.6 ( )2 500
32  Male vs. Female Births on the ( fit test) (5) pvalue is pvalue = χ 2 cdf(1.6,e99,1) = (6) Accept H 0. (7) proportion of male births is 50%.
33  Male vs. Female Births on the Perform the above test as a twotailed oneproportion z test. That is, let the alternative hypothesis be What is the pvalue? H 1 : p 1 p 2. What is the value of the test statistic z? Square that number. What do you get? See Univariate Relationships.
34 on the Homework Read Sections , pages Let s Do It! 14.2, Exercises 611, 14, 15, page 935.
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