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13.1 Test for Goodness of Fit.  Perform and analyze a chi-squared test for goodness of fit.

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Presentation on theme: "13.1 Test for Goodness of Fit.  Perform and analyze a chi-squared test for goodness of fit."— Presentation transcript:

1 13.1 Test for Goodness of Fit

2  Perform and analyze a chi-squared test for goodness of fit.

3  Chi-Squared ( ) test for goodness of fit- a single test that can be applied to see if the observed sample distribution is different from the hypothesized population distribution.  H₀:The distribution of the sample data is the SAME as the population  Ha:The distribution of the sample data is DIFFERENT than the population

4 Chi-Squared test statistic:

5  Random Sample  All expected counts are greater than 1  No more than 20% of expected counts are less than 5  Degrees of freedom = n-1 where, n= # of categories

6  1- L₁: (observed); L₂: (expected); L₃: (L₁- L₂)²/ L₂  2- sum(L ₃)= x²  3- x²cdf(x²,1000,df)

7  Area under the curve= 1  Skewed to the right  As the degrees of freedom increase, the closer to a normal distribution your curve becomes

8  Ex 1: The “graying of America” is the recent belief that with better medicine and healthier lifestyles, people are living longer, and consequently a larger percentage of the population is of retirement age. Is this perception accurate? (Is there evidence that the distribution of ages changed drastically in 1996 from 1980?)  US Population by age group, 1980 Age Group Population(in thousandths) Percent 0-2493,77741.39 25-4462,71627.68 45-6444,50319.64 65-older25,55011.28 Total 226,546 100.00

9  We select an SRS of 500. We first calculate our expected counts AgeCount1980 pop %Expected Counts 0-2417741.39500(.4139)206.95 25-4415827.68500(.2768)138.4 45-6410119.64500(.1964)98.2 65+6411.28500(.1128)56.4

10  H₀: the distribution of the ages in 1996 is the SAME as it was in 1980  Ha: the distribution of the ages in 1996 is DIFFERENT than it was in 1980  Assumptions: -random sample -all expected counts are ≥ 1 -no more than 20% of expected counts are < 5 (See chart of expected counts)  Chi-squared test (goodness of fit) w/ α=0.05  P(x²>8.214)= 0.042) df=4-1=3  Since p<α, it is statistically significant, therefore we reject H₀. There is enough evidence to say the distribution of ages in 1996 is different than it was in 1980

11  Ex 2: A wheel at a carnival game is divided into 5 equal parts. You suspect the wheel is unbalanced. The results of 100 spins are listed below. Perform a goodness of fit test. Is there evidence the wheel is not balanced?  H₀:The wheel is balanced  Ha: The wheel is unbalanced OutcomesFree SpinYou LoseTeddy Bear Baseball Card Airplane Frequency2428181614

12  Assumptions: -random sample -all expected counts are ≥ 1 -no more than 20% of expected counts are < 5 Chi-Squared Test (goodness of fit) w/ α=0.05 P(x²>6.8)=0.146 df=4 Since p∡α, it is not statistically significant, therefore we do not reject H₀. There is not enough evidence to say the wheel is unbalanced. Expected counts Free SpinYou LoseTeddy Bear Baseball Card Airplane Frequency20


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