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Chapter 11 Chi-Square Procedures. 11.1 Chi-Square Goodness of Fit. Characteristics of the Chi-Square Distribution 1. It is not symmetric. Characteristics of the Chi-Square Distribution 1. It is not symmetric.

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chapter 11 chi square procedures

Chapter 11Chi-Square Procedures

11.1

Chi-Square Goodness of Fit

slide3

Characteristics of the Chi-Square Distribution

1. It is not symmetric.

2. The shape of the chi-square distribution depends upon the degrees of freedom, just like Student’s t-distribution.

slide4

Characteristics of the Chi-Square Distribution

1. It is not symmetric.

2. The shape of the chi-square distribution depends upon the degrees of freedom, just like Student’s t-distribution.

3. As the number of degrees of freedom increases, the chi-square distribution becomes more symmetric as is illustrated in Figure 1.

slide5

Characteristics of the Chi-Square Distribution

1. It is not symmetric.

2. The shape of the chi-square distribution depends upon the degrees of freedom, just like Student’s t-distribution.

3. As the number of degrees of freedom increases, the chi-square distribution becomes more symmetric as is illustrated in Figure 1.

4. The values are non-negative. That is, the values of are greater than or equal to 0.

slide7

A goodness-of-fit test is an inferential procedure used to determine whether a frequency distribution follows a claimed distribution.

slide8

Expected Counts

  • Suppose there are n independent trials an experiment with k> 3 mutually exclusive possible outcomes. Let p1 represent the probability of observing the first outcome and E1 represent the expected count of the first outcome, p2 represent the probability of observing the second outcome and E2 represent the expected count of the second outcome, and so on. The expected counts for each possible outcome is given by
  • Ei = i = npi for i = 1, 2, …, k
slide9

EXAMPLE Finding Expected Counts

A sociologist wishes to determine whether the distribution for the number of years grandparents who are responsible for their grandchildren is different today than it was in 2000. According to the United States Census Bureau, in 2000, 22.8% of grandparents have been responsible for their grandchildren less than 1 year; 23.9% of grandparents have been responsible for their grandchildren 1or 2 years; 17.6% of grandparents have been responsible for their grandchildren 3 or 4 years; and 35.7% of grandparents have been responsible for their grandchildren for 5 or more years. If the sociologist randomly selects 1,000 grandparents that are responsible for their grandchildren, compute the expected number within each category assuming the distribution has not changed from 2000.

slide10

Test Statistic for Goodness-of-Fit Tests

Let Oi represent the observed counts of category i, Ei represent the expected counts of an category i, k represent the number of categories, and n represent the number of independent trials of an experiment. Then,

i = 1, 2, …, k

approximately follows the chi-square distribution with k – 1 degrees of freedom provided (1) all expected frequencies are greater than or equal to 1 (all Ei> 1) and (2) no more than 20% of the expected frequencies are less than 5. NOTE: Ei = npifor i = 1,2, ..., k.

slide11

The Chi-Square Goodness-of-Fit Test

  • If a claim is made regarding a distribution, we can use the following steps to test the claim provided
  • the data is randomly selected
slide12

The Chi-Square Goodness-of-Fit Test

  • If a claim is made regarding a distribution, we can use the following steps to test the claim provided
  • the data is randomly selected
  • 2. all expected frequencies are greater than or equal to 1.
slide13

The Chi-Square Goodness-of-Fit Test

  • If a claim is made regarding a distribution, we can use the following steps to test the claim provided
  • the data is randomly selected
  • 2. all expected frequencies are greater than or equal to 1.
  • 3. no more than 20% of the expected frequencies are less than 5.
slide14

Step 1: A claim is made regarding a distribution. The claim is used to determine the null and alternative hypothesis.

Ho: the random variable follows the claimed distribution

H1: the random variable does not follow the claimed distribution

slide15

Step 2: Calculate the expected frequencies for each of the k categories. The expected frequencies are npi for i = 1, 2, …, k where n is the number of trials and pi is the probability of the ith category assuming the null hypothesis is true.

slide16

Step 3: Verify the requirements fort he goodness-of-fit test are satisfied.

  • (1) all expected frequencies are greater than or equal to 1 (all Ei> 1)
  • (2) no more than 20% of the expected frequencies are less than 5.
slide20

EXAMPLE Testing a Claim Using the Goodness-of-Fit Test

A sociologist wishes to determine whether the distribution for the number of years grandparents who are responsible for their grandchildren is different today than it was in 2000. According to the United States Census Bureau, in 2000, 22.8% of grandparents have been responsible for their grandchildren less than 1 year; 23.9% of grandparents have been responsible for their grandchildren 1or 2 years; 17.6% of grandparents have been responsible for their grandchildren 3 or 4 years; and 35.7% of grandparents have been responsible for their grandchildren for 5 or more years. The sociologist randomly selects 1,000 grandparents that are responsible for their grandchildren and obtains the following data.

solution
Solution:
  • Step 1. Construct the Hypothesis
  • H0 : The distribution for the number of years grandparents who are responsible for their grandchildren is the same today as it was in 2000.

H1 : The distribution for the number of years grandparents who are responsible for their grandchildren is different today from what it was in 2000.

slide23
Step 2. Compute the expected counts for each category, assuming that the null hypothesis is true.
solution cont d
Solution(cont’d):
  • Step 3. Verify that the requirements for the goodness-of-fit test are satisfied.

1. All expected frequencies( or expected counts ) are bigger than or equal to 1?

2. No more than 20% of the expected frequencies are less than 5.

slide25
Step 4. Find the critical values, determine the critical region.

α=0.05, k = 4, degree of freedom = k-1 =3

Look in table IV,

χα2=7.815

C:=(7.815, infinity)

slide26
Step 5. Compute the test statistic

χ2 = (252-228)^2/228+(255-239)^2/239

+( 162-176)^2/176+(331-357)^2/357

=6.605

  • Step 6. Compare the test statistics with the critical values

the test statistic < the critical value

or the test statistic does not lie in th critical region.

  • Step 7. Conclusion?
  • There is no sufficient evidence at the α=0.05 level of significance to reject the null hypothesis, i.e., the claim of the distribution for the number of years grandparents who are responsible for their grandchildren is the same today as it was in 2000

Or ….