1. Section 10-1 – Goodness of Fit
Multinomial Experiment – a fixed number of trials in which there are more than two possible outcomes for each independent trial.
The probability of each outcome is fixed, and each outcome is classified into categories.
Chi-Square Goodness of Fit Test
Used to test whether a frequency distribution fits an expected distribution (claim).
Two conditions that must be met in order to conduct a Chi- Square Goodness of Fit test
1) Samples must be randomly selected
2) Each E (expected value) must be ≥ 5.
You may need to combine categories to achieve this.

2. Section 10-1 – Goodness of Fit
Guidelines for conducting the test.
1) Write the hypotheses and identify claim
H0: Simply state the claimed percentages for each category.
Ha: State that the distribution is different from what is listed above.
2) Identify α.
3) Identify degrees of freedom (k – 1) where k is the number of categories.
4) STAT – Edit
Enter the claimed percentages for each category into L (as decimals)
Enter the Observed Frequencies for each category into L2
Highlight L3 and enter L1*n (Use the actual number for n)
Check to be sure that all values in L3 are > 5.

3. Section 10-1 – Goodness of Fit
Guidelines for conducting the test.
5) Run STAT–TEST–D ( 𝑋 2 GOF-Test)
Enter L2 for Observed:
Enter L3 for Expected:
Enter the degrees of freedom (k – 1)
6) Make a decision to reject H0 or Fail to Reject H0.
If 𝑝≤𝛼, reject the null.
If 𝑝>𝛼, fail to reject the null.

4. Example 1 (Page 553)
A marketing executive randomly selects 500 radio music listeners from the broadcast region and asks each whether he or she prefers classical, country, gospel, oldies, pop, or rock music. The results are shown in the table. Find the observed frequencies and the expected frequencies for each type of music.
Survey Results (n = 500)
Type of Music
Claimed %
Observed
Expected
Classical 4% 8
Country
36%
210
Gospel
11% 72
Oldies 2% 10
Pop
18% 75
Rock
29%
125
The observed frequency for each type of music is the number of radio music listeners naming that particular type of music. This is what is given in the Observed column above.
The expected frequency for each type of music is the claimed percentage for that music type times the total number in the sample (n). (Table above, far right column)

5. Example 1 (Page 553)
A marketing executive randomly selects 500 radio music listeners from the broadcast region and asks each whether he or she prefers classical, country, gospel, oldies, pop, or rock music. The results are shown in the table. Find the observed frequencies and the expected frequencies for each type of music.
Survey Results (n = 500)
Type of Music
Claimed %
Observed
Expected
Classical 4% 8
Country
36%
210
Gospel
11% 72
Oldies 2% 10
Pop
18% 75
Rock
29%
125
On the calculator, enter the Claimed % column into L1 in decimal form.
.04, .36, .11, .02, .18, and .29
Enter the Observed column into L2
8, 210, 72, 10, 75, and 125
Highlight L3 and type in “L1*500”; this will give you the results shown in the Expected column. 20
180 55 10 90
145
CHECK to be sure that ALL values in the Expected column are ≥ 5!!

6. Example 2 (Page 555)
The music preferences of the listeners in a radio station’s broadcast region are distributed as shown in the table from Example 1. You randomly select 500 radio music listeners from the broadcast region and ask each whether he or she prefers classical, country, gospel, oldies, pop, or rock music. The survey results are shown in the table from Example 1 (2nd column). Using 𝛼 = 0.01, perform a chi-square goodness-of-fit test to test whether the distributions are different.
Write the hypotheses:
𝐻 0 : The distribution of music preferences in the broadcast region is 4% classical, 36% country, 11% gospel, 2% oldies, 18% pop, and % rock.
𝐻 𝑎 : The distribution of music preferences in the broadcast region differs from the claimed or expected distribution. (claim).

7. Example 2 (Page 555)
You’ve already put the data into STAT – Edit, and confirmed that all expected values are at least 5, so it’s time to run the test.
STAT – TEST – D
Designate L2 for your Observed
Designate L3 for your Expected
Designate 5 as your degrees of freedom (6 categories minus 1)
𝛸 2 = (this is your standardized test statistic).
𝑝=3.83 𝐸 − 4, or
Since 𝑝≤𝛼, reject the null hypothesis. This means that the actual distribution is different from the claimed distribution.
At the 1% significance level, there is enough evidence to conclude that the distribution of music preferences differs from the radio station’s claimed or expected distribution.

8. Opinions on what is more important to save for.
Example 3 (Page 556)
The display below (left) shows two distributions describing opinions on what is more important to save for. You work for a financial services company and want to test the distribution describing men’s opinions.
To test the distribution, you randomly select 400 men and ask each which is more important – saving for retirement or saving for children’s college education. The results are shown below (right). At 𝛼 = 0.05, test the claimed or expected distribution.
Opinions on what is more important to save for.
Retirement
Children’s College
Not sure
Men
44%
40%
16%
Women
41%
46%
13%
Men’s Survey Results
Retirement
186
Children’s College
143
Not sure 71

9. Example 3 (Page 556)
Write the hypotheses:
𝐻 0 : The distribution of men’s opinions on saving is 44% retirement, % college, and 16% not sure. (claim)
𝐻 𝑎 : The distribution of men’s opinions on savings differs from the claimed or expected distribution.
STAT – Edit
L1 – Enter the % claimed for each category, in decimal form.
(.44, .40, .16)
L2 – Enter the observed values for each category
(186, 143, 71)
L3 – Highlight the L3, then type in “L1*400” (n in this survey).
This gives you the expected values.
Check to be sure that all of these values are ≥ 5!!
Now, check to be sure that the Chi-square goodness-of-fit test can be used.
Your sample is random, and each expected frequency is greater than 5, so we can use this test.

10. Example 3 (Page 556)
STAT – TEST – D
Designate L2 for your Observed
Designate L3 for your Expected
Designate 2 as your degrees of freedom (3 categories minus 1)
𝛸 2 =3.14 (this is your standardized test statistic).
𝑝=.208
Since 𝑝>𝛼, fail to reject the null hypothesis. This means that the actual distribution is not different from the claimed distribution.
At the 5% significance level, there is not enough evidence to dispute the claimed distribution of men’s opinions on savings.

11. Example 4 (Page 558)
The manufacturer of M&M’s candies claims that the number of different-colored candies in bags of dark chocolate M&M’s is uniformly distributed. To test this claim, you randomly select a bag that contains 500 dark chocolate M&M’s. The results are shown in the table. Using 𝛼 = .10, perform a chi-square goodness-of-fit test to test the claimed or expected distribution. What can you conclude?
The claim is that the distribution is uniform; so the expected frequencies of the colors are equal. To find each expected percentage, divide 1 by the number of categories. In this case, divide 1 by 6. The expected percentage for each color is 1/6.
Color
Frequency
Brown 80
Yellow 95
Red 88
Blue 83
Orange 76
Green 78

12. The null and alternate hypotheses are as follows:
Example 4 (Page 558)
The null and alternate hypotheses are as follows:
𝐻 0 : The distribution of different colored M&M’s is uniform. (Claim)
𝐻 𝑎 : The distribution of different colored M&M’s is not uniform.
STAT – Edit
L1 – Enter the % claimed for each category, in decimal form.
(1/6, 1/6, 1/6, 1/6, 1/6, 1/6)
L2 – Enter the observed values for each category
(80, 95, 88, 83, 76, 78)
L3 – Highlight the L3, then type in “L1*500” (n in this survey).
This gives you the expected values.
Check to be sure that all of these values are ≥ 5!!
Color
Frequency
Brown 80
Yellow 95
Red 88
Blue 83
Orange 76
Green 78

13. Example 4 (Page 558)
Because each expected frequency is at least 5, and the M&M’s were randomly selected, you can use the chi-square goodness-of-fit test to test the claimed distribution.
STAT – TEST – D
Designate L2 for your Observed
Designate L3 for your Expected
Designate 5 as your degrees of freedom (6 categories minus 1)
𝛸 2 =3.016 (this is your standardized test statistic).
𝑝=.698
Since 𝑝>𝛼, fail to reject the null hypothesis. This means that the actual distribution is not different from the claimed distribution.
At the 10% significance level, there is not enough evidence to dispute the claim that the distribution of the different-colored candies in bags of dark chocolate M&M’s is uniform.

14. Classwork: Page 560 #1-8 All
For #7 and 8, use the p-value and 𝜶 to make your decision instead of rejection regions.
Homework: Pages #9-17 All
Use the p-value and 𝜶 to make your decision instead of rejection regions.