# 11.4 Test of independence  (Page 3/21)

 Page 3 / 21

De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. [link] shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events.

Need to succeed in school vs. anxiety level
Need to Succeed in School High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
Row Total
High Need 35 42 53 15 10 155
Medium Need 18 48 63 33 31 193
Low Need 4 5 11 15 17 52
Column Total 57 95 127 63 58 400

a. How many high anxiety level students are expected to have a high need to succeed in school?

a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400.

$E=\frac{\text{(row total)(column total)}}{\text{total surveyed}}=\frac{155\cdot 57}{400}=22.09$

The expected number of students who have a high anxiety level and a high need to succeed in school is about 22.

b. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety?

b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400.

c. $E=\frac{\text{(row total)(column total)}}{\text{total surveyed}}$ = ________

c. $E=\frac{\text{(row total)(column total)}}{\text{total surveyed}}=8.19$

d. The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________.

d. 8

## References

DiCamilo, Mark, Mervin Field, “Most Californians See a Direct Linkage between Obesity and Sugary Sodas. Two in Three Voters Support Taxing Sugar-Sweetened Beverages If Proceeds are Tied to Improving School Nutrition and Physical Activity Programs.” The Field Poll, released Feb. 14, 2013. Available online at http://field.com/fieldpollonline/subscribers/Rls2436.pdf (accessed May 24, 2013).

Harris Interactive, “Favorite Flavor of Ice Cream.” Available online at http://www.statisticbrain.com/favorite-flavor-of-ice-cream (accessed May 24, 2013)

“Youngest Online Entrepreneurs List.” Available online at http://www.statisticbrain.com/youngest-online-entrepreneur-list (accessed May 24, 2013).

## Chapter review

To assess whether two factors are independent or not, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least 5.

## Test of independence

• The number of degrees of freedom is equal to (number of columns - 1)(number of rows - 1).
• The test statistic is $\sum _{i\cdot j}\frac{{\left(O-E\right)}^{2}}{E}$ where O = observed values, E = expected values, i = the number of rows in the table, and j = the number of columns in the table.
• If the null hypothesis is true, the expected number $E=\frac{\text{(row total)(column total)}}{\text{total surveyed}}$ .

Determine the appropriate test to be used in the next three exercises.

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?
Kyle
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research.net
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sciencedirect big data base
Ernesto
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Bharti
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