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Using Yates鈥檚 Correction for Continuity The chi-square distribution is continuous, whereas the test statistic used in this section is discrete. Some statisticians use Yates鈥檚 correction for continuity in cells with an expected frequency of less than 10 or in all cells of a contingency table with two rows and two columns. With Yates鈥檚 correction, we replace

\(\sum \frac{{{{\left( {O - E} \right)}^2}}}{E}\)with \(\sum \frac{{{{\left( {\left| {O - E} \right| - 0.5} \right)}^2}}}{E}\)

Given the contingency table in Exercise 9 鈥淔our Quarters the Same as $1?鈥 find the value of the test \({\chi ^2}\)statistic using Yates鈥檚 correction in all cells. What effect does Yates鈥檚 correction have?

Short Answer

Expert verified

The value of the \({\chi ^2}\) test statistic using the Yates鈥檚 correction factor is equal to 10.717.

There is sufficient evidence to reject the claim that whether students purchased gum or kept the money is independent of whether they were given four quarters or a $1 bill.

The effect of Yates鈥檚 correction factor is that the test statistic value decreases as compared to the test statistic value computed without using any correction factor.

Step by step solution

01

Given information

Data are given on the number of students who spent/kept the money given that they were given four quarters and a $1 bill.

02

Hypotheses

Null Hypothesis: The act of spending/keeping the money is independent of whether the students were given four quarters or a $1 bill.

Alternative Hypothesis: The act of spending/keeping the money is not independent of whether the students were given four quarters or a $1 bill.

03

Observed frequencies

The following table shows the observed frequencies:

Purchased Gum

Kept the Money

Students given four Quarters

\({O_1} = \)27

\({O_2} = \)16

Students Given $1 Bill

\({O_3} = \)12

\({O_4} = \)34

04

Calculation of expected frequencies

The following table shows the row totals and column totals:

Purchased Gum

Kept the Money

Row Total

Students given four Quarters

27

16

43

Students Given $1 Bill

12

34

46

Column total

39

50

89

The expected frequencies are computed using the given formula:

\({\rm{Expected}}\;{\rm{Frequency}} = \frac{{{\rm{Row}}\;{\rm{Total}} \times {\rm{Column}}\;{\rm{Total}}}}{{{\rm{Grand}}\;{\rm{Total}}}}\)

The following table shows the expected frequencies:

Purchased Gum

Kept the Money

Students given four Quarters

\(\begin{aligned}{c}{E_1} = \frac{{43 \times 39}}{{89}}\\ = 18.84\end{aligned}\)

\(\begin{aligned}{c}{E_2} = \frac{{43 \times 50}}{{89}}\\ = 24.16\end{aligned}\)

Students Given $1 Bill

\(\begin{aligned}{c}{E_3} = \frac{{46 \times 39}}{{89}}\\ = 20.16\end{aligned}\)

\(\begin{aligned}{c}{E_4} = \frac{{46 \times 50}}{{89}}\\ = 25.84\end{aligned}\)

05

Calculation of the test statistic

The test statistic with Yates鈥檚 correction factor is computed as follows:

\(\begin{aligned}{c}{\chi ^2} = \sum \frac{{{{\left( {\left| {O - E} \right| - 0.5} \right)}^2}}}{E}\;\;\;\;\chi _{\left( {r - 1} \right)\left( {c - 1} \right)}^2\\ = \frac{{{{\left( {\left| {27 - 18.84} \right| - 0.5} \right)}^2}}}{{18.84}} + \frac{{{{\left( {\left| {16 - 24.16} \right| - 0.5} \right)}^2}}}{{24.16}} + \frac{{{{\left( {\left| {12 - 20.16} \right| - 0.5} \right)}^2}}}{{20.16}} + \frac{{{{\left( {\left| {34 - 25.84} \right| - 0.5} \right)}^2}}}{{25.84}}\\ = 10.717\end{aligned}\)

Therefore, the value of test statistic is 10.717.

The test statistic with Yates鈥檚 correction factor is computed as follows:

\(\begin{aligned}{c}{\chi ^2} = \sum \frac{{{{\left( {\left| {O - E} \right| - 0.5} \right)}^2}}}{E}\;\;\;\;\chi _{\left( {r - 1} \right)\left( {c - 1} \right)}^2\\ = \frac{{{{\left( {\left| {27 - 18.84} \right| - 0.5} \right)}^2}}}{{18.84}} + \frac{{{{\left( {\left| {16 - 24.16} \right| - 0.5} \right)}^2}}}{{24.16}} + \frac{{{{\left( {\left| {12 - 20.16} \right| - 0.5} \right)}^2}}}{{20.16}} + \frac{{{{\left( {\left| {34 - 25.84} \right| - 0.5} \right)}^2}}}{{25.84}}\\ = 10.717\end{aligned}\)

Therefore, the value of test statistic is 10.717.

06

Calculation of the critical value

The level of significance is 0.05.

The degrees of freedom is computed below:

\(\begin{aligned}{c}df = \left( {r - 1} \right)\left( {c - 1} \right)\\ = \left( {2 - 1} \right)\left( {2 - 1} \right)\\ = 1\end{aligned}\)

Now look at the \({\chi ^2}\) distribution table for 1 degrees of freedom with 0.05 significance level.

The critical value is 3.842.

07

Conclusion of the test

The calculated value of the test statistic is greater than the critical value. Therefore, the null hypothesis is rejected.

Thus, there is sufficient evidence to reject the claim that whether students purchased gum or kept the money is independent of whether they were given four quarters or a $1 bill.

It seems that there is a denomination effect.

08

Effect of Yates’s correction

The effect of Yates鈥檚 correction factor is that the tests statisticdecreases as compared to the test statistic computed without using any correction factor.

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Most popular questions from this chapter

Probability Refer to the results from the 150 subjects in Cumulative Review Exercise 5.

a.Find the probability that if 1 of the 150 subjects is randomly selected, the result is a woman who spent the money.

b.Find the probability that if 1 of the 150 subjects is randomly selected, the result is a woman who spent the money or was given a single 100-yuan bill.

c.If two different women are randomly selected, find the probability that they both spent the money.

Exercises 1鈥5 refer to the sample data in the following table, which summarizes the last digits of the heights (cm) of 300 randomly selected subjects (from Data Set 1 鈥淏ody Data鈥 in Appendix B). Assume that we want to use a 0.05 significance level to test the claim that the data are from a population having the property that the last digits are all equally likely.

Last Digit

0

1

2

3

4

5

6

7

8

9

Frequency

30

35

24

25

35

36

37

27

27

24

Given that the P-value for the hypothesis test is 0.501, what do you conclude? Does it appear that the heights were obtained through measurement or that the subjects reported their heights?

In his book Outliers,author Malcolm Gladwell argues that more

American-born baseball players have birth dates in the months immediately following July 31 because that was the age cutoff date for nonschool baseball leagues. The table below lists months of births for a sample of American-born baseball players and foreign-born baseball players. Using a 0.05 significance level, is there sufficient evidence to warrant rejection of the claim that months of births of baseball players are independent of whether they are born in America? Do the data appear to support Gladwell鈥檚 claim?


Born in America

Foreign Born

Jan.

387

101

Feb.

329

82

March

366

85

April

344

82

May

336

94

June

313

83

July

313

59

Aug.

503

91

Sept.

421

70

Oct.

434

100

Nov.

398

103

Dec.

371

82

In Exercises 5鈥20, conduct the hypothesis test and provide the test statistic and the P-value and, or critical value, and state the conclusion.

Kentucky Derby The table below lists the frequency of wins for different post positions through the 141st running of the Kentucky Derby horse race. A post position of 1 is closest to the inside rail, so the horse in that position has the shortest distance to run. (Because the number of horses varies from year to year, only the first 10 post positions are included.) Use a 0.05 significance level to test the claim that the likelihood of winning is the same for the different post positions. Based on the result, should bettors consider the post position of a horse racing in the Kentucky Derby?

Post Position

1

2

3

4

5

6

7

8

9

10

Wins

19

14

11

15

15

7

8

12

5

11

In a study of high school students at least 16 years of age, researchers obtained survey results summarized in the accompanying table (based on data from 鈥淭exting While Driving and Other Risky Motor Vehicle Behaviors Among U.S. High School Students,鈥 by O鈥橫alley, Shults, and Eaton, Pediatrics,Vol. 131, No. 6). Use a 0.05 significance level to

test the claim of independence between texting while driving and driving when drinking alcohol. Are those two risky behaviors independent of each other?


Drove when drinking Alcohol?


Yes

No

Texted while driving

731

3054

No Texting while driving

156

4564

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