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One Big Bill or Many Smaller Bills In a study of the “denomination effect,” 150 women in China were given either a single 100 yuan bill or a total of 100 yuan in smaller bills. The value of 100 yuan is about $15. The women were given the choice of spending the money on specific items or keeping the money. The results are summarized in the table below (based on “The Denomination Effect,” by Priya Raghubir and Joydeep Srivastava, Journal of Consumer Research, Vol. 36). Use a 0.05 significance level to test the claim that the form of the 100 yuan is independent of whether the money was spent. What does the result suggest about a denomination effect?

Spent the Money

Kept the Money

Women Given a Single 100-Yuan Bill

60

15

Women Given 100 Yuan in Smaller Bills

68

7

Short Answer

Expert verified

There is not enough evidence to conclude that the form of currency received is not independent of whether the money was spent /kept.

The claim of a denomination effect is not supported by substantial evidence. Whether 100 yuan is in the shape of a single bill or multiple smaller notes appears to have little effect on Chinese women.

Step by step solution

01

Given information

A table is devised showing the number of women who spent/kept the money depending on whether they received a single bill or smaller bills.

02

Hypotheses

The null hypothesis is as follows:

The form of currency received is independent of whether the money was spent /kept.

The alternative hypothesis is as follows:

The form of currency received is not independent of whether the money was spent /kept.

It is a right-tailed test.

If the test statistic value is greater than the critical value, then the null hypothesis is rejected, otherwise not.

03

Determine the observed and expected frequencies

Observed Frequency:

The frequencies provided in the table are the observed frequencies. They are denoted by\(O\).

The following contingency table shows the observed frequency for each cell:


Spent the Money

Kept the Money

Women Given a Single 100-Yuan Bill

\({O_1}\)=60

\({O_2}\)=15

Women Given 100 Yuan in Smaller Bills

\({O_3}\)=68

\({O_4}\)=7

Expected Frequency:

The formula for computing the expected frequency for each cell is shown below:

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The row total for the first row is computed below:

\(\begin{aligned}{c}Row\;Tota{l_1} = 60 + 15\\ = 75\end{aligned}\)

The row total for the second row is computed below:

\(\begin{aligned}{c}Row\;Tota{l_2} = 68 + 7\\ = 75\end{aligned}\)

The column total for the first column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_1} = 60 + 68\\ = 128\end{aligned}\)

The column total for the second column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_2} = 15 + 7\\ = 22\end{aligned}\)

The grand total can be computed as follows:

\(\begin{aligned}{c}Grand\;Total = \left( {75 + 75} \right)\\ = \left( {128 + 22} \right)\\ = 150\end{aligned}\)

Thus, the following table shows the expected frequencies for each of the corresponding observed frequencies:


Spent the Money

Kept the Money

Women Given a Single 100-Yuan Bill

\[\begin{aligned}{c}{E_1} = \frac{{\left( {75} \right)\left( {128} \right)}}{{150}}\\ = 64\end{aligned}\]

\[\begin{aligned}{c}{E_2} = \frac{{\left( {75} \right)\left( {22} \right)}}{{150}}\\ = 11\end{aligned}\]

Women Given 100 Yuan in Smaller Bills

\[\begin{aligned}{c}{E_3} = \frac{{\left( {75} \right)\left( {128} \right)}}{{150}}\\ = 64\end{aligned}\]

\[\begin{aligned}{c}{E_4} = \frac{{\left( {75} \right)\left( {22} \right)}}{{150}}\\ = 11\end{aligned}\]

The table below shows the necessary calculations:

O

E

O – E

\({\left( {O - E} \right)^2}\)

\(\frac{{{{\left( {O - E} \right)}^2}}}{E}\)

60

64

-4

16

0.250

15

11

4

16

1.4545

68

64

4

16

0.250

7

11

-4

16

1.4545

04

Compute the test statistic, critical value and p-value

The test statistic is computed as shown below:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}\;\;\; \sim } {\chi ^2}_{\left( {r - 1} \right)\left( {c - 1} \right)}\\ = 0.25 + 1.4545 + 0.25 + 1.4545\\ = 3.409\end{aligned}\]

Thus,\({\chi ^2} = 3.409\).

Let r denote the number of rows in the contingency table.

Let c denote the number of columns in the contingency table.

The degrees of freedom are computed below:

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

The critical value of\({\chi ^2}\)for 1 degree of freedom at 0.05 level of significance for a right-tailed test is equal to 3.8415.

The corresponding p-value is approximately equal to 0.0648.

05

Decision and conclusion of the test

Since the value of the test statistic is less than the critical value and the p-value is greater than 0.05, the null hypothesis fails to reject.

There is not enough evidence to conclude that the form of currency received is not independent of whether the money was spent /kept.

The claim of a denomination effect is not supported by substantial evidence. Whether 100 yuan is in the shape of a single bill or multiple smaller notes appears to have little effect on Chinese women.

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Most popular questions from this chapter

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Placebo

Atorvastatin 10 mg

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Atorvastatin 80 mg

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27

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No Infection

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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’s claim?


Born in America

Foreign Born

Jan.

387

101

Feb.

329

82

March

366

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April

344

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May

336

94

June

313

83

July

313

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Aug.

503

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Sept.

421

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Oct.

434

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Nov.

398

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Dec.

371

82

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

Baseball Player Births In his book Outliers, author Malcolm Gladwell argues that more baseball players have birth dates in the months immediately following July 31, because that was the age cutoff date for nonschool baseball leagues. Here is a sample of frequency counts of months of birth dates of American-born Major League Baseball players starting with January: 387, 329, 366, 344, 336, 313, 313, 503, 421, 434, 398, 371. Using a 0.05 significance level, is there sufficient evidence to warrant rejection of the claim that American-born Major League Baseball players are born in different months with the same frequency? Do the sample values appear to support Gladwell’s claim?

A case-control (or retrospective) study was conductedto investigate a relationship between the colors of helmets worn by motorcycle drivers andwhether they are injured or killed in a crash. Results are given in the table below (based on datafrom “Motorcycle Rider Conspicuity and Crash Related Injury: Case-Control Study,” by Wellset al., BMJ USA,Vol. 4). Test the claim that injuries are independent of helmet color. Shouldmotorcycle drivers choose helmets with a particular color? If so, which color appears best?

Color of helmet


Black

White

Yellow/Orange

Red

Blue

Controls (not injured)

491

377

31

170

55

Cases (injured or killed)

213

112

8

70

26

Benford’s Law. According to Benford’s law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. In Exercises 21–24, test for goodness-of-fit with the distribution described by Benford’s law.

Leading Digits

Benford's Law: Distributuon of leading digits

1

30.10%

2

17.60%

3

12.50%

4

9.70%

5

7.90%

6

6.70%

7

5.80%

8

5.10%

9

4.60%

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