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

Chocolate and Happiness In a survey sponsored by the Lindt chocolate company, 1708 women were surveyed and 85% of them said that chocolate made them happier.

a. Is there anything potentially wrong with this survey?

b. Of the 1708 women surveyed, what is the number of them who said that chocolate made them happier?

A randomized controlled trial was designed to compare the effectiveness of splinting versus surgery in the treatment of carpal tunnel syndrome. Results are given in the table below (based on data from “Splinting vs. Surgery in the Treatment of Carpal Tunnel Syndrome,” by Gerritsen et al., Journal of the American Medical Association,Vol. 288,

No. 10). The results are based on evaluations made one year after the treatment. Using a 0.01 significance level, test the claim that success is independent of the type of treatment. What do the results suggest about treating carpal tunnel syndrome?

Successful Treatment

Unsuccessful Treatment

Splint Treatment

60

23

Surgery Treatment

67

6

A study of people who refused to answer survey questions provided the randomly selected sample data shown in the table below (based on data from “I Hear You Knocking But You Can’t Come In,” by Fitzgerald and Fuller, Sociological Methods and Research,Vol. 11, No. 1). At the 0.01 significance level, test the claim that the cooperation of

the subject (response or refusal) is independent of the age category. Does any particular age group appear to be particularly uncooperative?

Age


18-21

22-29

30-39

40-49

50-59

60 and over

Responded

73

255

245

136

138

202

Refused

11

20

33

16

27

49

Loaded Die The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 27, 31, 42, 40, 28, and 32. Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die?

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.

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