Question

In a study of the “denomination effect,” 43 college students

were each given one dollar in the form of four quarters, while 46 other college students were each given one dollar in the form of a dollar bill. All of the students were then given two choices: (1) keep the money; (2) spend the money on gum. The results are given in the accompanying table (based on “The Denomination Effect,” by PriyaRaghubir and Joydeep Srivastava, Journal of Consumer Research,Vol. 36.) Use a 0.05 significance level to test the claim that whether students purchased gum or kept the money is independent of whether they were given four quarters or a \(1 bill. Is there a “denomination effect”?

	Purchased Gum	Kept the Money
Students Given Four Quarters	27	16
Students Given a \)1 Bill	12	34

Short answer

At 0.05 level of significance, it can be concluded that students purchased gum or kept the money is dependent on whether they were given four quarters or a $1 bill.

Thus, there is a denomination effect.

Step-by-step solution

Given information

The data for thepurchase of the gum and denominations is provided.

The level of significance is 0.05.

Compute the expected frequencies

Theexpected frequency is computed as,

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

The observed values with the totals for row and column are represented as,

	Purchased Gum	Kept the Money	Row total
Students Given Four Quarters	27	16	43
Students Given a $1 Bill	12	34	46
Column total	39	50	89

Theexpected frequency table is represented as,

	Purchased Gum	Kept the Money
Students Given Four Quarters	18.843	24.157
Students Given a $1 Bill	20.157	25.843

State the null and alternate hypothesis

The hypotheses to test the independence are formulated as follows:

\({H_0}:\)Students who purchased gum or kept the money is independent of the denomination received.

\({H_1}:\)Students who purchased gum or kept the money is dependent on the denomination received.

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{array}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {27 - 18.843} \right)}^2}}}{{18.843}} + \frac{{{{\left( {16 - 24.157} \right)}^2}}}{{24.157}} + ... + \frac{{{{\left( {34 - 25.843} \right)}^2}}}{{25.843}}\\ = 12.162\end{array}\]

Therefore, the value of the test statistic is 12.162.

Compute the degrees of freedom

The degrees of freedomare computed as,

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

Therefore, the degrees of freedom are 1.

Compute the critical value

From chi-square distribution table, the critical value for row corresponding to 1 degree of freedom and at 0.05 level of significance 3.841.

Therefore, the critical value is 3.841.

The p-value is obtained as 0.000.

State the decision

The critical value (3.841) is less than the value of the test statistic (12.162). In this case, the null hypothesis is rejected.

Therefore, the decision is to reject the null hypothesis.

State the conclusion

There issufficient evidence to reject the claimthat the students’ purchase is independent of the denomination they received(four quarters or a $1 bill). Thus, the two variables are dependent.

Therefore, it appears that there is a denomination effect as denomination affects expenditure.