91影视

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

The level of significance is 0.05.

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,

Theexpected frequency table is represented as,

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.

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.

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.

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.

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.

There issufficient evidence to reject the claimthat the students鈥� 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.