91影视

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.

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.

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:

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:

The table below shows the necessary calculations:

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.

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.