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In a sample of 75 women who were given a single 100-yuan bill, 60 of them spent some or all of the money. In another sample of 75 women, who were given smaller bills worth 100-yuan, 68 of them spent some or all of the money. The significance level is \(\alpha = 0.05\) .

It is claimed that the proportion of women who receive a single bill and spend it is less than the proportion of women who receive smaller bills and spend the money.

Since the given claim does not have an equality sign, the following hypotheses are set up:

Null Hypothesis: The proportion of women in China who have a single large bill and spend the money is equal to the proportion of women in China who have smaller bills and spend the money.

\({H_0}:{p_1} = {p_2}\)

Alternative Hypothesis: The proportion of women in China who have a single large bill and spend the money is less than the proportion of women in China who have smaller bills and spend the money.

\({H_1}:{p_1} < {p_2}\)

The test is left-tailed.

The sample size of women in China who were given a single large bill and spent some or all money\(\left( {{n_1}} \right)\)is equal to 75.

The sample size of women in China who were given smaller bills and spent some or all money\(\left( {{n_2}} \right)\)is equal to 75.

Let \({\hat p_1}\) denote the sampleproportion of women in China who were given a single large bill and spent some or all money

\(\begin{array}{c}{{{\rm{\hat p}}}_{\rm{1}}} = \frac{{{x_1}}}{{{n_1}}}\\ = \frac{{60}}{{75}}\\ = 0.8\end{array}\)

Let\({\hat p_2}\) denote the sampleproportion of women in China who were given smaller bills and spent some or all money.

\(\begin{array}{c}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}\\ = \frac{{68}}{{75}}\\ = 0.9067\end{array}\)

The value of the pooled sample proportion is computed as follows:

\(\begin{array}{c}\bar p = \frac{{\left( {{x_1} + {x_2}} \right)}}{{\left( {{n_1} + {n_2}} \right)}}\,\\ = \frac{{\left( {60 + 68} \right)}}{{\left( {75 + 75} \right)}}\\ = 0.853\end{array}\)

\(\begin{array}{c}{\rm{\bar q}} = 1 - {\rm{\bar p}}\\ = 1 - 0.853\\ = 0.147\end{array}\)

The value of test statistic is computed as follows:

\(\begin{array}{c}z = \frac{{\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\left( {\frac{{\bar p\bar q}}{{{n_1}}} + \frac{{\bar p\bar q}}{{{n_2}}}} \right)} }}\\ = \frac{{\left( {0.8 - 0.907} \right) - 0}}{{\sqrt {\left( {\frac{{0.853 \times 0.147}}{{75}} + \frac{{0.853 \times 0.147}}{{75}}} \right)} }}\\ = - 1.846\end{array}\)

Thus, the value of test statistic is -1.846

Referring to the standard normal distribution table, the critical value of z corresponding to\(\alpha = 0.05\)for a left-tailed test is equal to -1.645.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.0324.

Since the p-value is less than 0.05, the null hypothesis is rejected.

a.

There is sufficient evidence to support the claim that theproportion of women who receive a single bill and spend it is less than the proportion of women who receive smaller bills and spend the money.

b.

The general formula for the confidence interval of the difference between population proportions is written below:

\({\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E\,\,,\,\,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\,\,\,\,\,\,\,\,...\left( 1 \right)\)\(\)

Where, E is the margin of error and has the following formula:

\(E = {z_{\frac{\alpha }{2}}} \times \sqrt {\left( {\frac{{{{\hat p}_1} \times {{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2} \times {{\hat q}_2}}}{{{n_2}}}} \right)} \)

For computing the confidence interval, first find the critical value\({z_{\frac{\alpha }{2}}}\).

The confidence level is 90% if the level of significance used in the one-railed test is 0.05.

Thus, the value of the level of significance for the confidence interval becomes\(\alpha = 0.10\).

Hence,

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.10}}{2}\\ = 0.05\end{array}\)

The value of\({z_{\frac{\alpha }{2}}}\)form the standard normal table is equal to 1.645.

Now, the margin of error (E) is equal to:

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}} \times \sqrt {\left( {\frac{{{{\hat p}_1} \times {{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2} \times {{\hat q}_2}}}{{{n_2}}}} \right)} \\ = 1.645 \times \sqrt {\left( {\frac{{0.8 \times 0.2}}{{75}} + \frac{{0.9067 \times 0.0933}}{{75}}} \right)} \\ = 0.094\end{array}\)

Substitute the value of E in equation (1) as follows:

\(\begin{array}{c}{\rm{Confidence}}\,\,{\rm{Interval}} = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E\,\,,\,\,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\\ = \left( {\left( {0.8 - 0.907} \right) - 0.094\,\,,\,\,\left( {0.8 - 0.907} \right) + 0.094} \right)\\ = \left( { - 0.201\,\,,\,\,\, - 0.013} \right)\end{array}\)

Thus, 95% confidence interval for the difference between proportions is (-0.201, -0.013).

c.

Find the critical value and p-value for\(\alpha = 0.01\)

Referring to the standard normal distribution table, the critical value of z corresponding to\(\alpha = 0.01\)for a left-tailed test is equal to -2.3263 and the corresponding p-value is equal to 0.0322.

Since the p-value is greater than 0.01, the null hypothesis fails to be rejected. Thus, the conclusion is thatthere is not sufficient evidence to support the claim that theproportion of women who receive a single bill and spend it is less than the proportion of women who receive smaller bills and spend the money.

For\(\alpha = 0.05\), the null hypothesis is rejected, but for \(\alpha = 0.01\) the null hypothesis fails to be rejected. So,if the significance level is changed, the results and the conclusion of the claim change.