Question

Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Denomination Effect A trial was conducted with 75 women in China given a 100-yuan bill, while another 75 women in China were given 100 yuan in the form of smaller bills (a 50-yuan bill plus two 20-yuan bills plus two 5-yuan bills). Among those given the single bill, 60 spent some or all of the money. Among those given the smaller bills, 68 spent some or all of the money (based on data from “The Denomination Effect,” by Raghubir and Srivastava, Journal of Consumer Research, Vol. 36). We want to use a 0.05 significance level to test the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. If the significance level is changed to 0.01, does the conclusion change?

Short answer

a. The null hypothesis is rejected, and hence, there is sufficient evidence to claim that the proportion of the women in China who were given a single bill and who spent the money is smaller than the proportion of women who were given the smaller bills and who spent the money.

b. The 90% confidence interval is equal to (-0.201, -0.013).

c. Yes, the significance level changes the conclusion of the claim.

Step-by-step solution

Given information

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 $\alpha =0.05$ .

Describe the hypotheses

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.

Find the important values

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

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{60}{75} \\ =0.8 \end{gathered}$$

Let ${\hat{p}}_2$denote the sampleproportion of women in China who were given smaller bills and spent some or all money.

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{68}{75} \\ =0.9067 \end{gathered}$$

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

$$\begin{gathered} \overline{p}=\frac{\left(x_1+x_2\right)}{\left(n_1+n_2\right)} \\ =\frac{\left(60+68\right)}{\left(75+75\right)} \\ =0.853 \end{gathered}$$

and

$$\begin{gathered} \overline{q}=1-\overline{p} \\ =1-0.853 \\ =0.147 \end{gathered}$$

Find the test statistic

The value of test statistic is computed as follows:

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

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.

Conclusion of the test

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.

Find the confidence interval

b.

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

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

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,

$$\begin{gathered} \frac{\alpha }{2}=\frac{0.10}{2} \\ =0.05 \end{gathered}$$

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:

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

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

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

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

Changing the level of significance

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.