Question

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In today's economy, everyone has become savings savvy. It is still believed, though, that a higher percentage of women than men clip coupons. A random survey of 180 female shoppers indicated that 132 clipped coupons while 56 out of 100 men did so. At \(\alpha=0.01\), is there sufficient evidence that the proportion of couponing women is higher than the proportion of couponing men? Use the \(P\) -value method.

Short answer

There is sufficient evidence that a higher percentage of women clip coupons than men.

Step-by-step solution

State the Hypotheses

We need to formulate the null and alternative hypotheses. Let \( p_1 \) be the proportion of women who clip coupons and \( p_2 \) be the proportion of men who clip coupons. The null hypothesis \( H_0 \) is: \( p_1 \leq p_2 \).The alternative hypothesis \( H_a \) is: \( p_1 > p_2 \). This is the claim we are testing.

Find the Critical Value

Since the significance level \( \alpha = 0.01 \) and we are performing a one-tailed test, we find the critical value using a standard normal distribution table or a calculator. The critical value \( z_c \) for \( \alpha = 0.01 \) in a one-tailed test is approximately 2.33.

Calculate the Test Statistic

First, calculate the sample proportions: \( \hat{p}_1 = \frac{132}{180} = 0.7333 \) for women and \( \hat{p}_2 = \frac{56}{100} = 0.56 \) for men. Next, calculate the pooled sample proportion: \( \hat{p} = \frac{132 + 56}{180 + 100} = \frac{188}{280} = 0.6714 \).The test statistic is given by \[ z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}} \]where \( n_1 = 180 \) and \( n_2 = 100 \).Thus,\[ z = \frac{0.7333 - 0.56}{\sqrt{0.6714 \times 0.3286 \left( \frac{1}{180} + \frac{1}{100} \right)}} \approx 2.891 \].

Make the Decision

Compare the computed test statistic \( z \approx 2.891 \) with the critical value \( z_c = 2.33 \). Since \( z \) is greater than \( z_c \), we reject the null hypothesis \( H_0 \).

Summarize the Results

Based on our hypothesis test, we reject the null hypothesis that the proportion of women who clip coupons is less than or equal to the proportion of men. Thus, there is sufficient statistical evidence at the \( \alpha = 0.01 \) significance level to support the claim that a higher percentage of women clip coupons than men.

Key concepts

Critical Value

In hypothesis testing, the critical value is a threshold that helps in determining whether to reject the null hypothesis. It is based on the chosen level of significance, denoted by \( \alpha \). For instance, in our example, the significance level is 0.01, which means we are 99% confident in our decision. We use the standard normal distribution table, or a calculator, to find the critical value corresponding to this significance level in a one-tailed test.

Since \( \alpha = 0.01 \), the critical value \( z_c \) is approximately 2.33.

• A critical value acts as a cutoff point.
• It determines the boundary for the acceptance or rejection region.

By comparing the test statistic to the critical value, we can decide to reject or not reject the null hypothesis.

Test Statistic

The test statistic is a standardized value calculated from the sample data during a hypothesis test. It is used to determine how far the sample data diverges from the null hypothesis. In the case of hypothesis testing involving proportions, such as the proportion of coupon users in men and women, we use a formula involving sample proportions and a pooled proportion.

The test statistic \( z \) is calculated using the formula:

• \( \hat{p}_1 \) - Proportion of women clipping coupons.
• \( n_1 \) - Number of women.
• \( \hat{p}_2 \) - Proportion of men clipping coupons.
• \( n_2 \) - Number of men.

The formula looks like this:
\[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}} \] where \( \hat{p} \) is the pooled sample proportion.

In this example, the calculated \( z \) value is approximately 2.891, which will then be compared against the critical value to make a decision about the null hypothesis.

Null Hypothesis

The null hypothesis, denoted as \( H_0 \), is an assumption that there is no effect or no difference between the groups being compared. In other words, it's a statement of no change or no association. Hypothesis tests are designed to work against this assumption, with evidence either strong enough to reject \( H_0 \) or not.

For our specific example, the null hypothesis is \( p_1 \leq p_2 \), meaning the proportion of women who clip coupons is less than or equal to the proportion of men. This acts as a baseline assumption until we have enough evidence to suggest otherwise.

• The null hypothesis is not necessarily what the researcher believes to be true.
• Often, it simply represents the status quo or a position of no difference.

Rejecting the null hypothesis implies that the data provides sufficient evidence to support the alternative hypothesis.

Alternative Hypothesis

The alternative hypothesis, represented by \( H_a \), is what you might consider the bold claim or research hypothesis. It suggests that there is indeed an effect or a difference from the null hypothesis. It's a statement that we aim to provide evidence for through our hypothesis test.

In our example, the alternative hypothesis is \( p_1 > p_2 \), which claims that the proportion of women who clip coupons is greater than the proportion of men. This is the hypothesis we seek to support with our evidence.

The process of hypothesis testing determines:

• If the evidence is significant enough to "reject" the null hypothesis.
• The significance level (\( \alpha \)) chosen determines our threshold for rejecting \( H_0 \).

A successful test outcome would show sufficient evidence leaning towards the alternative hypothesis, as demonstrated in our example when \( z = 2.891 \) exceeded the critical value 2.33, leading to a rejection of \( H_0 \).