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. It has been found that many firsttime interviewees commit errors that could very well affect the outcome of the interview. An astounding \(77 \%\) are guilty of using their cell phones or texting during the interview! A researcher wanted to see if the proportion of male offenders differed from the proportion of female ones. Out of 120 males, 72 used their cell phone and 80 of 150 females did so. At the 0.01 level of significance is there a difference?

Short answer

No significant difference found at 0.01 significance level.

Step-by-step solution

State the Hypotheses

We begin by stating the null and alternative hypotheses. The null hypothesis (\(H_0\)) is: there is no difference in the proportions of males and females who use their cell phones during an interview. The alternative hypothesis (\(H_a\)) is: there is a difference in the proportions of males and females who use their cell phones during an interview. Thus, \(H_0 : p_1 = p_2\) and \(H_a : p_1 eq p_2\), where \(p_1\) and \(p_2\) represent the proportion of male and female offenders, respectively.

Identify the Claim

The claim we are testing is that there is a difference in the proportion of male and female offenders who use their cell phone during an interview. This claim is represented by the alternative hypothesis \(H_a : p_1 eq p_2\).

Find the Critical Value(s)

For a two-tailed test at a significance level of \(\alpha = 0.01\), we look up the critical z-values from the standard normal distribution table. The critical values are \(z = \pm 2.576\). These values determine the rejection region for our hypothesis test.

Compute the Test Value

First, we calculate the sample proportions for males and females: \(\hat{p}_1 = \frac{72}{120} = 0.6\) and \(\hat{p}_2 = \frac{80}{150} = 0.5333\). The combined sample proportion \(\hat{p}\) is calculated as follows: \(\hat{p} = \frac{72 + 80}{120 + 150} = 0.5643\). Next, we compute the standard error (SE): \(SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.5643 \times 0.4357 \times (\frac{1}{120} + \frac{1}{150})} = 0.0638\). The test statistic (z) is then calculated by: \(z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.6 - 0.5333}{0.0638} \approx 1.044\).

Make the Decision

Since the calculated test statistic \(z = 1.044\) does not fall in the critical region (i.e., it is not beyond \(z = \pm 2.576\)), we fail to reject the null hypothesis \(H_0\). This means there is not enough evidence to support the claim that there is a difference between the proportions of male and female offenders at the \(0.01\) significance level.

Summarize the Results

At the \(0.01\) significance level, there is insufficient evidence to conclude that there is a difference in the proportion of males and females who use their cell phone during an interview. The null hypothesis that there is no difference in proportions cannot be rejected.

Key concepts

Null Hypothesis

The null hypothesis is a critical part of hypothesis testing. It is a statement that suggests there is no effect or no difference between groups or variables. In our case, the null hypothesis states that there is no difference in the proportions of males and females who use their cell phones during an interview. We denote this as \( H_0: p_1 = p_2 \), where \( p_1 \) is the proportion of male offenders and \( p_2 \) is the proportion of female offenders.
Understanding the null hypothesis is important because the test revolves around either rejecting this hypothesis or not. It acts as the default or starting assumption that the research aims to challenge. When we say we "accept" the null hypothesis, it means we lack enough evidence to conclude that any other claim is valid.

Alternative Hypothesis

The alternative hypothesis is what you usually hope to prove through your research. It directly contradicts the null hypothesis. For this exercise, the alternative hypothesis suggests that there is a difference in the proportions of male and female offenders who use their cell phones during interviews. In mathematical notation, it's expressed as \( H_a: p_1 eq p_2 \).
This is the hypothesis that states there is an effect or a difference—essentially, what the researcher aims to support or prove. If our test results show sufficient evidence, we reject the null hypothesis in favor of this alternative. In practical research scenarios, the results of tests often lead to further investigations or interventions if the alternative hypothesis is accepted.

Significance Level

The significance level, denoted by \( \alpha \), is a threshold you set before the testing process to determine the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it is the risk of making a Type I error—finding a difference when there isn't one.
In this exercise, a common significance level of 0.01 is used, which means we are allowing a 1% risk of concluding that there is a difference between male and female offenders when, in fact, none exists. A smaller significance level means stricter criteria for detecting a real effect, thereby reducing the probability of type I errors, but possibly increasing the chance of type II errors—failing to detect a difference that actually exists.

Test Statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It measures how far away your sample statistic is from the null hypothesis expectation in terms of standard errors.
In this context, the test statistic is computed using the formula for a two-proportion z-test: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \]Where \( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions of male and female offenders, respectively, and \( SE \) is the standard error of the difference between the two proportions.
In our example, the calculated test statistic is approximately 1.044. Comparing this value to the critical value helps decide whether to reject the null hypothesis. If the test statistic falls into the critical region (beyond the critical z-values), we reject \( H_0 \). Otherwise, as in this exercise, we fail to reject it.