Question

For Exercises 7 through \(27,\) 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. Interview Errors It has been found that many first-time 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

At \( \alpha = 0.01 \), if the absolute z-value exceeds 2.576, there is a significant difference. Otherwise, there isn't.

Step-by-step solution

State the Hypotheses

Define the null and alternative hypotheses. The null hypothesis is that there is no difference between male and female offenders in terms of cell phone usage, i.e., \( H_0: p_1 - p_2 = 0 \). The alternative hypothesis is that there is a difference, i.e., \( H_1: p_1 - p_2 eq 0 \). The claim is the alternative hypothesis.

Find the Critical Value(s)

At a significance level of \( \alpha = 0.01 \) for a two-tailed test, find the critical z-values. Using the standard normal distribution table, the z-values corresponding to \( \alpha/2 = 0.005 \) are approximately \( \pm 2.576 \). These are the critical values. If the test statistic falls beyond these values, the null hypothesis will be rejected.

Compute the Test Value

Calculate the pooled sample proportion \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \), where \( x_1 = 72 \), \( n_1 = 120 \), \( x_2 = 80 \), and \( n_2 = 150 \). So, \( \hat{p} = \frac{72 + 80}{120 + 150} = \frac{152}{270} \approx 0.563 \).Next, compute the standard error \( SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \). Substitute \( \hat{p} = 0.563 \), \( n_1 = 120 \), and \( n_2 = 150 \) into the formula. Calculate the z-value: \[ z = \frac{\left(\frac{x_1}{n_1} - \frac{x_2}{n_2}\right)}{SE} = \frac{\left(\frac{72}{120} - \frac{80}{150}\right)}{SE} \approx \frac{0.6 - 0.533}{SE} \approx \frac{0.067}{SE} \] Substitute the computed SE to find \( z \).

Make the Decision

Compare the computed z-value to the critical values. If the z-value is beyond \( \pm 2.576 \), reject the null hypothesis. Otherwise, do not reject it.

Summarize the Results

If the null hypothesis is rejected, conclude there is a significant difference between the proportions of male and female offenders who used their cell phone during an interview. If not rejected, conclude there is no significant evidence to suggest a difference at the 0.01 level of significance.

Key concepts

Null Hypothesis

In hypothesis testing, the null hypothesis is a starting assumption that there is no effect, or no difference, in the population. It's denoted by \(H_0\). In the context of the exercise, the null hypothesis states that there is no difference between male and female offenders when it comes to using cell phones during interviews. Mathematically, this is expressed as \( p_1 - p_2 = 0 \), where \( p_1 \) is the proportion of male offenders and \( p_2 \) is the proportion of female offenders. The null hypothesis functions as a baseline that we test against to see if there is sufficient evidence to suggest a significant difference. It is always an equality statement, implying that whatever outcome you're testing, the initial assumption is that things remain unchanged or equal. When interpreting results, if your test statistic falls outside the critical region, you reject the null hypothesis.

Alternative Hypothesis

The alternative hypothesis, denoted as \(H_1\), is what you might consider the opposite of the null hypothesis. It represents a claim in hypothesis testing that there is indeed an effect or a difference. In our exercise, the alternative hypothesis claims that there is a difference in cell phone usage between male and female offenders during interviews. This is mathematically expressed as \( p_1 - p_2 eq 0 \). Here, this statement aligns with the researcher's claim that the proportions differ, not specifying which one is higher or lower, only that they are not equal. The purpose of proving or supporting an alternative hypothesis is to find evidence strong enough to reject the null hypothesis. Notably, the alternative hypothesis is what usually aligns with the research question or the presumed effect the researcher aims to demonstrate.

Significance Level

The significance level is a critical component in hypothesis testing that helps you decide whether to reject the null hypothesis. It is usually denoted by \( \alpha \) and represents the probability of making a Type I error, which is rejecting a true null hypothesis. In our exercise, the significance level is set at 0.01. This means there is a 1% risk of concluding that there is a difference between male and female offenders in their cell phone usage during interviews when there actually isn't. This significance level choice impacts the critical values that determine the cutoff points for rejecting the null hypothesis. Common significance levels are 0.05 and 0.01, with lower levels providing stronger evidence against the null hypothesis, as they show more conservative thresholds for claiming significant results.

Standard Error

Standard error is a measure of how much variability exists in a sample statistic relative to the underlying population parameter. It provides an estimate of the noise level in our data analysis, helping us understand how a sample proportion might oscillate around the real population proportion. For the cell phone usage example, standard error is essential in calculating the z-value during step 3. Its formula in the context of comparing two proportions is given by \[ SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \] where \( \hat{p} \) is the pooled sample proportion, \( n_1 \) and \( n_2 \) are the sample sizes.This calculation allows you to determine whether the observed difference between sample proportions is significant or simply due to sampling variability. The lower the standard error, the more reliable the test statistic will be, helping to make a decisive argument about the difference or similarity of proportions in hypothesis testing.