Question

Does It Pay to Haggle? In Exercise \(16,\) a survey done by Consumer Reports indicates that you should always try to negotiate for a better deal when shopping or paying for services. \({ }^{20}\) In fact, based on the survey, \(37 \%\) of the people under age 34 were more likely to "haggle," while only \(13 \%\) of those 65 and older. Suppose that this group included 72 people under the age of 34 and 55 people who are 65 or older. a. What are the values of \(\hat{p}_{1}\) and \(\hat{p}_{2}\) for the two groups in this survey? b. Find a \(95 \%\) confidence interval for the difference in the proportion of people who are more likely to "haggle" in the "under 34 " versus " 65 and older" age groups. c. What conclusions can you draw regarding the groups compared in part b?

Short answer

If yes, which age group is more likely to haggle? Answer: Yes, there is a significant difference between the likelihood of people in the two age groups to haggle. People under the age of 34 are more likely to haggle than those who are 65 and older.

Step-by-step solution

Calculate sample proportions \(\hat{p}_1\) and \(\hat{p}_2\)

The given probabilities of haggling for each age group are \(37 \%\) for under 34 and \(13 \%\) for 65 and older. We will use these percentages to calculate the number of people who haggle in each group by multiplying the respective probability by the number of people in each group. For under 34: $$ \hat{p}_1 = \frac{\text{Number of people haggling in under 34}}{\text{Total number of people in under 34}} = \frac{0.37 \times 72}{72} $$ For 65 and older: $$ \hat{p}_2 = \frac{\text{Number of people haggling in 65 and older}}{\text{Total number of people in 65 and older}} = \frac{0.13 \times 55}{55} $$ Now, let's calculate the values of \(\hat{p}_1\) and \(\hat{p}_2\). $$ \hat{p}_1 = 0.37 \\ \hat{p}_2 = 0.13 $$

Finding 95% Confidence Interval

We want to find the 95% confidence interval for the difference in the proportion of people who are more likely to haggle between the two age groups. The formula for the confidence interval can be written as: $$ (\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2}\sqrt{\frac{\hat{p}_1 (1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2 (1-\hat{p}_2)}{n_2}} $$ where \(n_1\) and \(n_2\) are the number of people in each age group, and \(z_{\alpha/2}\) is the critical value for the desired level of confidence (in our case, it is 95%, therefore \(z_{\alpha/2} = 1.96\)). Now, we can calculate the confidence interval using the given values: $$ (0.37 - 0.13) \pm 1.96\sqrt{\frac{0.37 (1-0.37)}{72}+\frac{0.13 (1-0.13)}{55}} $$ After simplifying, $$ 0.24 \pm 1.96\sqrt{\frac{0.2329}{72}+\frac{0.1131}{55}} = 0.24 \pm 1.96(0.0646) $$ Now, multiply 1.96 by 0.0646 to get: $$ 0.24 \pm 0.1265 $$ So, the 95% confidence interval for the difference in the proportion of people who are more likely to haggle between the two age groups is \((0.1135, 0.3665)\).

Drawing conclusions

The 95% confidence interval for the difference in the proportion of people who are more likely to haggle among under 34 and 65 and older groups is \((0.1135, 0.3665)\). Since the interval does not include \(0\), we can conclude that there is a significant difference between the two age groups in terms of their likelihood to haggle. Specifically, people under the age of 34 are more likely to haggle than those who are 65 and older.

Key concepts

Confidence Interval Calculation

Understanding the confidence interval is crucial for interpreting study results in statistics. It provides a range within which you can be a certain percentage sure that the true population parameter lies. For example, a 95% confidence interval implies that if we collected 100 different samples and calculated a confidence interval for each, 95 of those intervals would contain the true population parameter.
In the context of the exercise provided, calculating a confidence interval for the difference in proportions involves several steps. First, you identify the sample proportions, \(\hat{p}_1\) and \(\hat{p}_2\), for each group. Next, you determine the standard error of the difference in proportions. This involves calculating the variance for each sample proportion and then summing these variances. The standard error is the square root of this sum. Finally, multiplying the standard error by the z-score for the desired confidence level gives the margin of error. Adding and subtracting this margin of error from the difference in sample proportions yields the confidence interval.
When performing such calculations, it is critical to remind students to use consistent units and clearly interpret the results. It's also essential to explain the relevance of the confidence level selected (e.g., 95%) and the role of the z-score. In our example, the z-score corresponds to a confidence level of 95%, indicating that the z-score captures the middle 95% of the standard normal distribution.

• Identify the sample proportions (\(\hat{p}_1, \hat{p}_2\)).
• Calculate the standard error for the difference in proportions.
• Determine the z-score associated with the desired confidence level.
• Calculate the confidence interval using the formula.
• Interpret the results with a focus on the real-world implication of the interval.

Proportion Difference

The difference in proportions is a comparative statistic that allows us to analyze the variance between two groups with respect to a particular characteristic—in this case, the tendency to haggle. In simple terms, we're looking at whether one group is more likely to exhibit a behavior than another and by how much.
In the worked-out exercise, the proportion difference is highlighted by comparing the likelihood of bargaining between two distinct age groups. Mathematically, this difference is calculated as \(\hat{p}_1 - \hat{p}_2\). This gives us a preliminary idea of the disparity before accounting for the potential variability in the samples. It is important to emphasize that the proportion difference alone doesn't account for sample size or the distribution of the sample data, which is why we compute a confidence interval to understand the range of possible differences given a level of confidence.
Explicit communication with students about how sample size and variability can affect the proportion difference, and ultimately the confidence interval, is paramount. Larger samples, for instance, tend to provide more accurate estimates of the population parameter, as they reduce the impact of outliers and random variations. Encouraging students to critically evaluate these factors deepens their understanding of statistical analysis.

• Explain the formula for computing the difference in proportions (\(\hat{p}_1 - \hat{p}_2\)).
• Discuss the importance of sample sizes in determining the accuracy of the proportion difference.
• Emphasize the relevance of the proportion difference in a real-world context, such as decision-making or policy development.

Statistical Significance

Statistical significance plays a key role in hypothesis testing and helps researchers decide if a result is likely due to chance or if it reflects a true effect in the population. In the textbook solution, the finding that the confidence interval for the difference in haggling proportions between the two age groups does not include zero suggests that the observed difference is statistically significant. This means that the likelihood of observing such a difference if there were actually no difference in the population is very small, typically less than 5% given a 95% confidence interval.
Educationally, it's imperative to convey to students that statistical significance is not only about the numbers but also about the context. A difference could be statistically significant but not necessarily meaningful in a practical sense. For example, a very small difference in proportions could be statistically significant with a large sample size, but it may not be of practical importance.
Conversely, students should also be aware that a statistically significant result does not prove that one variable causes changes in another—only that there is an association. Correlation does not imply causation, and this distinction is critical in data interpretation.

• Define statistical significance and its connection to the confidence interval.
• Teach the difference between statistical significance and practical significance.
• Clarify why statistical significance does not establish causation.
• Explain the role of the p-value and the alpha level in assessing significance.