Question

Auto Accidents A recent year's records of auto accidents occurring on a given section of highway were classified according to whether the resulting damage was \(\$ 3000\) or more and to whether a physical injury resulted from the accident. The data follows: $$\begin{array}{lcc}\hline & \text { Under } \$ 3000 & \$ 3000 \text { or More } \\\\\hline \text { Number of Accidents } & 32 & 41 \\\\\text { Number Involving Injuries } & 10 & 23\end{array}$$ a. Estimate the true proportion of accidents involving injuries when the damage was \(\$ 3000\) or more for similar sections of highway and find the margin of error. b. Estimate the true difference in the proportion of accidents involving injuries for accidents with damage under \(\$ 3000\) and those with damage of \(\$ 3000\) or more. Use a \(95 \%\) confidence interval.

Short answer

Answer: The 95% confidence interval for the true difference in the proportion of accidents involving injuries with damage under $3000 and those with damage of $3000 or more is approximately (-0.4775, -0.0195).

Step-by-step solution

Estimating the Proportion of Accidents with Injuries When Damage is \$3000 or More

To estimate the true proportion of accidents involving injuries when the damage was \$3000 or more, we will use the sample proportion formula: $$ \hat{p} = \frac{x}{n} $$ Where: - \(\hat{p}\) is the sample proportion - \(x\) is the number of accidents involving injuries - \(n\) is the total number of accidents in the category From the given data, \(x = 23\) and \(n = 41\). So, the sample proportion is: $$ \hat{p} = \frac{23}{41} \approx 0.5610 $$

Calculating Margin of Error for Proportion

Now, we need to calculate the margin of error (ME) for this proportion, using the following formula: $$ ME = Z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} $$ Where: - \(Z\) is the z-score corresponding to the desired confidence interval (for a 95% confidence interval, \(Z=1.96\)) In our case, the margin of error is: $$ ME = 1.96 \times \sqrt{\frac{0.5610 \times (1 - 0.5610)}{41}} \approx 0.1509 $$

Estimating Proportion and Margin of Error for Accidents with Injuries and Damage Under $3000

To estimate the true proportion of accidents involving injuries with damage under \$3000, we will apply the same sample proportion formula: $$ \hat{p} = \frac{x}{n} $$ From the given data, \(x = 10\) and \(n = 32\). So, the sample proportion is: $$ \hat{p} = \frac{10}{32} \approx 0.3125 $$

Estimating the True Difference in Proportion of Accidents with Damage Under \$3000 and \$3000 or More

Now we want to estimate the true difference in the proportion of accidents involving injuries with damage under \$3000 and those with damage of \$3000 or more, using a 95% confidence interval. The formula for estimating this difference is as follows: $$ (\hat{p}_1 - \hat{p}_2) \pm Z \times \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} $$ Where: - \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions for accidents with damage under \$3000 and those with damage of \$3000 or more, respectively - \(n_1\) and \(n_2\) are the total number of accidents in each category - \(Z\) is the z-score corresponding to the desired confidence interval (for a 95% confidence interval, \(Z=1.96\)) We already calculated the sample proportions earlier: - \(\hat{p}_1 \approx 0.3125\) - \(\hat{p}_2 \approx 0.5610\) Thus, we can calculate the difference in proportions and the margin of error: $$ (0.3125 - 0.5610) \pm 1.96 \times \sqrt{\frac{0.3125 \times (1 - 0.3125)}{32} + \frac{0.5610 \times (1 - 0.5610)}{41}} \approx -0.2485 \pm 0.2290 $$ So, the 95% confidence interval for the true difference in the proportion of accidents involving injuries with damage under \$3000 and those with damage of \$3000 or more is approximately \((-0.4775, -0.0195)\).

Key concepts

Proportion Estimation

Understanding how to estimate proportions is a fundamental concept in statistics that allows us to infer about a larger population based on sample data. Proportion estimation involves determining the ratio of a subset of interest within a larger group. For instance, referring to the auto accidents data provided, we can estimate the true proportion of accidents that result in physical injuries when the damage is \(3000 or more. Using the sample proportion formula \[ \hat{p} = \frac{x}{n} \], where \(hat{p}\) is the sample proportion, \({x}\) represents the number of accidents with injuries, and \({n}\) is the total number of accidents surveyed, we calculate a sample proportion. With \({x} = 23\) and \({n} = 41\), we obtain a sample proportion of approximately 0.5610.

Such estimations are crucial as they provide insights into the characteristics of populations where collecting data from every individual is impractical. To improve the comprehension of this step, juxtapose these calculations with a real-world example, such as predicting election outcomes or estimating the infection rate in a population during an epidemic.

Margin of Error Calculation

The margin of error is an important concept used to define the range within which we can expect the true proportion of the population to lie, with a certain level of confidence. Calculating the margin of error (ME) helps to understand the precision of our proportion estimation. The formula for the margin of error is \[ ME = Z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] where \({Z}\) is the z-score, which corresponds to the desired confidence level (e.g., \({Z}\) is about 1.96 for a 95% confidence interval), \({\hat{p}}\) is the estimated proportion, and \({n}\) is the sample size.

For the auto accidents data, when calculating the margin of error for the proportion of accidents with damages of \)3000 or more that involve injuries, we find it to be approximately 0.1509. It's paramount for students to recognize how this calculation affects the interpretation of data—the smaller the margin of error, the closer we are to pinpointing the true population parameter. To further clarify this concept, it's helpful to relate the calculation to everyday scenarios like forecasting weather conditions, where the margin of error indicates the confidence in predicting rain or sunshine.

Difference in Proportions

Comparing proportions between two groups reveals the 'Difference in Proportions.' This comparison often explores whether there's a significant variation in traits or occurrences between the groups. For the provided auto accidents data, we're interested in the difference between the proportion of accidents involving injuries with damages under \(3000 and those with damages of \)3000 or more. Employing the formula \[ (\hat{p}_1 - \hat{p}_2) \pm Z \times \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} \], where \({\hat{p}_1}\) and \({\hat{p}_2}\) are the sample proportions, and \({n_1}\) and \({n_2}\) are the sample sizes for the two groups, we can compute this difference along with a confidence interval.

The calculated 95% confidence interval for the difference in proportions is approximately (-0.4775, -0.0195), which means there's a significant difference in the injury rates between the two levels of damages, considering our confidence level. To enrich the learning experience, one could illustrate this concept by analyzing the difference in voter turnout across different age groups or comparing the effectiveness of two medications.