Question

In Exercise 2.4, we presented the annual 2010 premium for a male, licensed for \(6-8\) years, who drives a Honda Accord 12,600 to 15,000 miles per year and has no violations or accidents. \({ }^{11}\) $$ \begin{array}{lcc} \text { City } & \text { GEICO (\$) } & \text { 21st Century (\$) } \\ \hline \text { Long Beach } & 2780 & 2352 \\ \text { Pomona } & 2411 & 2462 \\ \text { San Bernardino } & 2261 & 2284 \\ \text { Moreno Valley } & 2263 & 2520 \end{array} $$ a. Why would you expect these pairs of observations to be dependent? b. Do the data provide sufficient evidence to indicate that there is a difference in the average annual premiums between GEICO and 21 st Century insurance? Test using \(\alpha=.01\). c. Find the approximate \(p\) -value for the test and interpret its value. d. Find a \(99 \%\) confidence interval for the difference in the average annual premiums for GEICO and 2 1st Century insurance. e. Can we use the information in the table to make valid comparisons between GEICO and 21 st Century insurance throughout the United States? Why or why not?

Short answer

Answer: No, the information in the table is limited to only 4 cities and one specific type of driver. Thus, it is not sufficient to make valid comparisons between GEICO and 21st Century insurance throughout the United States. To make a valid comparison, a more comprehensive dataset comprising of several types of drivers and various locations would be needed.

Step-by-step solution

Analyze the relationship between premiums in the cities

The pairs of observations can be considered dependent because they are collected for the same type of driver in the same cities. That means the characteristics used by GEICO and 21st Century insurance to calculate premiums might be similar, such as driving habits, risk factors, and population density. The premium in one city may be influenced by the same factors for both the insurance companies. #b. Do the data provide sufficient evidence to indicate that there is a difference in the average annual premiums between GEICO and 21 st Century insurance? Test using \(\alpha=.01\).#

Perform a paired t-test for mean difference

To test for a difference in the average annual premiums, we will perform a paired t-test. First, calculate the differences in premiums for each city: 1. Long Beach: \(2780-2352=428\) 2. Pomona: \(2411-2462=-51\) 3. San Bernardino: \(2261-2284=-23\) 4. Moreno Valley: \(2263-2520=-257\) Now we will find the mean (\(\bar{d}\)) and standard deviation (\(s_d\)) of these differences: \(\bar{d} = \frac{428-51-23-257}{4} = \frac{97}{4} = 24.25\) \(s_d = \sqrt{\frac{(428-24.25)^2 +(-51-24.25)^2 +(-23-24.25)^2 +(-257-24.25)^2}{4-1}} \approx 251.07\) Next, we will calculate the t-value: \(t = \frac{\bar{d}}{s_d/\sqrt{4}} = \frac{24.25}{251.07/\sqrt{4}} \approx 0.305\) Using a t-distribution table or calculator, we find that the t-value of \(0.305\) corresponds to a two-sided p-value greater than \(0.05\) (exact value not found but it's ok since we're comparing it with \(\alpha=0.01\)). Since this p-value is greater than our significance level \(\alpha=.01\), we fail to reject the null hypothesis. There is not sufficient evidence to indicate a difference in the average annual premiums. #c. Find the approximate \(p\)-value for the test and interpret its value.#

Approximate p-value and interpretation

As we found during the t-test, the approximate p-value is larger than \(0.05\). This means that we do not have enough evidence to reject the null hypothesis. In other words, we cannot say for sure that there is a significant difference in the average annual premiums. #d. Find a \(99\%\) confidence interval for the difference in the average annual premiums for GEICO and 2 1st Century insurance.#

Calculate 99% confidence interval

To find the 99% confidence interval, we will calculate the margin of error using the t-distribution with 3 degrees of freedom (4 observations - 1) and \(\alpha=0.01\): \(ME= t_{\alpha/2, df} \times \frac{s_d}{\sqrt{n}} = t_{0.005, 3} \times \frac{251.07}{\sqrt{4}}\) \(t_{0.005, 3} \approx 5.841\) (from t-distribution tables) \(ME = 5.841 \times \frac{251.07}{\sqrt{4}} \approx 725.49\) Now we can calculate the confidence interval: \(CI = (\bar{d} - ME, \bar{d} + ME) = (24.25 - 725.49, 24.25 + 725.49) = (-701.24, 749.74)\) The 99% confidence interval for the difference in average annual premiums is \((-\raise.17ex\hbox{\)\scriptstyle 701\(}.24, \raise.17ex\hbox{\)\scriptstyle 749\(}.74)\). #e. Can we use the information in the table to make valid comparisons between GEICO and 21 st Century insurance throughout the United States? Why or why not?#

Validity of comparisons for the United States

The information in the table is limited to only 4 cities and one specific type of driver. Thus, it is not sufficient to make valid comparisons between GEICO and 21st Century insurance throughout the United States. Several factors might change depending on the location and type of driver, resulting in different comparative results. To make a valid comparison, a more comprehensive dataset comprising of several types of drivers and various locations would be needed.

Key concepts

Dependent Observations

When analyzing data involving paired t-tests, understanding dependent observations is crucial. In this context, dependent observations refer to pairs of data that are linked or matched in some way, such that the measurement in one element of the pair may directly affect the other.
For instance, in the given exercise, the premium values for GEICO and 21st Century insurance are gathered for the same type of driver in each of the cities listed.

• This pairing means that any factors affecting the insurance premium (such as traffic conditions, parking availability, or city population) are likely consistent across both companies in the same city.
• Thus, the observations are not independent; they are dependent on the conditions specific to the city and type of driver.

This dependency is central because a paired t-test can adjust for these shared characteristics, making it a suitable test for detecting differences in means between the two related groups.

Mean Difference

The mean difference is a core concept that emerges when conducting paired t-tests. In simple terms, it is the average difference between paired observations.
For the exercise, the differences between the premiums of GEICO and 21st Century in each city context provide a basis for analysis. By calculating these differences, one can explore whether there is a consistent premium difference across cities.
The computation involves:

• Finding the difference for each pair (e.g., Long Beach: \( 2780 - 2352 = 428 \)).
• Finding the mean of these differences. Here, the mean difference \( \bar{d} \) is calculated as \( \frac{428 - 51 - 23 - 257}{4} = 24.25 \).

This mean difference helps determine whether the premiums are consistently higher or lower for one company compared to the other across all locations.

Confidence Interval

Understanding the confidence interval is key to interpreting the results of a paired t-test.
A confidence interval gives a range within which the true mean difference between two related groups (like the premiums by two insurance companies) is expected to lie, for a given level of confidence.
In the exercise, a 99% confidence interval is constructed, meaning we can be 99% confident that the true mean difference in premiums falls within this interval.
The steps include:

• Calculating the margin of error \( ME = t_{\alpha/2, df} \times \frac{s_d}{\sqrt{n}} \), given a selected t-value for 99% confidence.
• Finding the confidence interval using\( CI = (\bar{d} - ME, \bar{d} + ME) \).

The resulting interval \((-701.24, 749.74)\) suggests that the mean difference in premiums could realistically be anywhere within this range, indicating a high level of uncertainty in pinpointing a precise mean difference.

Null Hypothesis Testing

Null hypothesis testing is a foundational method used to deduce whether a perceived effect or difference, like the premium differences, is statistically significant.
The null hypothesis (often denoted as \( H_0 \)) generally states that there is no effect or difference. In our case, \( H_0 \) postulates no significant difference in the average premiums between GEICO and 21st Century insurance.
During hypothesis testing:

• A paired t-test is conducted to assess \( H_0 \), based on the calculated t-value and associated p-value.
• The decision rule is straightforward:
• If the p-value is less than the significance level \( \alpha \) (here \( 0.01 \)), \( H_0 \) is rejected.
• If not, as found in the exercise \( (p > 0.05) \), \( H_0 \) is not rejected.

In this case, the conclusion drawn is that no statistically significant difference is evident at the 1% significance level. Consequently, the data does not provide sufficient evidence to claim the premiums are different, reinforcing the importance of rigorous testing in statistical analysis.