91Ó°ÊÓ

The president of the American Insurance Institute wants to compare the yearly costs of auto insurance offered by two leading companies. He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The data for the sample of families and their two insurance quotes are reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted? $$ \begin{array}{|lcc|} \hline \text { Family } & \begin{array}{c} \text { Midstates } \\ \text { Car Insurance } \end{array} & \begin{array}{c} \text { Gecko } \\ \text { Mutual Insurance } \end{array} \\ \hline \text { Becker } & \$ 2,090 & \$ 1,610 \\ \text { Berry } & 1,683 & 1,247 \\ \text { Cobb } & 1,402 & 2,327 \\ \text { Debuck } & 1,830 & 1,367 \\ \text { DuBrul } & 930 & 1,461 \\ \text { Eckroate } & 697 & 1,789 \\ \text { German } & 1,741 & 1,621 \\ \text { Glasson } & 1,129 & 1,914 \\ \text { King } & 1,018 & 1,956 \\ \text { Kucic } & 1,881 & 1,772 \\ \text { Meredith } & 1,571 & 1,375 \\ \text { Obeid } & 874 & 1,527 \\ \text { Price } & 1,579 & 1,767 \\ \text { Phillips } & 1,577 & 1,636 \\ \text { Tresize } & 860 & 1,188 \\ \hline \end{array} $$

Step by step solution

01

Formulate Hypotheses

We begin by setting up the null and alternative hypotheses for the test. The null hypothesis \( H_0 \) is that there is no difference in the mean costs of auto insurance between the two companies, \( \mu_1 - \mu_2 = 0 \). The alternative hypothesis \( H_a \) is that there is a difference, \( \mu_1 - \mu_2 eq 0 \).

02

Calculate Differences

For each family, calculate the difference in insurance quotes between Midstates Car Insurance and Gecko Mutual Insurance. Let \( d_i = x_{1i} - x_{2i} \), where \( x_{1i} \) and \( x_{2i} \) are the quotes from Midstates and Gecko, respectively.

03

Compute Mean and Standard Deviation of Differences

Find the mean \( \bar{d} \) and the standard deviation \( s_d \) of the differences \( d_i \). This is done by computing \( \bar{d} = \frac{\sum d_i}{n} \) and \( s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \), where \( n \) is the number of observations.

04

Perform t-Test

Calculate the t-statistic using \( t = \frac{\bar{d}}{s_d / \sqrt{n}} \). Compare the calculated t-value with the critical t-value from the t-distribution table at \( n-1 = 14 \) degrees of freedom for a significance level of \( \alpha = 0.10 \).

05

Make a Decision

If the absolute value of the calculated t-statistic is greater than the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-test

The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In the context of our auto insurance comparison, it's used to see if there is a difference in annual insurance quotes between Midstates and Gecko insurance companies. Here’s how it works:

• We first calculate the mean difference between the two sets of data (e.g., the insurance quotes from each company).
• We then determine the standard deviation of these differences. This helps us understand the spread or variability in the differences.
• With these parameters, we compute a t-statistic, which shows how far our sample difference is from the null hypothesis (no difference between means).

The t-statistic is then compared against a critical value from the t-distribution to decide whether the observed difference is statistically significant.

significance level

The significance level, denoted by \( \alpha \), is a threshold set by the researcher before conducting a test. It determines how much evidence we require before rejecting the null hypothesis. In our exercise, a significance level of 0.10 is used. This means we are willing to accept a 10% chance of wrongly rejecting the null hypothesis when it is actually true.

• A lower significance level (e.g., 0.05 or 0.01) indicates stricter criteria for rejecting the null hypothesis.
• A higher significance level allows more flexibility but increases the risk of a Type I error, where we incorrectly assume a difference exists when it doesn't.

Using a significance level is crucial as it guides the decision-making process in hypothesis testing, balancing between sensitivity and accuracy.

null hypothesis

The null hypothesis, symbolized as \( H_0 \), is a statement used in hypothesis testing that assumes no effect or no difference. In our scenario, the null hypothesis states that there is no difference in the mean insurance costs between Midstates and Gecko.

• It serves as a starting point for statistical testing and is what we're essentially 'testing against'.
• If we have enough evidence from our sample to reject the null hypothesis, it suggests that the observed effect (in this case, the difference in insurance costs) is unlikely to have occurred by random chance.

The goal is to gather data and perform statistical tests, like the t-test, to determine if we can confidently reject \( H_0 \) or not.

alternative hypothesis

The alternative hypothesis, denoted as \( H_a \), claims that there is a statistically significant effect or difference. In the context of the insurance study, it argues that Midstates and Gecko do indeed have different average costs for auto insurance.

• While the null hypothesis takes the position of no change or no difference, the alternative hypothesis is actively proposing an effect or difference.
• Our aim in hypothesis testing is often to provide evidence for the alternative hypothesis by demonstrating that the null hypothesis can be rejected.

The alternative hypothesis is what we might accept if our data can demonstrate such a relationship or difference with enough certainty, defined by our chosen significance level.