Questions: The records of a casualty insurance company show that, in the past, its clients have had a mean of 1.9 auto accidents per day with a variance of 0.0016. The actuaries of the company claim that the variance of the number of accidents per day is no longer equal to 0.0016. Suppose that we want to carry out a hypothesis test to see if there is support for the actuaries' claim. State the null hypothesis (H0) and the alternative hypothesis (H1) that we would use for this test. (H0: square) (H1: square)

Solution Steps

We are testing the variance of the number of auto accidents per day. The null and alternative hypotheses are defined as follows:

\[ H_0: \sigma^2 = 0.0016 \] \[ H_1: \sigma^2 \neq 0.0016 \]

The test statistic for the chi-square test is calculated using the formula:

\[ \chi^2 = \frac{(n-1) s^2}{\sigma_0^2} \]

Substituting the values:

\[ \chi^2 = \frac{(30 - 1) \cdot 0.0016}{0.0016} = 29.0 \]

To find the P-value, we calculate the cumulative distribution function (CDF) of the chi-square distribution for the test statistic:

\[ P = P(\chi^2(29) \geq 29.0) = 0.9301 \]

For a two-tailed test at a significance level of \( \alpha = 0.05 \), the critical values for the chi-square distribution with \( 29 \) degrees of freedom are:

\[ \text{Critical Values} = [16.0471, 45.7223] \]

Since the test statistic \( \chi^2 = 29.0 \) falls within the critical values \( [16.0471, 45.7223] \) and the P-value \( 0.9301 \) is greater than \( \alpha = 0.05 \), we fail to reject the null hypothesis.

Final Answer