Questions: An insurance agent says that the standard deviation of the total hospital charges for patients involved in a crash is 3900. A random sample of 20 crash victims has a standard deviation of 4000. At α = 0.10, can you support the agent's claim? Use the P-value method to test the claim. H₀: σ = 3900 H₁: σ ≠ 3900 Identify the standardized test statistic to be used: χ² = 19(4000/3900)² = 20.5319

Solution Steps

To test the claim of the insurance agent regarding the standard deviation of hospital charges, we first calculate the test statistic using the formula for the chi-square statistic:

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

Where:

• \( n = 20 \) (sample size)
• \( s = 4000 \) (sample standard deviation)
• \( \sigma_0 = 3900 \) (hypothesized population standard deviation)

Substituting the values, we have:

\[ \chi^2 = \frac{(20 - 1) \cdot (4000)^2}{(3900)^2} = \frac{19 \cdot 16000000}{15210000} = 19.9869 \]

Next, we determine the P-value corresponding to the calculated chi-square test statistic. The P-value is calculated based on the cumulative distribution function (CDF) of the chi-square distribution with \( n-1 = 19 \) degrees of freedom.

For our test statistic:

\[ P = P(\chi^2(19) \geq 19.9869) = 0.7907 \]

We compare the P-value with the significance level \( \alpha = 0.10 \):

• \( P = 0.7907 \)
• \( \alpha = 0.10 \)

Since \( P > \alpha \), we fail to reject the null hypothesis.

Based on the results, we conclude that there is not enough evidence to reject the agent's claim regarding the standard deviation of the total hospital charges for patients involved in a crash.

Final Answer