Question

an insurance agent says the standard deviation of the total hospital charges for patients involved in a vehicle struck a construction barricade is less than $3800. a random sample of 20 total hospital charg involved in this type of crash has a standard deviation of $4200. at α = 0.05 can you support the agent the p - value method to test the claim. identify the null and alternative hypotheses. let σ be the population standard deviation. a. h₀: σ < $3800, hₐ: σ ≥ $3800; b. h₀: σ > $3800, hₐ: σ ≤ $3800; c. h₀: σ ≤ $3800, hₐ: σ > $3800; d. h₀: σ ≥ $3800, hₐ: σ < $3800. identify the standardized test statistic. 23.21 (round to two decimal places as needed.) identify the p - value. (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the chi - square test statistic for standard deviation

The formula for the chi - square test statistic \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the population standard deviation under the null hypothesis.
We are given that \(n = 20\), \(s=\$4200\), and \(\sigma=\$3800\) (from the null hypothesis \(H_{0}:\sigma\geq3800\), we can use \(\sigma = 3800\) for the calculation of the test statistic).
First, calculate \((n - 1)s^{2}\):
\(n-1=20 - 1=19\)
\(s^{2}=(4200)^{2}=17640000\)
\((n - 1)s^{2}=19\times17640000 = 335160000\)
\(\sigma^{2}=(3800)^{2}=14440000\)

Step2: Calculate the chi - square test statistic

\(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}=\frac{335160000}{14440000}\approx23.21\) (which matches the given test statistic)

Step3: Determine the degrees of freedom and find the P - value

The degrees of freedom \(df=n - 1=20 - 1 = 19\)
We are performing a left - tailed test (since \(H_{a}:\sigma\lt3800\)) and the test statistic \(\chi^{2}=23.21\)
We can use a chi - square distribution table or a statistical software to find the P - value. For a chi - square distribution with \(df = 19\) and \(\chi^{2}=23.21\), the P - value is the probability that \(\chi^{2}\lt23.21\) (for a left - tailed test).
Using a chi - square calculator or software, we know that the cumulative distribution function for chi - square with \(df = 19\) gives \(P(\chi^{2}\lt23.21)=1 - P(\chi^{2}\geq23.21)\)
Looking up in the chi - square table or using a calculator, we find that \(P(\chi^{2}\geq23.21)\approx0.034\), so \(P(\chi^{2}\lt23.21)=1 - 0.034 = 0.966\)

Answer:

0.966