Question

an insurance agent says the standard deviation of the total hospital charges for patients involved in a crash in which the vehicle struck a construction barricade is less than $3800. a random sample of 20 total hospital charges for patients involved in this type of crash has a standard deviation of $4200. at α = 0.05 can you support the agent’s claim? use the p - value method to test the claim.

identify the null and alternative hypotheses. let σ be the population standard deviation.

a. ( h_0: sigma < $3800 ), ( h_a: sigma geq $3800 )

b. ( h_0: sigma > $3800 ), ( h_a: sigma leq $3800 )

c. ( h_0: sigma leq $3800 ), ( h_a: sigma > $3800 )

d. ( h_0: sigma geq $3800 ), ( h_a: sigma < $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 when testing a claim about the population standard deviation is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma\) is the hypothesized population standard deviation.
We are given that \(n = 20\), \(s=\$4200\), and \(\sigma=\$3800\).
First, calculate \((n - 1)s^{2}\) and \(\sigma^{2}\):
\(n-1=20 - 1=19\)
\(s^{2}=(4200)^{2}=17640000\)
\((n - 1)s^{2}=19\times17640000 = 335160000\)
\(\sigma^{2}=(3800)^{2}=14440000\)
Then, \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma^{2}}=\frac{335160000}{14440000}\approx23.21\) (which matches the given test statistic)

Step2: Determine the degrees of freedom and the type of test

The degrees of freedom for a chi - square test for standard deviation is \(df=n - 1=19\).
The null hypothesis is \(H_{0}:\sigma\geq3800\) and the alternative hypothesis is \(H_{a}:\sigma < 3800\), so this is a left - tailed test. But when using the chi - square distribution, we can find the P - value by considering the right - tailed probability and then subtracting from 1 (since for a left - tailed test with test statistic \(\chi^{2}\), \(P - value=1 - P(\chi^{2}<\chi_{calc}^{2})\) or we can use the cumulative distribution function of the chi - square distribution).
We can use a chi - square distribution table or a statistical software (like a TI - 84 Plus or R) to find the P - value.
Using the chi - square distribution with \(df = 19\) and \(\chi^{2}=23.21\), we want to find \(P(\chi^{2}<23.21)\) for a left - tailed test. But since the chi - square distribution is right - skewed, we can also find \(P(\chi^{2}>23.21)\) and then \(P - value = 1 - P(\chi^{2}>23.21)\) for a left - tailed test (because the total area under the curve is 1).
Using a chi - square calculator or software:
The cumulative distribution function for chi - square with \(df = 19\) at \(\chi^{2}=23.21\) gives \(P(\chi^{2}\leq23.21)\approx0.961\) (for a left - tailed test, the P - value is the probability that \(\chi^{2}\) is less than or equal to the test statistic when the null hypothesis is true. Wait, actually, for \(H_{0}:\sigma\geq\sigma_{0}\) and \(H_{a}:\sigma<\sigma_{0}\), the test statistic is \(\chi^{2}=\frac{(n - 1)s^{2}}{\sigma_{0}^{2}}\). The P - value is \(P(\chi^{2}<\frac{(n - 1)s^{2}}{\sigma_{0}^{2}})\) when \(H_{0}\) is true.
Using a chi - square table or a calculator (for example, in R, pchisq(23.21,19) gives approximately 0.961). So the P - value is approximately 0.961.

Answer:

0.961