Question

can calculate the p - value for a chi - square test using technology. after calculating the standardiz cumulative distribution function (cdf) to calculate the area under the curve. use the p - value me n. ospital spokesperson claims that the standard deviation of the waiting times experienced by patie ergency department is no more than 0.5 minutes. a random sample of 20 waiting times has a sta minutes. at α = 0.05, is there enough evidence to reject the spokesperson’s claim? entry the p - value. = 0.0803 (round to four decimal places as needed.) hoose the correct conclusion below. a. fail to reject h₀. there is not enough evidence at the 5% level of significance to reject the sp claim. b. reject h₀. there is enough evidence at the 5% level of significance to reject the spokesperso c. fail to reject h₀. there is enough evidence at the 5% level of significance to reject the spokes d. reject h₀. there is not enough evidence at the 5% level of significance to reject the spokespe

Explanation:

Step1: Recall P - value decision rule

The decision rule for hypothesis testing using the P - value is: If \(P - value<\alpha\), we reject the null hypothesis \(H_0\); if \(P - value\geq\alpha\), we fail to reject \(H_0\).
Here, \(\alpha = 0.05\) and the \(P - value=0.0803\).

Step2: Compare P - value and \(\alpha\)

Since \(0.0803>0.05\) (i.e., \(P - value\geq\alpha\)), we fail to reject the null hypothesis \(H_0\). The null hypothesis in this case is the spokesperson's claim. So, there is not enough evidence at the 5% level of significance to reject the spokesperson's claim.

Answer:

A. Fail to reject \(H_0\). There is not enough evidence at the 5% level of significance to reject the spokesperson's claim.