Question

you can calculate the p - value for a chi - square test using technology. after calculating the standardized test statistic, use the cumulative distribution function (cdf) to calculate the area under the curve. use the p - value method to test the claim.
a hospital spokesperson claims that the standard deviation of the waiting times experienced by patients in its minor emergency department is no more than 0.5 minutes. a random sample of 23 waiting times has a standard deviation of 0.7 minutes. at α = 0.05, is there enough evidence to reject the spokesperson’s claim?
...
identify the standardized test statistic.
χ² = 43.12 (round to two decimal places as needed.)
identify the p - value.
p = 0.0046 (round to four decimal places as needed.)
choose the correct conclusion below.
a. reject h₀. there is enough evidence at the 5% level of significance to reject the spokesperson’s claim.
b. reject h₀. there is not enough evidence at the 5% level of significance to reject the spokesperson’s claim.
c. fail to reject h₀. there is not enough evidence at the 5% level of significance to reject the spokesperson’s claim.
d. fail to reject h₀. there is enough evidence at the 5% level of significance to reject the spokesperson’s claim.

Explanation:

1. Recall the decision rule for hypothesis testing using the P - value method: If the P - value \( <\alpha \) (the significance level), we reject the null hypothesis \( H_0 \).
2. Here, the significance level \( \alpha = 0.05 \) and the calculated P - value is \( P=0.0046 \).
3. Compare the P - value and \( \alpha \): Since \( 0.0046<0.05 \), we reject the null hypothesis \( H_0 \). The null hypothesis \( H_0 \) is the spokesperson's claim that the standard deviation of waiting times is no more than 0.5 minutes. Rejecting \( H_0 \) means there is enough evidence at the 5% level of significance to reject the spokesperson's claim.

Answer:

A. Reject \( H_0 \). There is enough evidence at the 5% level of significance to reject the spokesperson’s claim.