Question

a hospital spokesperson claims that the standard deviation of the waiting times experienced by patients in its minor emergency department is no more than 0.7 minutes. a random sample of 28 waiting times has a standard deviation of 0.8 minutes. at α = 0.10, is there enough evidence to reject the spokesperson’s claim? assume the population is normally distributed. complete parts (a) through (e) below. click the icon to view the chi - square distribution table. (b) find the critical value(s). $\chi_{0}^{2} = 36.741$ (round to three decimal places as needed. use a comma to separate answers as needed.) identify the rejection region(s). choose the correct graph below. \\(\bigcirc\\) a. \\(\bigcirc\\) b. \\(\bigcirc\\) c. \\(\bigcirc\\) d. (c) find the standardized test statistic for the $\chi^{2}$ - test. $\chi^{2} = \square$ (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the chi - square test statistic for variance

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.

Step2: Identify the values

We are given that \(n = 28\), so \(n-1=28 - 1=27\), \(s = 0.8\) minutes, and \(\sigma=0.7\) minutes.

Step3: Substitute the values into the formula

First, calculate \(s^{2}=(0.8)^{2}=0.64\) and \(\sigma^{2}=(0.7)^{2}=0.49\)

Then, \(\chi^{2}=\frac{(27)\times(0.64)}{0.49}\)

Calculate the numerator: \(27\times0.64 = 17.28\)

Then, \(\chi^{2}=\frac{17.28}{0.49}\approx34.415\)

Answer:

34.415