Question

A hospital was concerned about reducing its wait time. A targeted wait time goal of 25 minutes was set. After implementing an improvement framework and process, a sample of 363 patients showed the mean wait time was 23.07 minutes, with a standard deviation of 16.41 minutes. Complete part (a) and (b) below. a. If you test the null hypothesis at the 0.05 level of significance, is there evidence that the population mean wait time is less than 25 minutes? State the null and alternative hypotheses O A. Ho u<25 H: 1225 OB. Ho u<25 Hu> 25 O E. Ho: #25 H: 25 0 C. Ho H225 Hu<25 OF. Hou > 25 Hu<25 OD. Ho: 1225 H: #25 Find the test statistic for this hypothesis test. tstar -(Type an integer or a decimal. Round to two decimal places as needed.) Find the p-value The p-value is (Type an integer or a decimal. Round to three decimal places as needed.)

Answer 1

(A) Answer: C (H0 ≥ 25 , H1 < 25)

Rationale: Null hypothesis is that the wait time is greater than or equal to 25 minutes.

Alternative hypothesis is that wait time is less than 25 minutes.

(B) Answer: -2.24

Rationale:

Mean ($\mu$) = 25

X-bar (
X
) = 23.07 minutes

Standard deviation (s) = 16.41

Sample size (n) = 363

$t-stat=\frac{\bar{X}-\mu}{S/\sqrt{n}}$

$t-stat=\frac{23.07-25}{16.41/\sqrt{363}}$

T- stat = -2.24

(C) Answer: 0.013

Degrees of freedom = n-1

Degrees of freedom = 363 - 1 = 362

To find probability, use the following Function on excel:

"=T.Dist(-2.24,362,TRUE)"

Value of probability = 0.013