Question

A hospital in New York commonly conducts stress tests to study the heart muscle after a person has a heart attack. Members of the diagnostic imaging department conducted a quality improvement project with the objective of reducing the turnaround time for stress tests. Turnaround time is defined as the time from when a test is ordered to when the radiologist signs off on the test results. Initially, the mean turnaround time for a stress test was 68 hours. After incorporating changes into the stress-test process, the quality improvement team collected a sample of 50 turnaround times. In this sample, the mean turnaround time was 32 hours, with a sample standard deviation of 9 hours.

(a) Using the critical value approach, if you test the null hypothesis at the 0.01 level of significance, is there evidence that the new process has reduced turnaround time? (Be sure to list the null and alternate hypotheses, the value of the test statistic, the decision rule (using the critical value(s) and conclude in a complete sentence including why you made the decision you did in terms of the decision rule.)  

(b) Using the p-value approach, if you test the null hypothesis at the 0.01 level of significance, is there evidence that the new process has reduced turnaround time? (You’ve already listed the null and alternate hypotheses and value of the test statistic in part (a). For part (b), include the p-value, the decision rule (using the p-value) and conclusion in a complete sentence including why you made the decision you did in terms of the decision rule.)

(c) Compare your conclusions in (a) and (b). Was it the same or different? Do you expect this to always or only sometimes be the case?

Answer 1

Given : Sample size=n=50

Sample mean=$\bar{X}=32$

Sample standard deviation=s=9

Hypothesized value=

Mo = 68

Significance level=

a=0.01

(a) Hypothesis :
HO :μ = 68
VS  $H_a:\mu<68$ ( Claim )

The value of the test statistic is ,

$t_{stat}=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}=\frac{32-68}{9/\sqrt{50}}=-28.28$

The critical value is ,

$t_{n-1,\alpha}=t_{50-1,0.01}=-2.68$ ; From excel "=TINV(0.01,49)"

Decision : Here , $t_{stat}=-28.28<t_{n-1,\alpha}=-2.68$

Therefore , reject Ho.

Conclusion : Hence , there is sufficient evidence to support the claim that the new process has reduced turnaround time.

(b) The p-value is ,

p-value=$P(t>|t_{stat}|)=P(t_{50-1}>28.28)=0.0000$ ; From excel "=TDIST(28.28,49,1)"

Decision : Here , p-value <

a=0.01

Therefore , reject Ho.

Conclusion : Hence , there is sufficient evidence to support the claim that the new process has reduced turnaround time.

(c) The conclusions from (a) and (b) are same.

It should be same for the both cases.