Bio-Statistics. Test of Hypotheses Exercises

Transcription

1 () Test of Hypotheses Exercises Dr. Ayman Eldeib

2 Problem I Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient s body, but the battery pack needs to be recharged about every four hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation σ= 0.2 hour. a. Is there evidence to support the claim that mean battery life exceeds 4 hours? Use α = b. What sample size would be required to detect a true mean battery life of 4.5 hours if we wanted the power of the test to be at least 0.9?

3 Problem I State the hypotheses Null hypothesis: µ = 4 Alternative hypothesis: µ > 4 Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too big. Formulate an analysis plan For this analysis, the significance level is The test method is a z-test. Analyze sample data Using sample data, we compute the standard error (SE), and the z-score test statistic.

4 Problem I Analyze sample data Using sample data, we compute the standard error (SE), and the z-score test statistic. SE = _ σ / sqrt(n) = 0.20 / sqrt(50) = 0.20/7.07 = z 0 = (x -µ) / SE = (4.05-4)/0.283 = 1.77 Reject H 0 if z 0 > z α where z 0.05 = 1.65 Interpret results Since 1.77>1.65, we reject the null hypothesis and there is sufficient evidence to conclude that the true average battery life exceeds 4 hours at the significance level α = 0.05.

5 Problem I b) Sample Size for an Upper-Tail α-level Test α=0.05, β=0.01, it follows that z α = z 0.05 = 1.645, and z β = z 0.01 = 1.29 Then, n = ( z α + ( µ a z β ) µ 0 2 ) σ 2 2 = ( ) (4.5 4) 2 2 (0.2) 2 =

6 Example Problem II A 1992 article in the Journal of the American Medical Association ( A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wundrlich ) reported body temperature, gender, and heart rate for a number of subjects. The body temperatures for 25 female subjects follow: 97.8, 97.2, 97.4, 97.6, 97.8, 97.9, 98.0, 98.0, 98.0, 98.1, 98.2, 98.3, 98.3, 98.4, 98.4, 98.4, 98.5, 98.6, 98.6, 98.7, 98.8, 98.8, 98.9, 98.9, and a. Test the hypotheses H 0 : µ= 98.6 versus H 1 : µ 98.6, using α = 0.05 b. Find the P-value

7 Problem II State the hypotheses Null hypothesis: µ = 98.6 Alternative hypothesis: µ 98.6 Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small. Formulate an analysis plan For this analysis, the significance level is The test method is a t-test. Analyze sample data Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t-score test statistic (t).

8 Problem II Analyze sample data Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t-score test statistic (t). SE = s / sqrt(n) = / sqrt(25) DF = _ n - 1 = 25-1 = 24 t 0 = (x -µ) / SE = ( )/SE = Reject H 0 if t 0 > t α/2,n-1 where t α/2,n-1 = Interpret results Since 3.48 > 2.064, reject the null hypothesis and there is sufficient evidence to conclude that the true mean female body temperature is not equal to 98.6 F at α = 0.05

9 Problem II b) P-Value Since we have a two-tailed test, the P-value is the probability that the t-score having 24 degrees of freedom is less than or greater than We use the t Distribution Table to find P(t < ) = 0.001, and P(t > 3.48) = Thus, the P-value = =

10 Problem III The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. a. Find a 95% two-sided CI on the fraction of defective circuits produced by this particular tool b. Do the data support the claim that the fraction of defective units produced is less than 0.05, using α = 0.05? c. Find the P-value

11 Problem III a) CI A 95% two-sided confidence interval for p can be computed as follows: 13/ *sqrt((13/300) (287/300)/300) p 13/ *sqrt((13/300) (287/300)/300)

12 Problem III b) State the hypotheses Null hypothesis: p = 0.05 Alternative hypothesis: p < 0.05 Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small. Formulate an analysis plan For this analysis, the significance level is The test method is a z-test. Analyze sample data Using sample data, we compute the standard error (SE) and the z-score test statistic (z).

13 Problem III Analyze sample data Z Interpret results ) Y np0 p p0 = np ( 1 p0) p0 0 Using sample data, we compute the standard error (SE) and the z-score test statistic (z). ( 1 p ) / n (0.05) 0 = = = 0 Reject H 0 if z 0 < -z α where -z 0.05 = (300)(0.05)(0.95) Since 0.53 > 1.65, do not reject null hypothesis and conclude that the true fraction of defective integrated circuits is not significantly less than 0.05, at α = 0.05 C) P-Value P Value = P( Z 0.53) =

14 Questions?