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Bio-Statistics (SBE 304) Test of Hypotheses Exercises Dr. Ayman Eldeib Fall 2010 Bio-Statistics Hypotheses Test SBE 304 Hypothesis Tests 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. b. Is there evidence to support the claim that mean battery life exceeds 4 hours? Use α = 0.05. 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? Fall 2010 Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests 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 Analyze sample data Fall 2010 For this analysis, the significance level is 0.05. The test method is a z-test. Using sample data, we compute the standard error (SE), and the z-score test statistic. Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests 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 = 0.283 z0 = (x - µ) / SE = (4.05 - 4)/0.283 = 1.77 Reject H0 if z0 > zα where z0.05 = 1.65 Interpret results Fall 2010 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. Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests Problem I b) Sample Size for an Upper-Tail α-Level Test α=0.05, β=0.01, it follows that zα = z0.05 = 1.645, and zβ = z0.01 = 1.29 Then, 2 n= ( zα + z β ) σ ( µa − µ 0) 2 Fall 2010 2 2 2 (1.645 + 1.29) (0.2) = = 34.7 ≅ 35 2 ( 4.5 − 4) Bio-Statistics Hypotheses Test SBE 304 Hypothesis Tests Example II Problem 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 99.0. a. b. Fall 2010 Test the hypotheses H0: µ= 98.6 versus H1: µ ≠ 98.6 , using α = 0.05 Find the P-value Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests 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 Analyze sample data Fall 2010 For this analysis, the significance level is 0.05. The test method is a t-test. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t-score test statistic (t). Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests 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) = 0.4821/ sqrt(25) DF =_ n - 1 = 25 - 1 = 24 t0 = (x - µ) / SE = (98.264 – 98.6)/SE = -3.48 Reject H0 if |t0| > tα/2,n-1 where t α/2,n-1 = 2.064 Since 3.48 > 2.064, reject the null hypothesis and there is Interpret results sufficient evidence to conclude that the true mean female body temperature is not equal to 98.6 °F at α = 0.05 Fall 2010 Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests 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 - 3.48 or greater than 3.48. We use the t Distribution Table to find P(t < - 3.48) = 0.001, and P(t > 3.48) = 0.001. Thus, the P-value = 0.001 + 0.001 = 0.002. Fall 2010 Bio-Statistics Hypotheses Test SBE 304 Hypothesis Tests 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 Fall 2010 Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests Problem III a) CI A 95% two-sided confidence interval for p can be computed as follows: 13/300 – 1.96*sqrt((13/300) (287/300)/300) ≤ p ≤ 13/300 + 1.96*sqrt((13/300) (287/300)/300) Fall 2010 Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests 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 Analyze sample data Fall 2010 For this analysis, the significance level is 0.05. The test method is a z-test. Using sample data, we compute the standard error (SE) and the z-score test statistic (z). Bio-Statistics Hypotheses Test SBE 304 Cont. Hypothesis Tests Problem III Analyze sample data Z0 = Y − np 0 = np 0(1 − p 0) Using sample data, we compute the standard error (SE) and the z-score test statistic (z). ) p − p0 13 − 300(0.05) = = −0.53 p0 (1 − p0 ) / n (300)(0.05)(0.95) Reject H0 if z0 < -zα where -z0.05 = -1.65. Interpret results C) P-Value Fall 2010 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 P − Value = P( Z ≤ −0.53) = 0.29806 Bio-Statistics Hypotheses Test SBE 304 Questions ? Fall 2010 Bio-Statistics Hypotheses Test SBE 304