STAT 400 Lecture AL1 Answers for 8.3, 8.1 Spring 2015 Dalpiaz 1. A scientist wishes to test if a new treatment has a better cure rate than the traditional treatment which cures only 60% of the patients. In order to test whether the new treatment is more effective or not, a test group of 20 patients were given the new treatment. Assume that each personal result is independent of the others. Trying to decide: cure rate p 0.60 vs. p > 0.60. a) If the new treatment has the same success rate as the traditional, what is the probability that at least 14 out of 20 patients (14 or more) will be cured? P ( X 14 | p = 0.60 ) = 1 – CDF ( 13 | p = 0.60 ) = 1 – 0.750 = 0.250. b) Suppose that 14 out of 20 patients in the test group were cured. Based on the answer for part (a), is there a reason to believe that the new treatment has a better cure rate than the traditional treatment? If p = 0.60, then 25% of all possible samples would have 14 or more patients cured (out of 20). Thus, it is not unusual to see 14 out of 20 patients cured for a treatment that cures 60% of the patients. We have no reason to believe that the new treatment has a better cure rate than the traditional treatment if X = 14 . c) If the new treatment has the same success rate as the traditional, what is the probability that at least 17 out of 20 patients (17 or more) will be cured? P ( X 17 | p = 0.60 ) = 1 – CDF ( 16 | p = 0.60 ) = 1 – 0.984 = 0.016. d) Suppose that 17 out of 20 patients in the test group were cured. Based on the answer for part (c), is there a reason to believe that the new treatment has a better cure rate than the traditional treatment? If p = 0.60, then only 1.6% of all possible samples would have 17 or more patients cured (out of 20). Thus, it is fairly unusual to see 17 out of 20 patients cured for a treatment that cures 60% of the patients. We have a good reason to believe that the new treatment has a better cure rate than the traditional treatment if X = 17 .2. A certain automobile manufacturer claims that at least 80% of its cars meet the tough new standards of the Environmental Protection Agency (EPA). Let p denote the proportion of the cars that meet the new EPA standards. The EPA tests a random sample of 400 its cars, suppose that 308 of the 400 cars in our sample meet the new EPA standards. a) Perform an appropriate test at a 10% level of significance ( = 0.10). Claim: p 0.80 H 0 : p 0.80 vs. H 1 : p < 0.80 Y = 308. n = 400. 400308Yˆnp = 0.77. Test Statistic: Z = 40020.080.0 80.077.01 000ˆnpppp = – 1.50. Rejection Region: Left – tailed. Reject H 0 if Z < – z = 0.10 z 0.10 = 1.282. Reject H 0 if Z < – 1.282. Decision: The value of the test statistic DOES fall into the Rejection Region. Reject H 0 at = 0.10. b) Perform an appropriate test at a 5% level of significance ( = 0.05). Rejection Region: Left – tailed. Reject H 0 if Z < – z = 0.05 z 0.05 = 1.645. Reject H 0 if Z < – 1.645. Decision: The value of the test statistic does NOT fall into the Rejection Region. Do NOT Reject H 0 at = 0.05. c) Find the p-value of the appropriate test. Left – tailed. P-value = P( Z ≤ – 1.50 ) = 0.0668. d) Using the p-value from part (c), state your decision (Accept H0 or Reject H0) at = 0.08. 0.0668 = p-value < = 0.08. Reject H 0 at = 0.08.3. Alex wants to test whether a coin is fair or not. Suppose he observes 477 heads in 900 tosses. Let p denote the probability of obtaining heads. a) Perform the appropriate test using a 10% level of significance. Claim: p = 0.50 H 0 : p = 0.50 vs. H 1 : p 0.50 Y = 477. n = 900. 900477Yˆnp = 0.53. Test Statistic: Z = 900500500 5005301 000....nppppˆ = 1.80. Rejection Region: Two – tailed. Reject H 0 if Z < – z /2 or Z > z /2 = 0.10 2 = 0.05. z 0.05 = 1.645. Reject H 0 if Z < – 1.645 or Z > 1.645. Decision: The value of the test statistic does fall into the Rejection Region. Reject H 0 at = 0.10. OR P-value: Two – tailed. P-value = P( | Z | > 1.80 ) = 2 0.0359 = 0.0718. Decision: 0.0718 = p-value < = 0.10. Reject H 0 at = 0.10. b) Find the p-value of the test in part (a). Two – tailed. P-value = P( | Z | > 1.80 ) = 2 0.0359 = 0.0718. c) Using the p-value from part (b), state your decision (Accept H 0 or Reject H 0 ) for = 0.05. 0.0718 = p-value > = 0.05. Do NOT Reject H 0 at = 0.05.4. H 0 : p ≤ 0.20 vs. H 1 : p > 0.20. Y = 72. n = 300. Compute the p-value. State your decision at = 0.05. 30072Yˆnp = 0.24. Test Statistic: 30080.020.0 20.024.0)1 ( Z000ˆnpppp = 1.73. P-value: Rightt – tailed. P-value = P( Z > 1.73 ) = 0.0418. 0.0418 = p-value < = 0.05. Reject H 0 at =
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