MATH 302-Quiz 5 Stats

MATH 302-Quiz 5 Stats Question 1 of 17 1.0/ 1.0 Points Serum ferritin is used in diagnosing iron deficiency. In a study conducted recently researchers discovered that in a sample of 28 elderly men the sample standard deviation of serum ferritin was 52.6 mg/L. For 26 younger men the sample standard deviation was 84.2 mg/L. At the .01 level of significance, do these data support the conclusion that the ferritin distribution in elderly men has a smaller variance than in younger men? A.Yes, because the test value 0.390 is less than the critical value 2.54 B.Yes, because the test value 2.56 is greater than the critical value 0.394 C.No, because the test value 1.60 is less than the critical value of 2.54 D.Yes, because the test value 2.56 is greater than the critical value of 2.54 Answer Key: D Question 2 of 17 1.0/ 1.0 Points A pharmaceutical company is testing the effectiveness of a new drug for lowering cholesterol. As part of this trial, they wish to determine whether there is a difference between the effectiveness for women and for men. Assume α = 0.05. What is the test value? Women Men Sample size 50 80 Mean effect 7 6.95 Sample variance 3 4 A.z = 0.455 B.t = 3.252 C.z = 0.081 D.t = 0.151 Answer Key: D Question 3 of 17 1.0/ 1.0 Points An investor wants to compare the risks associated with two different stocks. One way to measure the risk of a given stock is to measure the variation in the stock’s daily price changes. In an effort to test the claim that the variance in the daily stock price changes for stock 1 is different from the variance in the daily stock price changes for stock 2, the investor obtains a random sample of 21 daily price changes for stock 1 and 21 daily price changes for stock 2. The summary statistics associated with these samples are: n1 = 21, s1 = .725, n2 = 21, s2 = .529. If you compute the test value by placing the larger variance in the numerator, at the .05 level of significance, would you conclude that the risks associated with these two stocks are different? A.No, the p-value associated with this test is 0.0528 B.No, the test value of 1.879 does not exceed the critical value of 2.46 C.No, the test value of 1.371 does not exceed the critical value of 2.12 D.Yes, the p-value associated with this test is 0.0264