VIEWS: 9 PAGES: 5 POSTED ON: 6/23/2011 Public Domain
Name: __________________________ Date: _____________ 1. If the test value in the figure below, for a test of the difference between two large sample means, is 2.57 when the critical value is 1.96, what decision about the hypothesis should be made? A) reject the null hypothesis B) accept the null hypothesis C) reject the alternative hypothesis D) not enough information 2. Determine the 95% confidence interval of the true difference in the means. A sociologist wants to determine if the life expectancy of people in Africa is less than the life expectancy of people in Asia. The data obtained is shown in the table below. Use 0.05 . Africa Asia X 55.3 65.2 8.1 9.3 n 53 42 A) 12.16 1 2 6.86 B) 13.46 1 2 6.34 C) 11.35 1 2 7.58 D) 16.33 1 2 5.98 3. A marketing firm asked a random set of married and single men as to how much they were willing to spend for a vacation. At α = .05, is a difference in the two amounts? Married men Single men Sample size 60 70 Mean spending 260 245 Sample variance 5000 6000 A) No, because the test value 0.09 is inside the interval (-1.96, 1.96) B) No, because the test value 1.15 is inside the interval (-1.96, 1.96) C) No, because the test value 1.20 is inside the interval (-1.96, 1.96) D) No, because the test value 1.20 is outside the interval (-1.96, 1.96) Page 1 4. An field researcher is gathering data on the trunk diameters of mature pine and spruce trees in a certain area. The following are the results of his random sampling. Can he conclude, at α = .10, that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree? Pine trees Spruce trees Sample size 80 40 Mean trunk diameter (cm) 40 34 Sample variance 110 170 A) The data does not support the claim because the test value 1.07 is less than than 1.28. B) The data supports the claim because the test value 2.53 is greater than than 1.28. C) The data supports the claim because the test value 2.53 is greater than than 1.64. D) The data does not support the claim because the test value 1.07 is less than than 1.64. 5. In comparing the two variances below, what is the test value and what are the degrees of freedom that should be used? Variance Number of values Sample 1 5 15 Sample 2 10 29 A) test value = 0.50, degrees of freedom = 15 and 29 B) test value = 0.50, degrees of freedom = 14 and 28 C) test value = 2.00, degrees of freedom = 14 and 28 D) test value = 2.00, degrees of freedom = 15 and 29 6. A car salesman claims that the variance of prices on convertibles is higher than the variance on station wagons. The standard deviation of 16 convertibles is $6800 and the standard deviation of 24 station wagons is $3900. For 0.05 , what is the test value? A) 3.00 B) 3.04 C) 2.78 D) 2.33 7. A researcher hypothesizes that the variation in the car rental rates at a major cities' airport is less than the car rental rates in that city. The variance of 7 airport car rental rates was $30 and the variance of 6 city car rental rates was $60. What is the test value? A) 2.33 B) 1.41 C) 2.00 D) 4.00 Page 2 8. Determine the value of as shown in the figure below, if the degrees of freedom were seven and nine. A) 0.01 B) 0.025 C) 0.05 D) 0.1 9. In testing the equality of the two means below, what is the test statistic? (Use the unequal variances formula) Sample 1 Sample 2 Sample size 9 12 Sample mean 80 55 Sample variance 450 60 A) 2.31 B) 0.13 C) 0.45 D) 3.37 10. A medical researcher is interested in whether patients' left arms or right arms are longer. If 14 patients participate in this study (so that n left arms and n left arms are measured), how many degrees of freedom should the researcher use in her t-test critical value? A) 13 B) 14 C) 26 D) 27 Page 3 11. A running coach wanted to see whether runners ran faster after eating spaghetti the night before. 24 random runners were chosen for this study. They ran a 5 kilometer race after having a normal dinner the night before, and then a week later, reran the same race after having a spaghetti dinner the night before. Their results (in seconds) are in the table below. At α = .01, what is the test value to use for this test? Regular Dinner Spaghetti Difference Dinner by runner Sample mean 1100 1080 –20 Sample variance 2200 2800 350 A) –4.35 B) –1.07 C) –0.28 D) –5.24 12. One poll found that 47% of male voters will support a candidate while another found that 50% of female voters will be in support. To test whether this candidate has equal levels of support between male and female voters, the null hypothesis should be A) H0 : pmale pfemale B) H0 : pmale 50%, H0 : pfemale 50% C) H0 : pmale 47%, H0 : pfemale 50% D) H0 : pmale pfemale 13. One poll found that 34% of male voters will support a candidate while another found that 42% of female voters will be in support. To test whether this candidate has equal levels of support between male and female voters, the alternative hypothesis should be A) H0 : pmale pfemale 50%, H0 : pfemale 50% B) H0 : pmale C) H0 : pmale 34%, H0 : pfemale 42% D) H0 : pmale pfemale 14. A recent survey reported that in a sample of 300 students who attend two-year colleges, 105 work at least 20 hours a week. In a sample of 225 students attending private universities, only 20 students work at least 20 hours per week. What is the test value? A) 6.95 B) 7.61 C) 2.38 D) 4.18 Page 4 15. When testing the difference between two proportions, one sample had 30 out of 100 who were for capital punishment and the other sample had 60 out of 80 who were for capital punishment. Calculate the standard error. A) 0.075 B) 0.060 C) 0.042 D) 0.098 Page 5