www.LionTutors.com STAT 200 (Kankam) – Exam 3 – Practice Test Solutions 1. A – As the level of confidence increases, the interval gets wider. Increasing the level of confidence will increase the multiplier, which causes the interval to become wider. 2. B – Estimate a population parameter 3. C – 0.0256 * 2. The p-­‐value for a two-­‐tailed test (≠ test) will be the area to the left of the negative test statistic and the area to the right of the positive test statistic. P(Z < –1.95) = 0.0256 P(Z > 1.95) = 0.0256 4. C – 0.48 to 0.68 because the 99% confidence level will correspond to the widest confidence interval. 5. B – 95% confidence interval. The level of confidence determines the z multiplier in the confidence interval equation for a proportion. All we need to do to solve this problem is find the level of confidence to that correlates to the z multiplier of 1.96, which is 95%. 6. A – 0.9793. Since this is a left-­‐tailed test (< test), you need to find the area to the left of z-­‐score of 2.04. P(Z < 2.04) = 0.9793 7. A – Decrease the sample size. The margin of error determines the width of a confidence interval. Increasing the margin of error is the same thing as increasing the width of the confidence interval. The two ways to increase the width of a confidence interval are to 1) Decrease the sample size, or 2) Increase the level of confidence. 8. B – 175 ± 1.77*10√14 𝑥 = 175 s = 10 S.E. (𝑥) = 2.67 n = 14 df = 13 t (df = 13, 90%) = 1.77 CI = 𝑥 ± t*( S.E. (𝑥)) CI = 𝑥 ± t*(s/√n) 175 ± 1.77*2.67 would have also been a correct answer; however, it was not an answer choice. It is important that you don’t confuse standard error with the standard deviation of the sample. S.E. (𝑥) = s / √n 9. C – (1 – 0.2389). Since the alternative hypothesis is a greater than test, you are looking for the area to the right of the z-­‐statistic of –0.71. P(Z > –0.71) = 1 – 0.2389 10. A – Statistic because 70% was the result of a sample. 11. A – 0.0256. Since this is a left-­‐tailed test (< test), you are looking for the area to the left of the z-­‐statistic of –1.95. P(Z < –1.95) = 0.0256 12. D – Population 13. B – A random sample where n = 50 is taken from a right skewed population. The central limit theorem will apply when the distribution of a population is approximately normal or when n ≥ 30. None of the populations have a normal distribution; however, n ≥ 30 for answer choice B. 14. D – 0.2389*2. The alternative hypothesis is a ≠ test so we are looking for the area to the left of the negative test statistic and the area to the right of the positive test statistic. P(Z < –0.71) = 0.2389 P(Z > 0.71) = 0.2389 15. D – (0.25 – 0.45)/√[(0.45*(1–0.45))/64]. To solve for the value of the z-­‐
statistic for a population proportion, you need to know the null value, the sample statistic, and the sample size. Then you plug them into the equation below. Note that the sign of the alternative hypothesis does not matter for this problem. The sign of the alternative hypothesis is needed to determine the p-­‐value once the test statistic is calculated; however, the sign of the alternative hypothesis does play a role in the calculation of the test statistic. p0 = 0.45 𝑝 = 0.25 n = 64 𝑧 test statistic = Sample statitic – Null value
= 𝑁𝑢𝑙𝑙 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
𝑝 − 𝑝! 𝑝 (1 − 𝑝 )
! 𝑛 !
16. B – 0.9032. The alternative hypothesis is a less than test so we are looking for the area to the left of the z-­‐statistic of 1.30. P(Z < 1.30) = 0.9032 17. C – The alternative hypothesis includes values in either direction from a specific standard. This is another way of saying that the p-­‐value for a two-­‐
tailed test (≠ test) is the area to the left of the negative test statistic and the area to the right of the positive test statistic. 18. B – 175 ± 1.77*2.67 𝑥 = 175 s = 10 S.E. (𝑥) = 2.67 n = 14 df = 13 t (df = 13, 90%) = 1.77 CI = 𝑥 ± t*( S.E. (𝑥)) CI = 𝑥 ± t*(s/√n) 175 ± 1.77*(10/√14) would have also been a correct answer; however, it was not an answer choice. It is important that you don’t confuse standard error with the standard deviation of the sample. S.E. (𝑥) = s / √n 19. C – Ho: p ≠ 0.55 Ha: p = 0.55 Possible values for the null hypothesis: =, ≤, ≥ Possible value for the alternative hypothesis: ≠, <, > 20. D – The standard error increases as the sample size increases is false. The standard error will decrease as the sample size increases. This is because sample size, n, is in the denominator of the standard error equation. S.E. (𝑥) = s / √n 21. B – (1 – 0.9793). The p-­‐value for a right-­‐tailed test (> test) is the area to the right of the z-­‐statistic of 2.04. P(Z > 2.04) = 0.9793 22. C – 75 ± 2.46*8/√30 𝑥 = 75 s = 8 S.E. (𝑥) = 1.46 n = 30 df = 29 t (df = 29, 98%) = 2.46 CI = 𝑥 ± t*( S.E. (𝑥)) CI = 𝑥 ± t*(s/√n) 75 ± 2.46*1.46 would have also been a correct answer; however, it was not an answer choice. It is important that you don’t confuse standard error with the standard deviation of the sample. S.E. (𝑥) = s / √n 23. D – Values of a statistic will vary according to the sampling distribution for that statistic. However, the value of a population parameter is a fixed value that remains constant. 24. B – The new weight loss method has a success rate that is greater than 55%. We know the alternative hypothesis will be a greater than test for this problem because the doctors wants to test if the new method is more effective then the current method. 25. A – Confidence interval for a proportion. Percentages relate to proportions while averages relate to means. 26. B – 1.73 n = 20 df = 19 Confidence level = 90% 27. D – (1 – 0.9793) * 2. The p-­‐value for a two-­‐tailed test (≠ test) is the area to the left of the negative test statistic and to the right of the positive test statistic. P(Z < –2.04) = 1 – 0.9793 P(Z > 2.04) = 1 – 0.9793 28. C – 2.09 n = 20 df = 19 Confidence level = 95% 29. E – (1 –0.9032) * 2. Since the alternative hypothesis is a ≠ test, the p-­‐value is the area to the left of the negative test statistic and the area to the right of the positive test statistic. P(Z < –1.30) = 1 – 0.9032 P(Z > 1.30) = 1 – 0.9032 30. A – 90% confidence interval. The level of confidence determines the z multiplier in the confidence interval equation for a proportion. All we need to do to solve this problem is find the level of confidence to that correlates to the z multiplier of 1.65, which is 90%. 31. D – The alternative hypothesis includes values in one direction from a specific standard. 32. B – (1 – 0.0256). The p-­‐value for a right-­‐tailed test (> test) is the area to the right of the z-­‐statistic of –1.95. P(Z > –1.95) = 1 – 0.0256 33. C – Reject the hypothesis that the population average age is 20. We know we can reject the null hypothesis because the p-­‐value is less than the level of alpha. Hypothesis tests always relate to population values, not sample values. We are using sample data to test a statement about the larger population. 34. B – The confidence level determines the multiplier for a confidence interval. The only way to change the multiplier is to change the confidence level. As you increase the confidence level, the multiplier will also increase. 35. A – 5 because it is the smallest value for n in the answer choices. Standard error will be the greatest when sample size, n, is the smallest. We can see this by looking at the standard error equation. S.E. (𝑥) = s / √n 36. C – 10.00 S.E. (𝑥) = s / √n S.E. (𝑥) = 70 / √49 S.E. (𝑥) = 10 37. D – 2.05 n = 30 df = 29 Level of confidence = 95% 38. C – A normal distribution with mean equal to the population mean. 39. Correct answer is not listed – 71 ± (2.39)(8/√61) 𝑥 = 71 𝑠 = 8 n = 61 df = 60 Level of confidence = 98% t (df = 60, 98%) = 2.39 CI = 𝑥 ± t*( S.E. (𝑥)) CI = 𝑥 ± t*(s/√n) CI = 71 ± (2.39)(8/√61) 40. B – (5.5 – 5.77) / (0.44/√64). To solve for the value of the test statistic when dealing with means, you will need to know the null value, sample statistic, sample standard deviation, and sample size. Note that the sign of the alternative hypothesis does not affect the calculation of the test statistic. The sign of the alternative hypothesis is use to determine the p-­‐value once the test statistic is calculated. μ0 = 5.77 𝑥 = 5.5 𝑠 = 0.44 n = 64 𝑡 statistic = Sample statistic – Null value
𝑥 − 𝜇! = 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟
𝑠 𝑛
41. A – Ho: μ = 45 Ha: μ < 45. Immediately you can eliminate answer choice D because it is using sample notation in the hypothesis test, which is never correct. Answer C can be eliminated as well because the value 42.2 is the sample statistic, 𝑥 , and the population parameter, μ, should be used as the value in the hypothesis test. A is the correct answer because the problem tells us we want to test if the mean is less than 45. Thus the alternative hypothesis should be a less than test. 42. Please skip this problem. Additional problems have been posted to the files section to replace these problems. 43. Please skip this problem. Additional problems have been posted to the files section to replace these problems. 44. Please skip this problem. Additional problems have been posted to the files section to replace these problems. 45. Please skip this problem. Additional problems have been posted to the files section to replace these problems. 46. C – 90 ± 2.80*1.20 𝑥 = 90 s = 6 S.E. (𝑥) = 1.20 n = 25 df = 24 t (df = 24, 99%) = 2.80 CI = 𝑥 ± t*( S.E. (𝑥)) CI = 𝑥 ± t*(s/√n) 90 ± 2.80*(6/√25) would have also been a correct answer; however, it was not an answer choice. It is important that you don’t confuse standard error with the standard deviation of the sample. S.E. (𝑥) = s / √n 47. C – A random sample where n = 20 is taken from a normal population. The central limit theorem will apply when the distribution of a population is approximately normal or when n ≥ 30. None of the samples have n ≥ 30; however, the population for answer C is normally distributed. 48. C – Sample, population. Confidence intervals use sample data to estimate a population parameter. 49. B – Population size is a fixed value that does not impact the width of a confidence interval. 50. A – 90% the smallest level of confidence will result in the narrowest confidence interval. 51. B – A statistic can have only a single value once the population is defined is false. A statistic is the result of a sample. It is possible for samples taken from the same population to have different sample statistics. However, the population parameter can only have a single value once the population is defined. 52. A – Parameter because a census means that everyone in the entire population responded to the question. 53. B – Increase the size of the sample. 54. C – p. Answers A and B can be eliminated immediately because they are using sample notation. Answer C is correct because this problem deals with percentages, which relate to proportions. 55. B – 0.0150 Standard error = s. e. (𝑝) =
𝑝(1 − 𝑝)
𝑛
Standard error =
0.87(1 − 0.87)
= 0.015 500
56. D – μ = the mean amount spent on alcohol by all finance majors. Answers A and B can be immediately eliminated because they are using sample notation. D is correct because the population parameter relates to everyone in the population, not just the people included in the sample. 57. A – Confidence interval for a proportion because percentages relate to proportions while averages relate to means. 58. B – Alternative hypothesis because the statement is describing a ≠ test. 59. B – Type 2. If the alternative hypothesis is actually true and you have made an error, you must have decided the null hypothesis could not be rejected. If you have decided the null hypothesis cannot be rejected, the only type of error it is possible to make is a type 2 error. 60. C – The smaller the p-­‐value, the stronger the evidence against the null hypothesis. As the p-­‐value gets smaller, it becomes more likely we will reject the null hypothesis.