hw7 - STAT 350 – Spring 2009 Homework #7 – Solution...

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Unformatted text preview: STAT 350 – Spring 2009 Homework #7 – Solution through Lecture K Text Exercises Chapter 8, Problems #10, 12†, 18, 40* † There is a typographical error in the textbook. The text reads, ““…z = ( x - 34)/(s n )…” It should read, “…z = ( x - 34)/(s / n )…” * be sure to state the null and alternative hypotheses, give the value of the test statistic and the pvalue. 8.10 a. b. c. d. e. f. a. no – do not reject H0 no – do not reject H0 yes – reject H0 yes – reject H0 no – do not reject H0 yes – reject H0 34.43 − 34 = 2.868 1.06 50 p-value = 2P(Z > 2.868) = 2(1 - Φ(2.87)) = 2(1 – 0.9979) = 0.0042 z= 33.57 − 34 = −2.868 1.06 50 p-value = 2P(Z < -2.868) = 2Φ(-2.87)) = 0.0042 33.25 − 34 = −2.2447 1.89 32 p-value = 2P(Z < -2.24) = 2Φ(-2.24)) = 0.025 z= 34.66 − 34 = 1.565 2.53 36 p-value = 2P(Z > 1.565) = 2(1 - Φ(1.57)) = 2(1 – 0.9418) = 0.1164 8.12 b. z = c. d. z = Homework #7 – Solution Page 1 of 1 8.18 a. With n = 15, df = 14 and the critical value of t for a one-tailed, α = 0.05 test, tcritical = 1.761. Because the test statistic t is greater than the critical value, we would reject the null hypothesis in favor of the alternative hypothesis. The p-value for t is 0.003. b. With n = 9, df = 8 and the critical value of t for a one-tailed, α = 0.01 test, tcritical = 2.896. Because the test statistic t is less than the critical value, we would fail to reject the null hypothesis. The p-value for t is 0.055. c. With n = 24, df = 23. Recall that only positive values of the test statistic would support the alternative hypothesis (because the alternative hypothesis is only supported if x > 20). Here there is no support for the alternative (t < 0). The pvalue for t is 1 - 0.422 = 0.578. We would definitely fail to reject the null hypothesis. 8.40 player Expenditure Intake difference 1 14.4 14.6 -0.2 2 12.1 9.2 2.9 3 14.3 11.8 2.5 4 14.2 11.6 2.6 5 15.2 12.7 2.5 6 15.5 15 0.5 7 17.8 16.3 1.5 average stdev 1.757 1.197 H0: μd = 0, Ha: μd ≠ 0 d − Δ 1.757 − 0 t= = = 3.8835 sd n 1.197 7 df = 6, p-value = 0.004×2 = 0.008 We would reject the null hypothesis of no difference for a significance level of 0.05 or 0.01, but not for a significance level of 0.001. Homework #7 – Solution Page 2 of 2 Then answer the following problems 1. You are testing a hypothesis about a single population mean (that is, your null hypothesis is of the form μ = μ0). a. Assume you are using an α = 0.10 significance level and have 12 observations (x1, x2, . . . , x12). df = 12 – 1 = 11 i. Give the critical value of the test statistic if your alternative hypothesis is μ > μ0. Describe or sketch the rejection region(s) for this test. tcrit = +1.363 Rejection region: > +1.363 ii. Give the critical value of the test statistic if your alternative hypothesis is μ < μ0. Describe or sketch the rejection region(s) for this test. tcrit = -1.363 Rejection region: < -1.363 iii. Give the critical value of the test statistic if your alternative hypothesis is μ ≠ μ0. Describe or sketch the rejection region(s) for this test. tcrit = ±1.796 Rejection region: < -1.796 ∪ > +1.796 b. Assume you are using an α = 0.05 significance level and have 12 observations (x1, x2, . . . , x12). df = 12 – 1 = 11 i. Give the critical value of the test statistic if your alternative hypothesis is μ > μ0. Describe or sketch the rejection region(s) for this test. tcrit = Rejection region: > +1.796 ii. Give the critical value of the test statistic if your alternative hypothesis is μ < μ0. Describe or sketch the rejection region(s) for this test. tcrit = Rejection region: < -1.796 iii. Give the critical value of the test statistic if your alternative hypothesis is μ ≠ μ0. Describe or sketch the rejection region(s) for this test. tcrit = ±2.201 Rejection region: < -2.201 ∪ > +2.201 Homework #7 – Solution Page 3 of 3 c. Assume you are using an α = 0.05 significance level and have 19 observations (x1, x2, . . . , x19). df = 19-1 = 18 i. Give the critical value of the test statistic if your alternative hypothesis is μ > μ0. Describe or sketch the rejection region(s) for this test. tcrit = +1.734 Rejection region: > +1.734 ii. Give the critical value of the test statistic if your alternative hypothesis is μ < μ0. Describe or sketch the rejection region(s) for this test. tcrit = -1.734 Rejection region: < -1.734 iii. Give the critical value of the test statistic if your alternative hypothesis is μ ≠ μ0. Describe or sketch the rejection region(s) for this test. tcrit = ±2.101 Rejection region: < -2.101 ∪ > +2.101 Homework #7 – Solution Page 4 of 4 2. You are testing a hypothesis about a single population mean (that is, your null hypothesis is of the form μ = μ0). Give the p-value for the test under each of the following conditions: a. Test statistic = -1.6, n = 12, alternative hypothesis is μ > μ0. Test statistic is a "t", p-values are found using Appendix Table VI, df = 11 p-value = 1 – 0.068 = 0.932 b. Test statistic = -1.6, n = 12, alternative hypothesis is μ < μ0. Test statistic is a "t", p-values are found using Appendix Table VI, df = 11 p-value = 0.068 c. Test statistic = -1.6, n = 12, alternative hypothesis is μ ≠ μ0. Test statistic is a "t", p-values are found using Appendix Table VI, df = 11 p-value = 2(0.068) = 0.136 d. Test statistic = 2.3, n = 240, alternative hypothesis is μ > μ0. Large sample size, so test statistic can be called "z", p-values found using Appendix Table 1 p-value = 1 – 0.9893 = 0.0107 e. Test statistic = 2.3, n = 240, alternative hypothesis is μ < μ0. Large sample size, so test statistic can be called "z", p-values found using Appendix Table 1 p-value = 0.9893 f. Test statistic = 2.3, n = 240, alternative hypothesis is μ ≠ μ0. Large sample size, so test statistic can be called "z", p-values found using Appendix Table 1 p-value = 2(1 – 0.9893) = 2(0.0107) = 0.0214 Homework #7 – Solution Page 5 of 5 3. The lifetimes of 2 models of specialized batteries are compared for their lifetime in a prototype hearing aid. 20 hearing aids were loaded with Brand 1 batteries and 6 were loaded with Brand 2 batteries. The hearing aids were turned on and monitored until the batteries died. The mean lifetime of the 20 Brand 1 batteries was 118.6 hours with a standard deviation of 14.6 hours. The mean lifetime of the 6 Brand 2 batteries was 137.2 hours with a standard deviation of 20.1 hours. You wish to see if there is a difference between these 2 brands, thus you wish to test H0: µ1 - µ2 = 0 against Ha: µ1 - µ2 ≠ 0. Unless otherwise stated, in this class, we will use a significance level of α = 0.05 a. Answer the following assuming that the populations have a common variance. i. What are the degrees of freedom for the test statistic? df = n1 + n2 – 2 = 20 + 6 – 2 = 24 ii. What is the critical value for the test statistic and what is the rejection region? From Appendix Table IV, the t critical value is then ±2.064 The rejection region is <-2.064 ∪ >+2.064 iii. What is the value of the test statistic? sp = ( n1 − 1) s12 + ( n2 − 1) s22 n1 + n2 − 2 = ( 20 − 1)14.62 + ( 6 − 1) 20.12 20 + 6 − 2 = 252.9204167 = 15.9035 t= ( x1 − x2 ) − Δ ⎛1 1⎞ s2 ⎜ + ⎟ p ⎝ n1 n2 ⎠ = (118.6 − 137.2 ) − 0 ⎛ 1 1⎞ 252.9204167 ⎜ + ⎟ ⎝ 20 6 ⎠ = -2.5126 iv. What is the p-value of the test? From Appendix Table VI: p-value = 2(0.010) = 0.020 v. Based on your analysis, would you reject the null hypothesis? yes, I would reject the null hypothesis. Homework #7 – Solution Page 6 of 6 b. Answer the following assuming that the populations DO NOT have a common variance. i. What are the degrees of freedom for the test statistic? s12 n1 14.62 20 = 0.1366532894 c= 2 = 2 s1 s2 14.62 20.12 + + n1 n2 20 6 df = ( n1 − 1)( n2 − 1) 2 ( n2 − 1) c 2 + ( n1 − 1)(1 − c ) = 6.664163 → round down to 26 ii. What is the critical value for the test statistic and what is the rejection region? From Appendix Table IV, the t critical value is then ±2.447 The rejection region is <-2.447∪ >+2.447 iii. What is the value of the test statistic? (x − x )−Δ (118.6 − 137.2 ) − 0 = -2.106 = t= 1 2 2 2 s1 s2 14.62 20.12 + + 20 6 n1 n2 iv. What is the p-value of the test? From Appendix Table VI: p-value = 2(0.040) = 0.080 v. Based on your analysis, would you reject the null hypothesis? No, I would NOT reject the null hypothesis. Homework #7 – Solution Page 7 of 7 4. Shortly before giving a second exam, Dr. Random, Professor of Statistics, was asked if the second exam in his class was usually harder than the first. Being a good statistician, he wanted to test this. He took a random sample of 20 students from the past several semesters that he taught the class (these scores are given below). Use the data below to test the null hypothesis that there is no difference in the mean scores for the two exams against the alternative that the second exam is harder than the first. Student Exam 1 Score Abby Albert Betty Monica Gus Denise Shannon Radu Charles Matt Lisa Sally Penny John Rebecca Michael Charlie Joe John Yan 68 60 46 31 71 57 80 48 82 45 31 80 78 84 51 56 45 54 31 67 Exam 2 Score 85 73 76 65 49 34 81 58 92 57 73 86 83 83 81 85 82 58 51 74 difference ‐17 ‐13 ‐30 ‐34 22 23 ‐1 ‐10 ‐10 ‐12 ‐42 ‐6 ‐5 1 ‐30 ‐29 ‐37 ‐4 ‐20 ‐7 mean stdev 58.25 71.3 ‐13.05 17.48947 15.44123 17.57233 a. State the null and alternative hypothesis using proper statistical notation. H0: µ1 - µ2 = 0 against Ha: µ1 - µ2 > 0. Because the data are paired, it would be more appropriate to state the hypothesis as: H0: µd = 0 against Ha: µd > 0, where µd = µ1 - µ2. I will accept either as a correct answer b. What are the degrees of freedom for the test statistic? df = n – 1 = 20 – 1 = 19 Homework #7 – Solution Page 8 of 8 c. What is the critical value for the test statistic and what is the rejection region? For alpha = 0.05, the critical value (from Appendix Table IV) is +1.729 The rejection region is > +1.729 d. What is the value of the test statistic? d −Δ −13.05 − 0 t= = = -3.3212 17.57233 20 sd n e. What is the p-value of this test? = 1 – 0.002 = 0.998 f. Based on your analysis, can you conclude that the second exam is harder than the first? No! We would not reject the null hypothesis! We would certainly not conclude that the second exam is harder than the first! Even without statistics, there is zero evidence for this – the sample mean score from the second exam was actually HIGHER than that for the first exam, so it looks like the second exam is easier (but then we’d have to do a different test to conclude that the second is actually “significantly” easier. 5. Your company has developed to help people fall asleep faster. It is your job to test whether this new drug is effective. 200 volunteers are recruited, half will randomly be assigned to take the new drug and the other half will be assigned to take a placebo (also known as a “sugar pill”, a pill that looks the same, but does not have any active ingredients). None of the patients will know which they are getting. For each volunteer, the time (in minutes) until they fall asleep will be measured. You have been informed that the drug will only be approved for sale if patients will fall asleep at least 30 minutes faster on the drug than on the placebo. Let μd = the true mean time (in min.) until someone who has taken the drug falls asleep. Let μp = the true mean time (in min.) until someone who has taken the placebo falls asleep. a. Give the null an alternative hypothesis for the appropriate test in terms of μd and μp. Ho: μd - μp = -30 OR Ho: μp - μd = 30 vs. Ha: μp - μd > 30 vs. Ha: μd - μp < -30 b. Explain and justify your choice of a 2-tailed or 1-tailed alternative hypothesis. Conceptually our null hypothesis is that the drug is not effective and our alternative is that the drug is effective. The drug is only effective if it makes you fall asleep faster than the placebo, not just that the time to fall asleep is different for the drug than the placebo. This is why a 1-tailed test is appropriate. Homework #7 – Solution Page 9 of 9 ...
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