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1 WebAssign Mirka Martinez Math 3680 Homework 7 Devore Fall 2013 (Homework) Applied Statistics, Math 3680-Fall 2013, section 2, Fall 2013 Instructor: John Quintanilla Current Score : / 130 Due : Friday, November :59 PM CDT The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Important! Before you view the answer key, decide whether or not you plan to request an extension. Your Instructor may not grant you an extension if you have viewed the answer key. Automatic extensions are not granted if you have viewed the answer key. Request Extension View Key /10 points Previous Answers DevoreStat8 8.E.002. For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples): (a) H 0 : μ = 100, H a : μ > 100 These hypotheses comply with our rules. H 0 cannot include equality, so these hypotheses are not in compliance. Each μ is a statistic, so these hypotheses do not comply with our rules. The asserted value in H 0 should not appear in H a, so these hypotheses are not in compliance. (b) H 0 : σ = 20, H a : σ 20 These hypotheses comply with our rules. H a cannot include equality, so these hypotheses are not in compliance. Each σ is a statistic, so these hypotheses do not comply with our rules. The asserted value in H 0 should not appear in H a, so these hypotheses are not in compliance. (c) H 0 : p 0.25, H a : p = 0.25 Page 1 of 17

2 These hypotheses comply with our rules. H a cannot include equality, so these hypotheses are not in compliance. Each p is a statistic, so these hypotheses do not comply with our rules. The asserted value in H 0 should not appear in H a, so these hypotheses are not in compliance. (d) H 0 : μ 1 μ 2 = 25, H a : μ 1 μ 2 > 100 These hypotheses comply with our rules. H 0 cannot include equality, so these hypotheses are not in compliance. Each μ is a statistic, so these hypotheses do not comply with our rules. The asserted value in H 0 should also appear in H a, so these hypotheses are not in compliance. (e) H 0 : S 1 2 = S 2 2, H a : S 1 2 S 2 2 These hypotheses comply with our rules. H 0 cannot include equality, so these hypotheses are not in compliance. Each S is a statistic, so these hypotheses do not comply with our rules. The asserted value in H 0 should not appear in H a, so these hypotheses are not in compliance. (f) H 0 : μ = 120, H a : μ = 150 Page 2 of 17

3 These hypotheses comply with our rules. H a cannot include equality, so these hypotheses are not in compliance. Each μ is a statistic, so these hypotheses do not comply with our rules. If μ appears in H 0, then it should not appear in H a, so these hypotheses are not in compliance. (g) H 0 : σ 1 /σ 2 = 1, H a : σ 1 /σ 2 1 These hypotheses comply with our rules. H 0 cannot include equality, so these hypotheses are not in compliance. Each σ is a statistic, so these hypotheses do not comply with our rules. The asserted value in H 0 should not appear in H a, so these hypotheses are not in compliance. (h) H 0 : p 1 p 2 = 0.1, H a : p 1 p 2 < 0.1 These hypotheses comply with our rules. H 0 cannot include equality, so these hypotheses are not in compliance. Each p is a statistic, so these hypotheses do not comply with our rules. The asserted value in H 0 should not appear in H a, so these hypotheses are not in compliance. Page 3 of 17

4 /10 points Previous Answers DevoreStat8 8.E.003. To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected, and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose the specifications state that mean strength of welds should exceed 100 lb/in 2 ; the inspection team decides to test H 0 : μ = 100 versus H a : μ > 100. Explain why it might be preferable to use this H a rather than μ < 100. We want to determine if there is significant evidence that the mean strength of welds is less than 100 lb/in 2. The current hypotheses correctly place the burden of proof on those who wish to assert that the specification is not satisfied. We want to determine if there is significant evidence that the mean strength of welds exceeds 100 lb/in 2. The current hypotheses correctly place the burden of proof on those who wish to assert that the specification is satisfied. We want to determine if there is significant evidence that the mean strength of welds differs from 100 lb/in 2. The current hypotheses correctly place the burden of proof on those who wish to assert that the specification is not satisfied. We want to determine if there is significant evidence that the mean strength of welds equals 100 lb/in 2. The current hypotheses correctly place the burden of proof on those who wish to assert that the specification is satisfied /10 points Previous Answers DevoreStat8 8.E.006. Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40-amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40. If the mean amperage is lower than 40, customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction. To verify the amperage of the fuses, a sample of fuses is to be selected and inspected. If a hypothesis test were to be performed on the resulting data, what null and alternative hypotheses would be of interest to the manufacturer? Page 4 of 17

5 H 0 : μ = 40 H a : μ 40 H 0 : μ = 40 H a : μ > 40 H 0 : μ < 40 H a : μ = 40 H 0 : μ > 40 H a : μ = 40 H 0 : μ = 40 H a : μ < 40 Describe type I error in the context of this problem situation. A type I error would be declaring a fuse to be satisfactory when in fact it is defective. A type I error would be declaring a fuse to be satisfactory when in fact it is satisfactory. A type I error would be declaring a fuse as being defective when in fact there is nothing wrong with the fuse. A type I error would be declaring a fuse as being defective when in fact the fuse is defective. Describe type II error in the context of this problem situation. A type II error would be declaring a fuse to be satisfactory when in fact it is defective. A type II error would be declaring a fuse to be satisfactory when in fact it is satisfactory. A type II error would be declaring a fuse as being defective when in fact there is nothing wrong with the fuse. A type II error would be declaring a fuse as being defective when in fact the fuse is defective /10 points Previous Answers DevoreStat8 8.E.008. Page 5 of 17

6 A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce warpage. The regular laminate will be used on one sample of specimens and the special laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses. H 0 : μ regular = μ special H a : μ regular μ special H 0 : μ regular = μ special H a : μ regular > μ special H 0 : μ regular = μ special H a : μ regular < μ special H 0 : μ regular > μ special H a : μ regular = μ special H 0 : μ regular < μ special H a : μ regular = μ special Describe the type I error in the context of this situation. The type I error would be declaring the special laminate as having more warpage when in fact the true warpages are equal. The type I error would be declaring the true warpages of the laminates to be equal when in fact the special laminate has more warpage. The type I error would be declaring the special laminate as having less warpage when in fact the true warpages are equal. The type I error would be declaring the true warpages of the laminates to be equal when in fact the special laminate has less warpage. Describe the type II error in the context of this situation. Page 6 of 17

7 The type II error would be declaring the special laminate as having more warpage when in fact the true warpages are equal. The type II error would be declaring the true warpages of the laminates to be equal when in fact the special laminate has more warpage. The type II error would be declaring the special laminate as having less warpage when in fact the true warpages are equal. The type II error would be declaring the true warpages of the laminates to be equal when in fact the special laminate has less warpage /10 points Previous Answers DevoreStat8 8.E.015. Let the test statistic Z have a standard normal distribution when H 0 is true. Give the significance level for each of the following situations. (Round your answers to four decimal places.) (a) H a : μ > μ 0, rejection region z 1.76 (b) H a : μ < μ 0, rejection region z 2.64 (c) H a : μ μ 0, rejection region z 2.84 or z /10 points Previous Answers DevoreStat8 8.E.018. Consider a paint-drying situation in which drying time for a test specimen is normally distributed with σ = 8. The hypotheses H 0 : μ = 73 and H a : μ < 73 are to be tested using a random sample of n = 25 observations. Page 7 of 17

8 (a) How many standard deviations (of X) below the null value is x = 72.3? (Round your answer to two decimal places.) 0.44 standard deviations (b) If x = 72.3, what is the conclusion using α = 0.01? (Round your answers to two decimal places.) test statistic z = critical value z = What can you conclude? Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 73. Reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 73. Reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 73. Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 73. (c) What is α for the test procedure that rejects H 0 when decimal places.) α = z 2.5? (Round your answer to four (d) For the test procedure of part (c), what is β(70)? (Round your answer to four decimal places.) β(70) = (e) If the test procedure of part (c) is used, what n is necessary to ensure that β(70) = 0.01? (Round your answer up to the next whole number.) n = 166 specimens (f) If a level 0.01 test is used with n = 100, what is the probability of a type I error when μ = 76? (Round your answer to four decimal places.) Page 8 of 17

9 /10 points Previous Answers DevoreStat8 8.E.025. The desired percentage of SiO 2 in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. Suppose that the percentage of SiO 2 in a sample is normally distributed with σ = 0.32 and that x = (a) Does this indicate conclusively that the true average percentage differs from 5.5? Carry out the analysis using the sequence of steps suggested in the text. State the appropriate null and alternative hypotheses. H 0 : μ = 5.5 H a : μ 5.5 H 0 : μ = 5.5 H a : μ 5.5 H 0 : μ = 5.5 H a : μ < 5.5 H 0 : μ = 5.5 H a : μ > 5.5 State the rejection region(s) for an α = 0.01 test. If the critical region is one-sided, enter NONE for the unused region. Round your answers to two decimal places. z z 2.58 Compute the test statistic value. Round your answer to two decimal places. z = -3.5 State the conclusion in the problem context. Page 9 of 17

10 Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage. Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage. Reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage. Reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage. (b) If the true average percentage is μ = 5.6 and a level α = 0.01 test based on n = 16 is used, what is the probability of detecting this departure from H 0? (Round your answer to four decimal places.) (c) What value of n is required to satisfy α = 0.01 and β(5.6) = 0.01? (Round your answer up to the next whole number.) n = 247 samples /10 points Previous Answers DevoreStat8 8.E.026. To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of x = 53.7 and a sample standard deviation of s = 4.4. The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude? State the appropriate null and alternative hypotheses. Page 10 of 17

11 H 0 : μ = 50 H a : μ > 50 H 0 : μ = 50 H a : μ 50 H 0 : μ > 50 H a : μ = 50 H 0 : μ 50 H a : μ > 50 State the rejection region(s) for an α = 0.05 test. If the critical region is one-sided, enter NONE for the unused region. Round your answers to two decimal places. z NONE z 1.64 Compute the test statistic value. Round your answer to two decimal places. z = 5.64 State the conclusion in the problem context. Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils. Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils. Reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils. Page 11 of 17

12 /10 points Previous Answers DevoreStat8 8.E.028. Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. An article reports that for a sample of n = 75 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was min and the standard deviation was 8.7 min. Does this data suggest that true average lateral recumbency time under these conditions is less than 20 min? Test the appropriate hypotheses at level of significance State the appropriate null and alternative hypotheses. H 0 : μ 20 H a : μ > 20 H 0 : μ = 20 H a : μ < 20 H 0 : μ 20 H a : μ = 20 H 0 : μ = 20 H a : μ 20 State the rejection region(s) for an α = 0.10 test. If the critical region is one-sided, enter NONE for the unused region. Round your answers to two decimal places. z z NONE Compute the test statistic value. Round your answer to two decimal places. z = State the conclusion in the problem context. Page 12 of 17

13 Reject the null hypothesis. There is not sufficient evidence that the average lying-down time is less than 20 minutes. Do not reject the null hypothesis. There is sufficient evidence that the average lying-down time is less than 20 minutes. Do not reject the null hypothesis. There is not sufficient evidence that the average lying-down time is less than 20 minutes. Reject the null hypothesis. There is sufficient evidence that the average lying-down time is less than 20 minutes /10 points Previous Answers DevoreStat8 8.E.032. The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age years: n = 117, x = 11.2, and s = Does this data indicate that average daily zinc intake in the population of all males age falls below the recommended allowance? State the appropriate null and alternative hypotheses. H 0 : μ = 15 H a : μ 15 H 0 : μ = 15 H a : μ > 15 H 0 : μ = 15 H a : μ 15 H 0 : μ = 15 H a : μ < 15 State the rejection region(s) for an α = 0.05 test. If the critical region is one-sided, enter NONE for the Page 13 of 17

14 unused region. Round your answers to two decimal places. z z NONE Compute the test statistic value. Round your answer to two decimal places. z = State the conclusion in the problem context. Do not reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day. Reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. Do not reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. Reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day. Page 14 of 17

15 /10 points Previous Answers DevoreStat8 8.AE.005. Example 8.5 Let μ denote the true average nicotine content of brand B cigarettes. The objective is to test H 0 : μ = 1.5 versus H a : μ > 1.5 based on a random sample X 1, X 2,..., X 32 of nicotine content. Suppose the distribution of nicotine content is known to be normal with σ = Then X is normally distributed with mean value μ X = μ and standard deviation σ X = 0.20 / 32 = (rounded to four decimal places). H 0 is true. Rather than use X itself as the test statistic, let's standardize X, assuming that Test Statistic: X 1.5 Z = = σ/ n Z expresses the distance between X X and its expected value when H 0 is true as some number of standard deviations. For example, z = 3 results from an x that is 3 standard deviations larger than we would have expected it to be were H 0 true. Rejecting H 0 when x "considerably" exceeds 1.5 is equivalent to rejecting H 0 when z "considerably" exceeds 0. That is, the form of the rejection region is z c. Let's now determine c so that α = When H 0 is true, Z has a standard normal distribution. Thus α = P(type I error) = P(rejecting H 0 when H 0 is true) = P(Z c when Z ~ N(0, 1)) The value c must capture upper-tail area 0.05 under the z curve. From a standard normal curve table, c = z 0.05 = Notice that z is equivalent to x 1.5 (0.0354) (rounded to three decimal places), that is, x 1.56 X < 1.56 (rounded to two decimal places). Then β is the probability that and can be calculated for any μ greater than 1.5. Page 15 of 17

16 /10 points Previous Answers DevoreStat8 8.AE.007. Example 8.7 Let μ denote the true average tread life of a certain type of tire. Consider testing H 0 : μ = 30,000 versus H a : μ > 30,000 based on a sample of size n = 16 from a normal population distribution with σ = A test with α = 0.01 requires zα = z 0.01 = The probability of making a type II error when μ = 31,000 is β = Φ = Φ ,000 31, / 16 (rounded to two decimal places) = Since z 0.1 = 1.28, the requirement that the level 0.01 test also have β(31,000) = 0.1 necessitates n = 1500( ) 2 = ( 5.42) 2 = (rounded to two 30,000 31,000 decimal places). The sample size must be an integer, so n = 30 (rounded up to the next whole number) tires should be used. Page 16 of 17

17 /10 points Previous Answers DevoreStat8 8.AE.017. Example 8.17 The target thickness for silicon wafers used in a certain type of integrated circuit is 245 µm. A sample of 50 wafers is obtained and the thickness of each one is determined, resulting in a sample mean thickness of µm and a sample standard deviation of 3.60 µm. Does this data suggest that true average wafer thickness is something other than the target value? 1. Parameter of interest: μ = true average wafer thickness 2. Null hypothesis: H 0 : μ = Alternative hypothesis: H a : μ Formula for test statistic value: z = x 245 s/ n Calculation of test statistic value: z = = 2.32 (rounded to 3.60/ 50 two decimal places) 6. Determination of P-value: Because the test is two-tailed, P-value = 2(1 Φ(2.32)) = Conclusion: Using a significance level of 0.01, H 0 would not be rejected since > At this significance level, there is insufficient evidence to conclude that true average thickness differs from the target value. Page 17 of 17