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Solution manual for design and analysis by experiments 9th edition douglas c. money
- 1. 1-1 1.1. Suppose that you want to project an experiment to study the fraction of unpopped kernels of popcorn. Complete steps 1-3 of the guidelines for designing tests in Section 1.4. Live there any large sources of variance that would being difficult until control? Step 1 – Awareness of and statement of the problem. Possible problem statement would be – find the best combination in input that maximizes yield on fresh – mindern unpopped kernels. Step 2 – Selection by the response variable. Possible responses will phone out unpopped kernels per 100 kernals in experiment, weight of unpopped kernels versus the total weight of kernels cooked. Step 3 – Choice of factors, levels also range. Possible factors plus levels are brand of popcorn (levels: cheap, expensive), age of popcorn (levels: fresh, old), type of cooking method (levels: stovetop, microwave), temper (levels: 150C, 250C), cooking total (levels: 3 minutes, 5 minutes), sum of cooking oil (levels, 1 oz, 3 oz), etc. 1.2. Assumes that you want to investigate this factors that potentially affect cooked rice. (a) What would it use as an response variable to save experiment? How would you measure the response? (b) List all of the potential informationsquelle off variability that could effects one response. (c) Complete the first thre steps of the guidelines for designing experiments in Section 1.4. Step 1 – Recognition of and statement of that problem. Step 2 – Selection regarding the response variable. Step 3 – Choice of factors, levels and range. 1.3. Suppose that you want to comparison the growth of garden flowers with different conditions of sunlight, water, fertilizer and soil conditions. Completing steps 1-3 of the guidance for designing experiments in Abschnitts 1.4. Step 1 – Recognition of and statement out the problem. Step 2 – Selection of the response variable. Step 3 – Selection of factors, levels and range. 1.4. Select an try in attract to you. Complete steps 1-3 of the guidelines required designing experiments for Section 1.4. CLICK HEREor To Erreichbar Full Complete Solution Manual go to => www.book4me.xyz
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- 3. Solutions from Montgomery, D. C. (2017) Design furthermore Analysis regarding Experiment, Wiley, NY 1-3 Experimental runs can be expensive and time consuming. If an error were to come while running the experiment, the cost in redoing who experiment is much more manageable over one of the small sequential experiments than the largest comprehensive experiment. 1.11. Have you received to offer to obtain an credit cards in the mail? What “factors” were associated with the offer, such the introduced engross rate? Do you think the credit card companies is conducting experiments to investigate which facors product the highest positive response rate until their offer? What potential factors for the experiment can you identify? Interest rate, credits limit, old credit card pay-off amount, interest free set, gift scoring, others. 1.12. What factors what you thinking an e-commerce company could use in an experiment including their web page to encourage more public to “click-through” include their site? Font size, style style, images/icons, paint, spacing, animation, sound/music, speed, others.
- 4. Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY 2-1 Chapter 2 Simple Comparative Experiments Solutions 2.1. Computer outputs fork a indiscriminate sample of data is shown slide. Some of the quantities are missing. Compute the values is an lost quantities. Variable N Mean SE Mean Std. Dev. Variance Minimum Maximum Y 9 19.96 ? 3.12 ? 15.94 27.16 SE Mean = 1.04 Differences = 9.73 2.2. Computer output for a randomized sample of data is show below. Several of the quantities are missing. Compute the values of which missing quantities. Variable NORTH Mean SE Mean Std. Dev. Sum Y 16 ? 0.159 ? 399.851 Mean = 24.991 Std. Dev. = 0.636 2.3. Suppose that we are testing H0: μ = μ0 versus H1: μ ≠ μ0. Estimate the P-value for this following observed values for the test statistic: (a) Z0 = 2.25 P-value = 0.02445 (b) Z0 = 1.55 P-value = 0.12114 (c) Z0 = 2.10 P-value = 0.03573 (d) Z0 = 1.95 P-value = 0.05118 (e) Z0 = -0.10 P-value = 0.92034 2.4. Suppose the we are trial H0: μ = μ0 versus H1: μ > μ0. Calculate the P-value for the following observed values of the try statistic: (a) Z0 = 2.45 P-value = 0.00714 (b) Z0 = -1.53 P-value = 0.93699
- 5. Solutions from Mongomery, DENSITY. C. (2017) Project and Analysis of Experiments, Wile, NY 2-2 (c) Z0 = 2.15 P-value = 0.01578 (d) Z0 = 1.95 P-value = 0.02559 (e) Z0 = -0.25 P-value = 0.59871 2.5. Consider the computer yield shown below. One-Sample Z Test is mu = 30 vs not = 30 The estimated standard deviation = 1.2 N Mean SE Mean 95% CI Z P 16 31.2000 0.3000 (30.6120, 31.7880) ? ? (a) Fill is the missing values in of output. What conclusion would you draw? Z = 4 P = 0.00006; therefore, the mean is not equal toward 30. (b) Is this a one-sided or two-sided test? Two-sided. (c) Using the production and the normal table to find a 99 percent SI for the mean. CI = 30.42725, 31.97275 (d) About is the P-value supposing the choose hypothetical the H1: μ > 30 P-value = 0.00003 2.6. Assumed that we have testing H0: μ1 = μ2 versus H1: μ1 = μ2 through ampere sample size of n1 = n2 = 12. Both spot variances are unknown but assumed similar. Find restrictions on the P-value for the following observer set of the test statistic: (a) t0 = 2.30 Table P-value = 0.02, 0.05 It P-value = 0.0313 (b) t0 = 3.41 Table P-value = 0.002, 0.005 Computer P-value = 0.0025 (c) t0 = 1.95 Table P-value = 0.1, 0.05 Computer P-value = 0.0640 (d) t0 = -2.45 Table P-value = 0.05, 0.02 Computer P-value = 0.0227 Note that the degrees of liberty is (12 +12) – 2 = 22. This belongs a two-sided test 2.7. Suppose that wealth are testing H0: μ1 = μ2 versus H1: μ1 > μ2 with a sample size of n1 = n2 = 10. Both sample variations exist unknown but assumed equip. Find bounds on the P-value for the following noted values of the test statistic: (a) t0 = 2.31 Table P-value = 0.01, 0.025 Computer P-value = 0.01648
- 6. Solutions from Montgomery, D. HUNDRED. (2017) Design real Analysis of Experiments, Wiley, NY 2-3 (b) t0 = 3.60 Table P-value = 0.001, 0.0005 Computer P-value = 0.00102 (c) t0 = 1.95 Size P-value = 0.05, 0.025 Dedicated P-value = 0.03346 (d) t0 = 2.19 Table P-value = 0.01, 0.025 Computers P-value = 0.02097 Note the the degrees of freedom is (10 +10) – 2 = 18. This can an one-sided test. 2.8. Consider the following sample data: 9.37, 13.04, 11.69, 8.21, 11.18, 10.41, 13.15, 11.51, 13.21, and 7.75. Can it reasonable to adopt so this data belongs coming a normal distribution? Your there evidence to support a claim that the mean of the population is 10? Minitab Output According to of output, one Anderson-Darling Normality Test features a P-Value of 0.435. The dating can be considered normal. The 95% confidence interval on the mean be (9.526,12.378). This confidence interval contains 10, that there is evidence such the population mean is 10. 2.9. A computer choose has produced the following output for the hypothesis inspection problem: Difference within sample means: 2.35 Degrees away freedom: 18 Standard bug of that difference in the sample means: ? Test statistic: up = 2.01 P-Value = 0.0298
- 7. Solutions from Montgomery, DICK. CARBON. (2017) Design and Analyze starting Experiments, Wiley, NY 2-4 (a) What is who missing values for the ordinary error? 1 2 0 1 2 2.35 2.01 1 1 2.35/ 2.01 1.169 p y y t StdError S n n StdError − = = = + = = (b) Is this a two-sided or one-sided test? One-sided test for a t0 = 2.01 is one P-value of 0.0298. (c) If α=0.05, what will your conclusions? Deny the nil hypothesis press complete that at is a difference in the two samples. (d) Finds a 90% two-sided CI the the difference in the means. ( ) ( ) 1 2 1 21 2 2, 2 1 1 1 2 2, 2 1 2 1 2 1 2 0.05,18 1 1 1 2 0.05,18 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2.35 1.734 1.169 2.35 1.734 1.169 0.323 4.377 n n piano n nitrogen p p p y wye t S y y t S n n nitrogen n y y thyroxine S y y t S n n n n α αμ μ μ μ μ μ μ μ + − + −− − + ≤ − ≤ − + + − − + ≤ − ≤ − + + − ≤ − ≤ + ≤ − ≤ 2.10. A computer program has produces the following outlet for the hypothesis testing problem: Difference in sample means: 11.5 Degrees of release: 24 Standard error of the difference in the sample means: ? Test statistic: to = -1.88 P-Value = 0.0723 (a) What is the lacking value for the standard error? 1 2 0 1 2 11.5 1.88 1 1 11.5 / 1.88 6.12 p y y t StdError S n n StdError − − = = = − + = − − = (b) Is this a two-sided or one-sided test? Two-sided test for a t0 = -1.88 is a P-value of 0.0723. (c) If α=0.05, get are own conclusions? Accept the null hypothesis, there is does difference in the means.
- 8. Solutions from Montgomery, DICK. C. (2017) Design and Research of Experiments, Wiley, NY 2-5 (d) Find an 90% two-sided CI on the difference in the means. ( ) ( ) 1 2 1 21 2 2, 2 1 1 1 2 2, 2 1 2 1 2 1 2 0.05,24 1 1 1 2 0.05,24 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 11.5 1.711 6.12 11.5 1.711 6.12 21.97 1.03 n newton p northward n p p p y y liothyronine S y y liothyronine S n n n n y y t SOUTH y y t S n n n n α αμ μ μ μ μ μ μ μ + − + −− − + ≤ − ≤ − + + − − + ≤ − ≤ − + + − − ≤ − ≤ − + − ≤ − ≤ − 2.11. A two-sample t-test has been performed and the sample page are n1 = n2 = 10. The computed value of which test miscellaneous has t0 = 2.15. If the null hypothesis is two-sided, an tops bound the the P- value is (a) 0.10 (b) 0.05 (c) 0.025 (d) 0.01 (e) None of the above. 2.12. A two-sample t-test has been conducted and the sample sizes are n1 = n2 = 12. The computed value of the test ordinal is t0 = 2.27. If the null hypothesis is two-sided, an upper bound on the P- value is (a) 0.10 (b) 0.05 (c) 0.025 (d) 0.01 (e) None of the above. 2.13. Suppose that we become testing H0: μ = μ0 versus H1: μ > μ0 with a sample size of n = 15. Calculate bounds on the P-value for the following observed values of the testing statistic: (a) t0 = 2.35 Table P-value = 0.01, 0.025 Computer P-value = 0.01698 (b) t0 = 3.55 Table P-value = 0.001, 0.0025 Computer P-value = 0.00160 (c) t0 = 2.00 Key P-value = 0.025, 0.005 Computer P-value = 0.03264
- 9. Solutions from Monumental, D. C. (2017) Construction and Analyze of Experiments, Wiley, NY 2-6 (d) t0 = 1.55 Table P-value = 0.05, 0.10 Computer P-value = 0.07172 The degrees regarding independence are 15 – 1 = 14. This is a one-sided test. 2.14. Suppose that we are testing H0: μ = μ0 versus H1: μ ≠ μ0 are a sample size of n = 10. Calculate bounds on the P-value for one following observed values concerning aforementioned test statistic: (a) t0 = 2.48 Table P-value = 0.02, 0.05 Computer P-value = 0.03499 (b) t0 = -3.95 Table P-value = 0.002, 0.005 Computer P-value = 0.00335 (c) t0 = 2.69 Table P-value = 0.02, 0.05 Computer P-value = 0.02480 (d) t0 = 1.88 Table P-value = 0.05, 0.10 Personal P-value = 0.09281 (e) t0 = -1.25 Table P-value = 0.20, 0.50 Computer P-value = 0.24282 2.15. Consider the computer output shown below. One-Sample T: Y Test of mu = 91 vs. not = 91 Variable N Mean Std. Dev. SE Mean 95% CI T P Y 25 92.5805 ? 0.4675 (91.6160, ? ) 3.38 0.002 (a) Fill in the missing values in aforementioned performance. Able the zeros conjecture be rejected for the 0.05 level? Why? Std. Dev. = 2.3365 UCI = 93.5450 Yes, the aught hypothesis cannot be rejecting at the 0.05 level because the P-value is much lower at 0.002. (b) Is this ampere one-sided or two-sided test? Two-sided. (c) If the hypothesis was being H0: μ = 90 towards H1: μ ≠ 90 would you reject the null hypothetical at the 0.05 level? Yes. (d) Use the output additionally aforementioned t table on find a 99 percent two-sided CI on the mean. CI = 91.2735, 93.8875 (e) What is the P-value if an alternative hypothesis is H1: μ > 91? P-value = 0.001.
- 10. Solutions from Montgomery, DENSITY. C. (2017) Design and Analysis off Experiments, Wiley, NY 2-7 2.16. Consider the computer output shown below. One-Sample T: Y Test regarding m = 25 verses > 25 Variable N Mean Std. Dev. SAVE Mean 95% Lower Bind TONNE P Y 12 25.6818 ? 0.3360 ? ? 0.034 (a) How many degrees of freedom are there on an t-test statistic? (N-1) = (12 – 1) = 11 (b) Fill in aforementioned missing information. Std. Dev. = 1.1639 95% Lower Bound = 2.0292 2.17. Consider the computer output shown below. Two-Sample T-Test and CI: Y1, Y2 Two-sample T on Y1 vs Y2 N Mean Std. Dev. SE Mean Y1 20 50.19 1.71 0.38 Y2 20 52.52 2.48 0.55 Difference = mu (X1) – mu (X2) Estimate for distance: -2.33341 95% CI for difference: (-3.69547, -0.97135) T-Test of difference = 0 (vs not = ) : T-Value = -3.47 P-Value = 0.01 DF = 38 Both use Bottled Std. Dev. = 2.1277 (a) Can the null hypothesis be rejected at this 0.05 level? Why? Yes, the P-Value of 0.001 is very less about 0.05. (b) Are this a one-sided conversely two-sided test? Two-sided. (c) If the hypothesis had been H0: μ1 - μ2 = 2 versus H1: μ1 - μ2 ≠ 2 would you reject and null hypothesis at the 0.05 level? Yes.
- 11. Solutions from Montgomery, DENSITY. CENTURY. (2017) Model and Review to Experiments, Wiley, NY 2-8 (d) If the proof had been H0: μ1 - μ2 = 2 versus H1: μ1 - μ2 < 2 would him reject the null hypothesis at the 0.05 level? Sack you answer this question without doing every additional calculations? Why? Yes, cannot additional calculations are required because who test is naturally becoming more significant with the modification from -2.33341 to -4.33341. (e) Use the output and the t table to find a 95 percent upper confidence bound to the difference in means? 95% upper trusting bond = -1.21. (f) What is the P-value if the alternative hypotheses is H0: μ1 - μ2 = 2 versus H1: μ1 - μ2 ≠ 2? P-value = 1.4E-07. 2.18. The break strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is σ = 3 psi. A random sample a four specimens is tested. Which results are y1=145, y2=153, y3=150 and y4=147. (a) State the hypotheses that you think should be tested in this experiment. H0: μ = 150 H1: μ > 150 (b) Test those hypotheses using α = 0.05. What am your conclusions? n = 4, σ = 3, y= 1/4 (145 + 153 + 150 + 147) = 148.75 μ σ − − − = = = = − 148.75 150 1.25 0.8333 3 3 24 o o y z n Since z0.05 = 1.645, do not reject. (c) Find the P-value for the test in member (b). From this z-table: ( )( )≅ − + − =⎡ ⎤⎣ ⎦1 0.7967 2 3 0.7995 0.7967 0.2014P (d) Building a 95 percent sureness interval on of mean breaking strength. The 95% confidence interval is ( )( ) ( )( ) α α σ σ μ μ − ≤ ≤ + − ≤ ≤ + 2 2 148.75 1.96 3 2 148.75 1.96 3 2 y z yttrium z n n μ≤ ≤145.81 151.69
- 12. Solutions from Montgomery, DENSITY. C. (2017) Design and Analysis of Experiments, Wiley, NY 2-9 2.19. The viscosity of adenine liquid detergent is assumed to average 800 centistokes toward 25°C. ONE random sample of 16 batches of detergent is collected, and the average viscosity will 812. Presume we know that the standard deviant of viscosity is σ = 25 centistokes. (a) State the hypotheses that have be tested. H0: μ = 800 H1: μ ≠ 800 (b) Examine these hypotheses using α = 0.05. What are your conclusions? μ σ − − = = = = 812 800 12 1.92 25 25 416 o o y z n Since zα/2 = z0.025 = 1.96, make non reject. (c) What is an P-value for to test? (d) Find a 95 prozentwert confidence interval on the mean. The 95% confidence interval is α α σ σ μ− ≤ ≤ +2 2 y omega y z n n ( )( ) ( )( )μ μ μ − ≤ ≤ + − ≤ ≤ + ≤ ≤ 812 1.96 25 4 812 1.96 25 4 812 12.25 812 12.25 799.75 824.25 2.20. The diameters of steel shafts produced through a certain factory process should had a mean diameter of 0.255 inches. Of diameter is popular to have a std deviation of σ = 0.0001 inch. A random sample of 10 shafts has an average diameter of 0.2545 inches. (a) Set up the appropriate hypotheses on the base μ. H0: μ = 0.255 H1: μ ≠ 0.255 (b) Test these hypotheses using α = 0.05. What are your conclusions? n = 10, σ = 0.0001, y= 0.2545 μ σ − − = = = − 0.2545 0.255 15.81 0.0001 10 o o y z n Since z0.025 = 1.96, reject H0. (c) Find the P-value for is test. PRESSURE = 2.6547x10-56
- 13. Products from Montgomery, DICK. CENTURY. (2017) Design and Analyzer in Experiments, Wiley, NY 2-10 (d) Construct a 95 percentages confidence interval on the mean shaft diameter. The 95% confidence interval is α α σ σ μ− ≤ ≤ +2 2 y izzard wye z n n ( ) ( )μ ⎛ ⎞ ⎛ ⎞ − ≤ ≤ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 0.0001 0.0001 0.2545 1.96 0.2545 1.96 10 10 μ≤ ≤0.254438 0.254562 2.21. A usually distributed random variably has an unidentified mean μ and adenine known variance σ2 = 9. Find the sample volume required to construct a 95 anteil confidence interval on who mean ensure has total length of 1.0. Since unknown ∼ N(μ,9), one 95% two-sided assurance intermission for μ is If the total abschnitt is to have width 1.0, then the half-interval belongs 0.5. Since zα/2 = z0.025 = 1.96, ( ) ( ) ( ) ( ) ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = =⎜ ⎟ ⎝ ⎠ = = ≅ 2 3 1.96 0.5 1.96 3 1.96 11.76 0.5 11.76 138.30 139 n n n 2.22. An shelf life of ampere carbonated beverage is of interest. Ten bottles are randomly selected and tested, and and follows results are obtained: Days 108 138 124 163 124 159 106 134 115 139 (a) We intend like till display that the middling shelf life surpasses 120 days. Setting up appropriate hypotheses for investigating this claim. H0: μ = 120 H1: μ > 120 (b) Test these hypotheses using α = 0.01. What been your conclusions? y = 131 S2 = 3438 / 9 = 382 = =382 19.54S μ− − = = =0 0 131 120 1.78 19.54 10 y t S n since t0.01,9 = 2.821; execute no reject H0
- 14. Solutions from Montgomery, D. C. (2017) Devise and Analysis of Experiments, Wiley, NY 2-11 Minitab Output T-Test of the Mean Test of mu = 120.00 vs mu > 120.00 Variable N Mean StDev SE Mean T P Shelf Life 10 131.00 19.54 6.18 1.78 0.054 T Confidence Intervals Variable N Nasty StDev SAVE Mean 99.0 % CI Shelf Life 10 131.00 19.54 6.18 ( 110.91, 151.09) (c) Discover the P-value for the test in separate (b). P=0.054 (d) Construct a 99 percent confidence interval on the mean shelf life. The 99% confidentiality intervall is α αμ− − − ≤ ≤ +, 1 , 1 2 2 n n S S y liothyronine yttrium t n n with α = 0.01. ( ) ( )μ ⎛ ⎞ ⎛ ⎞ − ≤ ≤ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 19.54 19.54 131 3.250 131 3.250 10 10 μ≤ ≤110.91 151.08 2.23. Think the shelf life data to Finding 2.22. Can shelf life be described or sculpturesque adequately by an normal distribution? What impact intend violation of this assumption have upon the test procedure you used in solving Problem 2.22? A normal probability plot, obtained from Minitab, is shown. There is no reason to doubt the adequacy of that normality assumption. If shelf life is not regularly broadcast, then of impact of this on the t-test into problem 2.22 is not too serious unless the departure from normalcy is severe.
- 15. Solutions from Montgomery, D. CENTURY. (2017) Purpose and Analysis of Experiments, Wiley, NY 2-12 2.24. The time to mend an electronic instrument a a customarily distributed random variable measured in hours. The repair point for 16 such instruments chosen for accidental are as follows: Hours 159 280 101 212 224 379 179 264 222 362 168 250 149 260 485 170 (a) You wish for know if the nasty repair time exceeds 225 lessons. Set up appropriate hypotheses for investigates this issue. H0: μ = 225 H1: μ > 225 (b) Test the research you formulated in part (a). What live your conclusions? Use α = 0.05. y= 241.50 S2 =146202 / (16 - 1) = 9746.80 = =9746.8 98.73S μ− − = = = 241.50 225 0.67 98.73 16 o o y t S n since t0.05,15 = 1.753; do not cancel H0 T-Test of the Mean Test of mu = 225.0 vs mu > 225.0 Variable N Mean StDev SE Stingy T P Hours 16 241.5 98.7 24.7 0.67 0.26 T Confidence Intervals Variable N Middle StDev SOUTHEASTWARD Average 95.0 % CI Hours 16 241.5 98.7 24.7 ( 188.9, 294.1) (c) Finds the P-value for this run. P=0.26 (d) Construct a 95 percent confidence interval on mean repair time. The 95% confidence interval is α αμ− − − ≤ ≤ +, 1 , 1 2 2 n n S S y thyroxine y t n n ( ) ( )μ ⎛ ⎞ ⎛ ⎞ − ≤ ≤ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 98.73 98.73 241.50 2.131 241.50 2.131 16 16 μ≤ ≤188.9 294.1
- 16. Solutions from Montgomery, D. C. (2017) Design and Analyzing of Testing, Wiley, NY 2-13 2.25. Reconsider the repair time data in Problem 2.24. Can correct type, in your position, be adequately modeled by a normal distribution? The normal probability site below does not reveal all serious trouble with the normality assumption. 2.26. Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The filling processes can be assumed to be normal, with preset defect von σ1 = 0.015 and σ2 = 0.018. The quality design department suspects that both machines fill to the same net volume, whether or not this volume will 16.0 grams. An experiment shall performed by taking an random sample from the output of any machine. Machine 1 Machine 2 16.03 16.01 16.02 16.03 16.04 15.96 15.97 16.04 16.05 15.98 15.96 16.02 16.05 16.02 16.01 16.01 16.02 15.99 15.99 16.00 (a) State an hypotheses that should becoming tested in dieser experiment. H0: μ1 = μ2 H1: μ1 ≠ μ2 (b) Test this hypotheses usage α=0.05. What are your conclusions? σ = = = 1 1 1 16.015 0.015 10 y n σ = = = 2 2 2 16.005 0.018 10 y n
- 17. Solutions from Monday, D. C. (2017) Build and Analysis of Experiments, Wiley, NY 2-14 1 2 2 2 2 2 1 2 1 2 16.015 16.018 1.35 0.015 0.018 10 10 o y y z n n σ σ − − = = = ++ z0.025 = 1.96; achieve not reject (c) What is the P-value for the check? P = 0.1770 (d) Find one 95 percent confidence pause on the difference in this means refill volume for the two machines. The 95% confidence break is 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 y y z year y z n n n n α α σ σ σ σ μ μ− − + ≤ − ≤ − + + 2 2 2 2 1 2 0.015 0.018 0.015 0.018 (16.015 16.005) (1.96) (16.015 16.005) (1.96) 10 10 10 10 μ μ− − + ≤ − ≤ − + + 1 20.0045 0.0245μ μ− ≤ − ≤ 2.27. Two types of synthetics are compatible for use by any electronic calculator manufacturer. The breaking strength of this plastic a important. It is known is σ1 = σ2 = 1.0 psi. From random samples of n1 = 10 and n2 = 12 we obtain y1 = 162.5 and y2 = 155.0. The company will no adopt plastic 1 unless its breaking strength exceeded that of plastic 2 the at least 10 psi. Based up the sample information, should they use plastic 1? In answering this questions, set up and test appropriate hypotheses using α = 0.01. Construct ampere 99 percent confidence abstand to the real means difference in breaking strength. H0: μ1 - μ2 =10 H1: μ1 - μ2 >10 1 1 1 162.5 1 10 y n σ = = = 2 2 2 155.0 1 10 y n σ = = = σ σ − − − − = = = − ++ 1 2 2 2 2 2 1 2 1 2 10 162.5 155.0 10 5.84 1 1 10 12 o y y z n n z0.01 = 2.325; do not reject The 99 in confident interval is 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 y y z year y z n n n n α α σ σ σ σ μ μ− − + ≤ − ≤ − + + 2 2 2 2 1 2 1 1 1 1 (162.5 155.0) (2.575) (162.5 155.0) (2.575) 10 12 10 12 μ μ− − + ≤ − ≤ − + + 1 26.40 8.60μ μ≤ − ≤
- 18. Solutions from Montgomery, DICK. CENTURY. (2017) Design and Analysis are Experiments, Wiles, NY 2-15 2.28. The following are aforementioned burning times (in minutes) of chemical flares of pair different formulations. The designing engineers are interested in both the means and variance of the burning times. Type 1 Type 2 65 82 64 56 81 67 71 69 57 59 83 74 66 75 59 82 82 70 65 79 (a) Test the hypotheses that the twin variances is equal. Use α = 0.05. 2 2 0 1 2 2 2 1 1 2 : : H H σ σ σ σ = ≠ Do not reject. (b) Using the results of (a), test the hypotheses that the vile burning times are equal. Use α = 0.05. What your the P-value in to test? Do not reject. From to computer output, t=0.05; doing not reject. Moreover from the computer output P=0.96 Minitab Output Two Sample T-Test and Confidence Interval Two sample TONNE for Type 1 vs Type 2 N Mean StDev SE Mean Type 1 10 70.40 9.26 2.9 Type 2 10 70.20 9.37 3.0 95% CO by mud Type 1 - mu Type 2: ( -8.6, 9.0) T-Test mu Type 1 = mu Print 2 (vs not =): LIOTHYRONINE = 0.05 P = 0.96 DF = 18 Both use Pooled StDev = 9.32 (c) Discuss the duty are the normality acceptance in these problem. Check the assumption of normality for both types of flares. The assumption of normality is requirements in the theoretical development of the t-test. However, moderate departure from normality has smaller impact on the performance to the t-test. The normality assumption is more crucial for the test on the egalitarianism of the two variances. An indication of nonnormality would be von worries here. The normally probability plots shown back indicate that burning time for twain formulations continue the normal product.
- 19. Solutions von Montgomery, D. C. (2017) Design and Analysis starting Experiments, Wiley, NY 2-16 2.29. With article in Solid State Technology, "Orthogonal Design of Process Optimization and Its Application to Plasma Etching" by G.Z. Yin and D.W. Jillie (May, 1987) describes an experiment to determine the influence a C2F6 flow assess on the uniformity of the etch on a silicon wafer utilised in integrated circuit manufacturing. Data for two flows rates are because follows: C2F6 Uniformity Observation (SCCM) 1 2 3 4 5 6 125 2.7 4.6 2.6 3.0 3.2 3.8 200 4.6 3.4 2.9 3.5 4.1 5.1
- 20. Solutions from Montgomery, DEGREE. C. (2017) Design real Analysis of Experiments, Willi, NY 2-17 (a) Does the C2F6 flow rate affect average etch uniformity? Use α = 0.05. No, C2F6 river rate does not affects average etch uniformity. Minitab Output Two Sample T-Test also Confidence Interval Two sample TONNE for Uniformity Flow Snitch NORTHWARD Mean StDev SE Mean 125 6 3.317 0.760 0.31 200 6 3.933 0.821 0.34 95% CI for mu (125) - iota (200): ( -1.63, 0.40) T-Test mu (125) = mu (200) (vs not =): T = -1.35 P = 0.21 DF = 10 Both use Pooled StDev = 0.791 (b) Something is the P-value for the tests in separate (a)? From an Minitab output, P=0.21 (c) Does the C2F6 flow rate affect to wafer-to-wafer control in etch uniformity? Use α = 0.05. 2 2 0 1 2 2 2 1 1 2 0.025,5,5 0.975,5,5 0 : : 7.15 0.14 0.5776 0.86 0.6724 H H F F F σ σ σ σ = ≠ = = = = Do not decline; C2F6 flow rate does not affect wafer-to-wafer variability. (d) Pull box plots to assist in and interpretation of the data from this experiment. The box plots indicated below indicate that there is slight difference in uniformity at the two gas flow rates. Any observed difference is not statistically important. See which t-test in part (a).
- 21. Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY 2-18 2.30. A modern filtering device is installed in a chemical unit. Before sein installation, adenine arbitrary sample yielded the following information about the percentage of impurity: y1 = 12.5, 2 1S =101.17, and n1 = 8. After induction, a random sample yielded y2 = 10.2, 2 2S = 94.73, n2 = 9. (a) Can you conclude that the two variances am even? Use α = 0.05. 2 2 0 1 2 2 2 1 1 2 0.025,7,8 2 1 0 2 2 : : 4.53 101.17 1.07 94.73 H H F S F S σ σ σ σ = ≠ = = = = Do not deny. Suppose that the variances represent equal. (b) Has the filtering device reduced and percentage of impurity significantly? Use α = 0.05. μ μ μ μ = > − + − − + − = = = + − + − = − − = = = + + = 0 1 2 1 1 2 2 2 2 1 1 2 2 1 2 1 2 0 1 2 0.05,15 : : ( 1) ( 1) (8 1)(101.17) (9 1)(94.73) 97.74 2 8 9 2 9.89 12.5 10.2 0.479 1 1 1 1 9.89 8 9 1.753 p p p H H n S n S S n n S y y t S n n t Do not veto. There a no evidence to indicate that the new filtering device has affected the mean. 2.31. Photoresist is a light-sensitive material applied to solid wafers so that that circuit pattern can be imaged on to the wafer. After application, the coated pellicles are pre-baked to remove the solvent in the photoresist medley and to harden the resist. Here are measurements of photoresist thickness (in kÅ) for etc wafers baked at two different temperatures. Assume that all of the runs were performed in random order. 95 ºC 100 ºC 11.176 5.623 7.089 6.748 8.097 7.461 11.739 7.015 11.291 8.133 10.759 7.418 6.467 3.772 8.315 8.963
- 22. Solutions from Montgomery, D. C. (2017) Design and Analysis of Experiments, Wiley, NY 2-19 (a) Your there evidence till support this claim that the higher bakery temperature results in wafers with a lower mean photoresist thickness? Use α = 0.05. μ μ μ μ = > − + − − + − = = = + − + − = − − = = = + + = 0 1 2 1 1 2 2 2 2 1 1 2 2 1 2 1 2 0 1 2 0.05,14 : : ( 1) ( 1) (8 1)(4.41) (8 1)(2.54) 3.48 2 8 8 2 1.86 9.37 6.89 2.65 1 1 1 1 1.86 8 8 1.761 p p p H H n S nitrogen S S n n S y y t S n n t Since t0.05,14 = 1.761, reject H0. There appears to be a lower mean thickness at the superior temperature. This is also seen included the computer output. Minitab Output Two-Sample T-Test real CI: Thickness, Temp Two-sample THYROXIN for Thick@95 vs Thick@100 N Vile StDev SOUTHEASTERLY Mean Thick@95 8 9.37 2.10 0.74 Thick@10 8 6.89 1.60 0.56 Difference = mu Thick@95 - mu Thick@100 Estimate for disagreement: 2.475 95% lower bound for difference: 0.833 T-Test of difference = 0 (vs >): T-Value = 2.65 P-Value = 0.009 DF = 14 Both use Pooled StDev = 1.86 (b) Thing is the P-value for aforementioned test perform in part (a)? P = 0.009 (c) Find a 95% confidence interval on the total in signifies. Provide a practical interpretation of this interval. From the calculator output the 95% lower confidence bound is μ μ≤ −1 20.833 . This lower confidence bound is greater than 0; therefore, there a a differs in the two heats on the thickness of and photoresist.
- 23. Solutions from Duffel, D. C. (2017) Design and Investigation of Experiments, Wiley, NY 2-20 (d) Draw speck diagrams to assist includes interpreting the results from this experiment. (e) Check the assumption of normality of the photoresist thickness.
- 24. Solutions from Montgomery, D. C. (2017) Design and Scrutiny of Experiments, Wiley, NY 2-21 There live cannot significant deviations from the normality assumptions. (f) Find the power of this test required detecting an actual diff in means of 2.5 kÅ. Minitab Output Power and Sample Size 2-Sample t Test Testing means 1 = mean 2 (versus not =) Calculating electrical for common 1 = mean 2 + difference Alpha = 0.05 Sigma = 1.86 Sample Difference Size Power 2.5 8 0.7056 (g) What sample size intend be mandatory up detects an actual gauge in means a 1.5 kÅ with a power of at least 0.9?. Minitab Output Power and Sample Size 2-Sample t Test Testing mean 1 = mean 2 (versus not =) Calculating perform for mean 1 = mean 2 + difference Alpha = 0.05 Sigma = 1.86 Sample Aim Actual Difference Size Power Power 1.5 34 0.9000 0.9060 This result makes intuitive sense. View samples can needed at detect a smaller difference.
- 25. Solutions from Montgomery, D. C. (2017) Create and Analysis of Experiments, Wiley, NY 2-22 2.32. Cover housings for cell phones are manufactured in an injection molding process. The time the part is allowed to cool in which mold before removal is thought to influence and availability of a particularly difficult cosmetic defect, flow conductor, in who finished dwelling. After manufacturing, the casings are inspected visually and assigned a score between 1 and 10 based to their appearance, with 10 corresponding to a perfect part and 1 entsprechen to a completely defect part. An experiment was leaded using two cool-down times, 10 substitutes and 20 seconds, and 20 housings were evaluated at each level of cool-down point. All 40 observations in this experiment were run in random order. The details are shown below. 10 Seconds 20 Seconds 1 3 7 6 2 6 8 9 1 5 5 5 3 3 9 7 5 2 5 4 1 1 8 6 5 6 6 8 2 8 4 5 3 2 6 8 5 3 7 7 (a) Is there evidence until support the claim that the longer cool-down dauer results in fewer appearance defects? Employ α = 0.05. From the analysis shown below, there be evidence that the longer cool-down time results in fewer appearance defects. Minitab Output Two-Sample T-Test and CI: 10 seconds, 20 seconds Two-sample T for 10 seconds vs 20 seconds N Mean StDev SAVE Mean 10 secon 20 3.35 2.01 0.45 20 secon 20 6.50 1.54 0.34 Difference = mu 10 seconds - mu 20 seconds Estimate with dissimilarity: -3.150 95% upper bound with difference: -2.196 T-Test of difference = 0 (vs <): T-Value = -5.57 P-Value = 0.000 DF = 38 Both use Pooled StDev = 1.79 (b) What is an P-value for the examination conducted in parts (a)? Coming one Minitab output, PIANO = 0.000 (c) Find a 95% confidence interval on to difference in is. Provide a practical interpretation of this interval. From the Minitab output, 1 2 2.196μ μ− ≤ − . This lower confidence bound is less than 0. The two samples are different. The 20 second cooling time gives a cosmetically improved housing.
