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chpt7and8ReviewExercise

# chpt7and8ReviewExercise - 243 Review Exercises which has...

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Unformatted text preview: 243 Review Exercises which has, approximately, a standard normal distribution. When the pop- ulation is normal, the same statistic has a 2‘ distribution with n — 1 degrees of freedom. . Understand the interpretation of a level or test. If the null hypothesis is true, before the data are collected, the probability is a that the experiment will produce observations that lead to the rejection of the null hypothesis. Consequently, after many independent experiments, the proportion that lead to rejection of the null hypothesis will be nearly 07. Don’ts . Don’t routinely apply the statistical procedures above if the sample is not random but collected from convenient units or the data show a trend in time. Review Exercises 7.76 Specify the null hypothesis and the alternative hypoth- 7.77 7.78 7.79 7.80 7.81 esis in each of the following cases. (a) An engineer hopes to establish that an additive will increase the viscosity of an oil. (b) An electrical engineer hopes to establish that a modiﬁed circuit board will give a computer a higher average operating speed. With reference to the example on page 19, ﬁnd a 95% conﬁdence interval for the mean strength of the aluminum alloy. While performing a certain task under simulated weightlessness, the pulse rate of 32 astronaut trainees increased on the average by 26.4 beats per minute with a standard deviation of 4.28 beats per minute. What can one assert with 95% conﬁdence about the maximum error if f = 26.4 is used as a point estimate of the true average increase in the pulse rate of astronaut trainees performing the given task? With reference to the preceding exercise, construct a 95% conﬁdence interval for the true average increase in the pulse rate of astronaut trainees performing the given task. It is desired to estimate the mean number of days of continuous use until a certain kind of computer will ﬁrst require repairs. If it can be assumed that a = 48 hours, how large a sample is needed so that one will be able to assert with 90% conﬁdence that the sample mean is off by at most 10 hours? A sample of 12 camshafts intended for use in gasoline engines has an average eccentricity of 1.02 and a stan- dard deviation of 0.044 inch. Assuming the data may be treated as a random sample from a normal popula- tion, determine a 95% conﬁdence interval for the actual mean eccentricity of the camshaft. 7.82 7.83 7.84 7.85 7.86 In order to test the durability of a new paint, a highway department had test strips painted across heavily trav— eled roads in 15 different locations. If on the average the test strips disappeared after they had been crossed by 146,692 cars with standard deviation of 14,380 cars, construct a 99% conﬁdence interval for the true average number of cars it takes to wear off the paint. Assume a normal population. Referring to Exercise 7.82 and using 14,380 as an es- timate of a, ﬁnd the sample size that would have been needed to be able to assert with 95% conﬁdence that the sample mean is off by at most 10,000. [Hint First estimate n1 by using 2 = 1.96, then use t0.025 for n] ——1 degrees of freedom to obtain a second estimate n2, and repeat this procedure until the last two values of n thus obtained are equal] A laboratory technician is timed 20 times in the perfor— mance of a task, getting f = 7.9 and s = 1.2 minutes. If the probability of a Type I error is to be at most 0.05, does this constitute evidence against the null hy- pothesis that the average time is less than or equal to 7.5 minutes? Suppose that in the drying time example on page 222, n is changed from 36 to 50 while the other quantities remain [1.0 = 20, a = 2.4, and a = 0.03. Find (a) the new dividing line of the test criterion; (b) the probability of Type II errors for the same values of ,u as shown in the table on page 238. In an air-pollution study, ozone measurements were taken in a large California city at 5.00 RM. The eight readings (in parts per million) were 7.9 11.3 6.9 12.7 13.2 8.8 9.3 10.6 Assuming the population sampled is normal, construct a 95% conﬁdence interval for the corresponding true mean. 244 Chapter 7 Inferences Concerning a Mean 7.87 An industrial engineer concerned with service at a large medical clinic recorded the duration of time from the time a patient called until a doctor or nurse returned the call. A sample of size 180 calls had a mean of 1.65 hours and a standard deviation of 0.82. (a) Obtain a 95% conﬁdence interval for the popula— tion mean of time to return a call. (b) Does ,u lie in your interval obtained in part (a)? Explain. (C) In a long series or repeated experiments. with new random samples collected for each experi- ment, what proportion of the resulting conﬁdence intervals will contain the true population mean? Explain your reasoning. 7.88 Refer to Exercise 7.87. (a) Perform a test with the intention of establishing that the mean time to return a call is greater than 1.5 hours. Use or = 0.05. 7.89 (b) In light of your conclusion in part (a), what error could you have made? Explain in the context of this problem. (c) In a long series of repeated experiments, with new random samples collected for each experiment, what proportion of the resulting tests would reject the null hypothesis if it prevailed? Explain your reasoning. The compressive strength of parts made from a composite material are known to be nearly normally distributed. A scientist, using the testing device for the ﬁrst time, obtains the tensile strength (psi) of 20 specimens 95 102 105 107 109 110 111 112 114 115 134 135 136 138 139 141 142 144 150 155 shown in Figure 7.12. Should the scientist report the 95% conﬁdence interval based on the t—distribution? Explain your reasoning. O O O O .000 O. m 00 O. O O O I | I I | L | I Figure 7. I 2 90 100 110 120 130 140 150 160 Dot diagram of tensile strength Strength (psi) Key Terms Alternative hypothesis 225 Large sample Z test 232 Point estimation 204 Classical theory of testing Level of signiﬁcance 225 Power 238 hypotheses 226 Likelihood function 217 P value 231 Composite hypothesis 227 Maximum likelihood estimator 217 Simple hypothesis 227 Conﬁdence 208 Neyman—Pearson theory 226 Tail probability 230 Conﬁdence interval 210 Null hypothesis 227 Two-sided alternative 225 Conﬁdence limits 210 One sample I test 233 Two—sided criterion 226 Critical regions 229 One-sided alternative 225 Critical Values 230 One-sided criterion 226 Degree of conﬁdence 210 one-sided test 226 Estimated standard error 204 one-taﬂed test 226 Hat notation 205 Interval estimate 209 curve 238 Operating characteristic (0C) Two—sided test 226 Two—tailed test 226 223 223 Unbiased estimator Type I error Type 11 error 205 266 Chapter 8 Comparing Two Treatments Review Exercises 8.26 8.27 8.28 8.29 8.30 8.31 With reference to Exercise 2.64, test that the mean charge of the electron is the same for both tubes. Use or = 0.05. With reference to the previous exercise, ﬁnd a 90% conﬁdence interval for the difference of the two means. Two chemical additives for drying paint are to be com- pared. Five spray cans are prepared using Additive A and six are prepared using Additive B. Then 11 differ— ent boards are sprayed, one can per board. (a) The response is the time in minutes for the surface to dry, and the summary statistics are Sample Standard size Mean deviation Additive A 5 16.3 2 7 12.1 1 1 Should you pool or not pool the estimates of variance in order to conduct a test of hypothe- ses that is intended to show that there is a difference in means? Explain how you would proceed. (b) Conduct the test for part (a) using 01 = 0.05. (c) Describe how you would randomize the assign- ment of paints when conducting this experiment. With reference to the example on page 14, test that the mean copper content is the same for both heats. With reference to the previous exercise, ﬁnd a 90% conﬁdence interval for the difference of the two means. Random samples are taken from two normal popula— tions with 01 = 10.8 and 02 = 14.4 to test the null hypothesis u] — [22 = 53.2 against the alternative hypothesis m — M2 > 53.2 at the level of signiﬁ— cance a = 0.01. Determine the common sample size n = n1 = mg that is required if the probability of not rejecting the null hypothesis is to be 0.09 when ,LL} — M2 = 66.7. Key Terms 8.32 8.33 8.34 8.35 With reference to the example on page 254, ﬁnd a 90% conﬁdence interval for the difference of mean strengths of the alloys (a) using the pooled procedure; (b) using the large samples procedure. How would you randomize, for a two sample test, in each of the following cases? (a) Twenty cars are available for a mileage study and you want to compare a modiﬁed spark plug with the regular. (b) A new oven will be compared with the old. Fifteen ceramic specimens are available for baking. With reference to part (a) of Exercise 8.33, how would you pair and then randomize for a paired test? Two samples in C1 and C2 can be analyzed using the MINITAB commands Dialog box: Stat > Basic Statistics > 2-Sample t Type C l in First C2 in Second. Click Samples in different columns. Click Assume equal variances. Click OK. If you do not click Assume equal variances, the Smith-Satterthwaite test is performed. The output relating to the example on page 254 is TWO SAMPLE T FOR ALLOY 1 VS ALLOY 2 N MEAN STDEV SE MEAN ALLOY l 58 70.70 1.80 0.24 ALLOY 2 27 76.13 2.42 0.47 95 PCT Cl FOR MU ALLOY l — MU ALLOY 2: (~6.36, —4.50) T TEST MU ALLOY l = MU ALLOY 2 (VS NE): T = ——11.58 P = 0.000 DF = 83.0 Perform the test for the data in Exercise 8.10. 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chpt7and8ReviewExercise - 243 Review Exercises which has...

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