In an experiment to test the effectiveness of a new sleeping aid, a sample of 12

A crayon manufacturer is comparing the effects of two kinds of yellow dye on the brittleness of crayons. Dye B is more expensive than dye A, but it is thought that it might produce a stronger crayon. Four crayons are tested with each kind of dye, and the impact strength (in joules) is measured for each. The results are as follows: Dye A: 1.0 2.0 1.2 3.0 Dye B: 3.0 3.2 2.6 3.4 a. Can you conclude that the mean strength of crayons made with dye B is greater than that of crayons made with dye A? b. Can you conclude that the mean strength of crayons made with dye B exceeds that of crayons made with dye A by more than 1 J?

In a study of the relationship of the shape of a tablet to its dissolution time, 6 disk-shaped ibuprofen tablets and 8 oval-shaped ibuprofen tablets were dissolved in water. The dissolve times, in seconds, were as follows: Disk: 269.0 249.3 255.2 252.7 247.0 261.6 Oval: 268.8 260.0 273.5 253.9 278.5 289.4 261.6 280.2 Can you conclude that the mean dissolve times differ between the two shapes?

The article “Influence of Penetration Rate on Penetrometer Resistance” (J. Oliveira, M. Almeida, et al., Journal of Geotechnical and Geoenvironmental Engineering, 2011:695–703) presents measures of penetration resistance, expressed as a multiple of a standard quantity, for a certain fine-grained soil. Fifteen measurements taken at a depth of 1mhad a mean of 2.31 with a standard deviation of 0.89. Fifteen measurements taken at a depth of 2m had a mean of 2.80 with a standard deviation of 1.10. Can you conclude that the penetration resistance differs between the two depths?

The article “Time Series Analysis for Construction Productivity Experiments” (T. Abdelhamid and J. Everett, Journal of Construction Engineering and Management, 1999:87–95) presents a study comparing the effectiveness of a video system that allows a crane operator to see the lifting point while operating the crane with the old system in which the operator relies on hand signals from a tagman. Three different lifts, A, B, and C, were studied. Lift A was of little difficulty, lift B was of moderate difficulty, and lift C was of high difficulty. Each lift was performed several times, both with the new video system and with the old tagman system. The time (in seconds) required to perform each lift was recorded. The following tables present the means, standard deviations, and sample sizes. a. Can you conclude that the mean time to perform a lift of low difficulty is less when using the video system than when using the tagman system? Explain. b. Can you conclude that the mean time to perform a lift of moderate difficulty is less when using the video system than when using the tagman system? Explain. c. Can you conclude that the mean time to perform a lift of high difficulty is less when using the video system than when using the tagman system? Explain.

The Mastic tree (Pistacia lentiscus) is used in reforestation efforts in southeastern Spain. The article “Nutrient Deprivation Improves Field Performance of Woody Seedlings in a Degraded Semi-arid Shrubland” (R. Trubata, J. Cortina, and A. Vilagrosaa, Ecological Engineering, 2011:1164–1173) presents a study that investigated the effect of adding slow release fertilizer to the usual solution on the growth of trees. Following are the heights, in cm, of 10 trees grown with the usual fertilizer (the control group), and 10 trees grown with the slow-release fertilizer (treatment). These data are consistent with the mean and standard deviation reported in the article. Can you conclude that the mean height of plants grown with slow-release fertilizer is greater than that of plants with the usual fertilizer? Usual 17.3 22.0 19.5 18.7 19.5 18.5 18.6 20.3 20.3 20.3 Slow-release 25.2 23.2 25.2 26.2 25.0 25.5 25.2 24.1 24.8 23.6

Two weights, each labeled as weighing 100 g, are each weighed several times on the same scale. The results, in units of \(\mu g\) above 100 g, are as follows: First weight: 53 88 89 62 39 66 Second weight: 23 39 28 2 49 Since the same scale was used for both weights, and since both weights are similar, it is reasonable to assume that the variance of the weighing does not depend on the object being weighed. Can you conclude that the weights differ? Equation Transcription: Text Transcription: {mu}g

It is thought that a new process for producing a certain chemical may be cheaper than the currently used process. Each process was run 6 times, and the cost of producing 100 L of the chemical was determined each time. The results, in dollars, were as follows: New Process: 51 52 55 53 54 53 Old Process: 50 54 59 56 50 58 Can you conclude that the mean cost of the new method is less than that of the old method?

The article “Effects of Aerosol Species on AtmosphericVisibility in Kaohsiung City, Taiwan” (C. Lee, C. Yuan, and J. Chang, Journal of Air andWaste Management, 2005:1031–1041) reported that for a sample of 20 days in the winter, the mass ratio of fine to coarse particles averaged 0.51 with a standard deviation of 0.09, and for a sample of 14 days in the spring the mass ratio averaged 0.62 with a standard deviation of 0.09. Let \(\mu_{1}\) represent the mean mass ratio during the winter and let \(\mu_{2}\) represent the mean mass ratio during the summer. It is desired to test \(H_{0}: \mu_{2}-\mu_{1}=0\) versus \(H_{1}: \mu_{2}-\mu_{1} \neq 0\). a. Someone suggests that since the sample standard deviations are equal, the pooled variance should be used. Do you agree? Explain. b. Using an appropriate method, perform the test. Equation Transcription: Text Transcription: {mu}_1 {mu}_2 H_0:{mu}_2-{mu}_1=0 H_1:{mu}_2-{mu}_1{not=}0

The article “Wind-Uplift Capacity of Residential Wood Roof-Sheathing Panels Retrofitted with Insulating Foam Adhesive” (P. Datin, D. Prevatt, and W. Pang, Journal of Architectural Engineering, 2011:144–154) presents a study of the failure pressures of roof panels. A sample of 15 panels constructed with 8-inch nail spacing on the intermediate framing members had a mean failure pressure of 8.38 kPa with a standard deviation of 0.96 kPa. A sample of 15 panels constructed with 6-inch nail spacing on the intermediate framing members had a mean failure pressure of 9.83 kPa with a standard deviation of 1.02 kPa. Can you conclude that 6-inch spacing provides a higher mean failure pressure?

The article “Magma Interaction Processes Inferred from Fe-Ti Oxide Compositions in the D¨olek and Sari¸ci¸cek Plutons, Eastern Turkey” (O. Karsli, F. Aydin, et al., Turkish Journal of Earth Sciences, 2008:297–315) presents chemical compositions (in weight-percent) for several rock specimens. Fourteen specimens (two outliers were removed) of limenite grain had an average iron oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) content of 9.30 with a standard deviation of 2.71, and seven specimens of limenite lamella had an average iron oxide content of 9.47 with a standard deviation of 2.22. Can you conclude that the mean iron oxide content differs between limenite grain and limenite lamella? Equation Transcription: Text Transcription: (Fe_2O_3)

The article “Structural Performance of Rounded Dovetail Connections Under Different Loading Conditions” (T. Tannert, H. Prion, and F. Lam, Can J Civ Eng, 2007:1600–1605) describes a study of the deformation properties of dovetail joints. In one experiment, 10 rounded dovetail connections and 10 double rounded dovetail connections were loaded until failure. The rounded connections had an average load at failure of 8.27 kN with a standard deviation of 0.62 kN. The double-rounded connections had an average load at failure of 6.11 kN with a standard deviation of 1.31 kN. Can you conclude that the mean load at failure is greater for rounded connections than for double-rounded connections?

The article “Variance Reduction Techniques: Experimental Comparison and Analysis for Single Systems” (I. Sabuncuoglu, M. Fadiloglu, and S. Celik, IIE Transactions, 2008:538–551) describes a study of the effectiveness of the method of Latin Hypercube Sampling in reducing the variance of estimators of the mean time-in-system for queueing models. For the M/M/1 queueing model, ten replications of the experiment yielded an average reduction of 6.1 with a standard deviation of 4.1. For the serial line model, ten replications yielded an average reduction of 6.6 with a standard deviation of 4.3. Can you conclude that the mean reductions differ between the two models?

In an experiment to test the effectiveness of a new sleeping aid, a sample of 12 patients took the new drug, and a sample of 14 patients took a commonly used drug. Of the patients taking the new drug, the average time to fall asleep was 27.3 minutes with a standard deviation of 5.2 minutes, and for the patients taking the commonly used drug the average time was 32.7 minutes with a standard deviation of 4.1 minutes. Can you conclude that the mean time to sleep is less for the new drug?

Refer to Exercise 11 in Section 5.6. Can you conclude that the mean sodium content is higher for brand B than for brand A?

Refer to Exercise 12 in Section 5.6. Can you conclude that the mean permeability coefficient at \(60^{\circ} \mathrm{C}\) differs from that at \(61^{\circ} \mathrm{C}\)? Equation Transcription: Text Transcription: 60^oC 61^oC

The following MINITAB output presents the results of a hypothesis test for the difference \(\mu_{X}-\mu_{Y}\) between two population means. Two-sample T for X vs Y N Mean StDev SE Mean X 10 39.31 8.71 2.8 Y 10 29.12 4.79 1.5 Difference = mu (X) ? mu (Y) Estimate for difference: 10.1974 95% lower bound for difference: 4.6333 T-Test of difference = 0 (vs >): T-Value = 3.25 P-Value = 0.003 DF = 13 a. Is this a one-tailed or two-tailed test? b. What is the null hypothesis? c. Can \(H_{0}\) be rejected at the 1% level? How can you tell? Equation Transcription: Text Transcription: {mu}_X-{mu}_Y H_0

The following MINITAB output presents the results of a hypothesis test for the difference \(\mu_{X}-\mu_{Y}\) between two population means. Some of the numbers are missing. Fill in the numbers for (a) through (d). Two-sample T for X vs Y N Mean StDev SE Mean X 6 1.755 0.482 (a) Y 13 3.239 (b) 0.094 Difference = mu (X) ? mu (Y) Estimate for difference: (c) 95% CI for difference: (?1.99996, ?0.96791) T-Test of difference = 0 (vs not =): T-Value = (d) P-Value = 0.000 DF = 7 Equation Transcription: Text Transcription: {mu}_X-{mu}_Y