A copper wiring is being tested for its strength (MPa) with N(μ.O=0.42). Multiple samples have obtained strengths as shown in the databelow.68.7968.9668.2769.2269.5269.3169.03 68.49 68.85 68.5369.2869.2068.8268.9968.9568.5869.2168.9568.9169.19 69.14 69.0069.1468.4368.8668.8669.4468.90 69.47 69.6969.6669.0469.5868.2369.2168.60From the data, obtain an Event 1 by selecting 5 random groups with asample size of 5. From the same data. obtain an Event 2 by selecting arandom set of 3 three groups with an n of 3.• What is the probability of the difference of the sample Events 1 and 2?• In two sentences or less which of the two Events is a better estimator ofthe population mean (support this answer with computations).

A copper wiring is being tested for its strength (MPa) with N(μ,σ=0.42).Multiple samples have…

Answered: A copper wiring is being tested for its…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A researcher reports an F-ratio with dfbetween = 2 and dfwithin = 30 for an independent-measures ANOVA. a. How many treatment conditions were compared in the experiment? b. How many subjects participated ion the experiment?
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It is commonly thought that SUVs and vans are safer than cars. If an SUV and a car are in a collision, does the SUV sustain less damage and thus need a cheaper repair? We performed 7 sample crashes of SUVs at 10mph into the rear ends of stopped cars. See the table of normal data below. SUV into car SUV damage ($) Car damage ($) CRV into Civic 1721 1274 Rav4 into corolla 1434 2327 Tucson into Forte 850 3223 Tiguan into Golf 2329 2058 Patriot into Caliber 1415 3095 Escape into Focus 1470 3386 Rogue into Sentra 2884 4560 Would you be able to say that SUVs are safer since they need less repairs? Should I use: confidence intervals z-score A cause/correlation/regression or Chi-Squared hypothesis tests
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In a comparison of two methods for the determination of chromium in rye grass, the following results were obtained: Method A, mean(mg/kg Cr)=1.48 Standard deviation(mg/kg Cr)=0.28 n= 6 Method B, mean (mg/kg Cr)=2.33 Standard deviation(mg/kg Cr)=0.31 n=10 Is the precision of these two methods significantly different? Do these two methods give results with means which differ significantly? a=0.05
In experiments to determine the effectiveness of drugs such as anti-fungal ointments that are applied to the skin, the concentration of the drug in a patient’s skin must be measured. The article “Determining Bioequivalence of Topical Dermatological Drug Products by Tape- Stripping” (W. Navidi, A. Hutchinson, et al., Journal of Pharmacokinetics and Pharmacodynamics, 2008:213–220) describes an improved method of measuring that is designed to ensure that all the drug in the skin is recovered. Assume that in an experiment using the old method, the mean amount of drug recovered from a sample of 49 patients was 105 ng with a standard deviation of 20 ng, and that in an experiment using the new method the mean amount of drug recovered in a sample of 35 patients was 117 ng with a standard deviation of 15 ng. Find a 98% confidence interval for the difference in the mean amounts recovered between the two methods.
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