Student Solutions Manual for Devore's Probability and Statistics for Engineering and the Sciences, 9th
9th Edition
ISBN: 9798214004020
Author: Jay L. Devore
Publisher: Cengage Learning US
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Chapter 10.1, Problem 1E
In an experiment to compare the tensile strengths of I = 5 different types of copper wire, J = 4 samples of each type were used. The between-samples and within-samples estimates of
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One company's bottles of grapefruit juice are filled by a machine that is set to dispense an average of 180 milliliters (ml) of
liquid. A quality-control inspector must check that the machine is working properly. The inspector takes a random sample of 40
bottles and measures the volume of liquid in each bottle.
We want to test
Hg: μ = 180
Ha: 180
where μ = the true mean volume of liquid dispensed by the machine. The mean amount of liquid in the bottles is 179.6 ml and
the standard deviation is 1.3 ml. A significance test yields a P-value of 0.0589.
Interpret the P-value.
Assuming the true mean volume of liquid dispensed by the machine is 180 ml, there is a 0.0589 probability of getting a
sample mean of 179.6 just by chance in a random sample of 40 bottles filled by the machine.
Assuming the true mean volume of liquid dispensed by the machine is 180 ml, there is a 0.0589 probability of getting a
sample mean at least as far from 180 as 179.6 (in either direction) just by chance in a…
In an experiment to compare the tensile strengths of I = 6 different types of copper wire, J = 5 samples of each type were used. The between-samples and within-samples estimates of o? were computed as MSTr = 2654.3 and
MSE = 1154.2, respectively. Use the F test at level 0.05 to test Ho: 4, = µ, = ... = lg versus H: at least two u's are unequal.
You can use the Distribution Calculators page in SALT to find critical values and/or p-values to answer parts of this question.
Calculate the test statistic. (Round your answer to two decimal places.)
f =
What can be said about the P-value for the test?
O P-value > 0.100
O 0.050 < P-value < 0.100
O 0.010 < P-value < 0.050
O 0.001 < p-value < 0.010
O P-value < 0.001
State the conclusion in the problem context.
O Reject H. The data indicates a difference in the mean tensile strengths.
O Fail to reject Ho: The data indicates a difference in the mean tensile strengths.
O Fail to reject Ho: The data indicates there is not a difference in the mean…
An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in x = 18.18 kgf/cm2 for the
modified mortar (m = 42) and y = 16.86 kgf/cm for the unmodified mortar (n = 30). Let µ1 and Hz be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions
are both normal.
(a) Assuming that o1 = 1.6 and o2 = 1.3, test Ho: µ1 - 42 = 0 versus H3: µ1 – 42 > 0 at level 0.01.
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)
z =
P-value =
State the conclusion in the problem context.
Fail to reject Ho: The data does not suggest that the difference in average tension bond strengths exceeds from 0.
o Reject Ho: The data does not suggest that the difference in average tension bond…
Chapter 10 Solutions
Student Solutions Manual for Devore's Probability and Statistics for Engineering and the Sciences, 9th
Ch. 10.1 - In an experiment to compare the tensile strengths...Ch. 10.1 - Suppose that the compression-strength observations...Ch. 10.1 - The lumen output was determined for each of I = 3...Ch. 10.1 - It is common practice in many countries to destroy...Ch. 10.1 - Consider the following summary data on the modulus...Ch. 10.1 - The article Origin of Precambrian Iron Formations...Ch. 10.1 - An experiment was carried out to compare...Ch. 10.1 - A study of the properties of metal plate-connected...Ch. 10.1 - Six samples of each of four types of cereal grain...Ch. 10.1 - In single-factor ANOVA with I treatments and J...
Ch. 10.2 - An experiment to compare the spreading rates of...Ch. 10.2 - In Exercise 11, suppose x3. = 427.5. Now which...Ch. 10.2 - Prob. 13ECh. 10.2 - Use Tukeys procedure on the data in Example 10.3...Ch. 10.2 - Exercise 10.7 described an experiment in which 26...Ch. 10.2 - Reconsider the axial stiffness data given in...Ch. 10.2 - Prob. 17ECh. 10.2 - Consider the accompanying data on plant growth...Ch. 10.2 - Prob. 19ECh. 10.2 - Refer to Exercise 19 and suppose x1 = 10, x2 = 15,...Ch. 10.2 - The article The Effect of Enzyme Inducing Agents...Ch. 10.3 - The following data refers to yield of tomatoes...Ch. 10.3 - Apply the modified Tukeys method to the data in...Ch. 10.3 - The accompanying summary data on skeletal-muscle...Ch. 10.3 - Lipids provide much of the dietary energy in the...Ch. 10.3 - Samples of six different brands of diet/imitation...Ch. 10.3 - Although tea is the worlds most widely consumed...Ch. 10.3 - For a single-factor ANOVA with sample sizes Ji(i =...Ch. 10.3 - When sample sizes are equal (Ji = J). the...Ch. 10.3 - Reconsider Example 10.8 involving an investigation...Ch. 10.3 - When sample sizes are not equal, the non...Ch. 10.3 - In an experiment to compare the quality of four...Ch. 10.3 - Prob. 33ECh. 10.3 - Simplify E(MSTr) for the random effects model when...Ch. 10 - An experiment was carried out to compare flow...Ch. 10 - Cortisol is a hormone that plays an important role...Ch. 10 - Numerous factors contribute to the smooth running...Ch. 10 - An article in the British scientific journal...Ch. 10 - Prob. 39SECh. 10 - Prob. 40SECh. 10 - Prob. 41SECh. 10 - The critical flicker frequency (cff) is the...Ch. 10 - Prob. 43SECh. 10 - Four types of mortarsordinary cement mortar (OCM)....Ch. 10 - Prob. 45SECh. 10 - Prob. 46SE
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