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 a2 were computed as MSTr=2665.3 and MSE= 1104.2, respectively. Use the F test at level 0.05 to test Ho: H₁ H₂ Hs versus H.: at least two μ's are unequal. parts of this question. You can use the Distribution Calculators page in SALT to find critical values and/or p-values to answer Calculate the test statistic. (Round your answer to two decimal places.) f=
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- 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…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 = 18.11 kgf/cm² for the modified mortar (m = 42) and y = 16.82 kgf/cm² for the unmodified mortar (n = 30). Let μ₁ and μ₂ 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 ₁ = 1.6 and ₂ = 1.3, test Ho: ₁ - ₂ = 0 versus H₂ : ₁ - ₂ > 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. O Reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds 0. O Fail to reject Ho. The data does not suggest that the difference in average tension bond strengths…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…
- 1.56 Longleaf pine trees. The Wade Tract in Thomas County, Georgia, is an old-growth forest of longleaf pine trees (Pinus palustris) that has survived in a relatively undisturbed state since before the settlement of the area by Europeans. A study collected data on 584 of these trees. One of the variables measured was the diameter at breast height (DBH). This is the diameter of the tree at 4.5 feet, and the units are centimeters (cm). Only trees with DBH greater than 1.5 cm were sampled. Here are the diameters of a random sample of 40 of these trees: PINES 27 10.5 13.3 26.0 18.3 52.2 9.2 26.1 17.6 40.5 31.8 47.2 11.4 2.7 69.3 44.4 16.9 35.7 5.4 44.2 2.2 4.3 7.8 38.1 2.2 11.4 51.5 4.9 39.7 32.6 51.8 43.6 2.3 44.6 31.5 40.3 22.3 43.3 37.5 29.1 27.9 (a) Find the five-number summary for these data. (b) Make a boxplot. (c) Make a histogram. (d) Write a short summary of the major features of this distribution. Do you prefer the boxplot or the histogram for these data?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 o² were computed as MSTR = 2676.3 and MSE = 1107.2, respectively. Use the F test at level 0.05 to test Ho: M₁ = ₂ = ... = μ5 versus H₂: at least two μ'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 Ho. The data indicates there is not a difference in the mean tensile strengths. O Fail to reject Ho. The data indicates there is not a difference in the mean tensile strengths. O Fail to reject Ho. The data indicates a difference in…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 = 18.11 kgf/cm² for the modified mortar (m = 42) and y = 16.82 kgf/cm² for the unmodified mortar (n = 32). Let μ₁ and μ₂ 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 0₁ = 1.6 and ₂ = 1.3, test Ho: ₁ - ₂ = 0 versus H₂: M₁-M₂ > 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. O Reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds 0. O Fail to reject Ho. The data suggests that the difference in average tension bond strengths exceeds…
- The strength of concrete depends, to some extent, on the method used for drying. Two different drying methods showed the following results for independently tested specimens (measurements in psi): Do the methods appear to produce concrete with different mean strengths? Use α = 0.05 .Riboflavin (Vitamin B2) is determined in a cereal sample by measuring its fluorescence intensity(형광세기) in 5% acetic acid solution. A calibration curve was prepared by measuring the fluorescence intensities of a series of standards of increasing concentrations. The following data were obtained. Riboflavin (μg/mL) 0.000 0.100 0.200 0.400 0.800 Unknown sample Fluorescence intensity 0.0 5.8 12.2 22.3 43.3 15.4 (a) Use the method of least squares to obtain the best straight line through these five points (n=5). (b) Make a graph showing the experimental data and the calculated straight line. (c) An unknown sample gave an observed fluorescence intensity of 15.4. Calculate the concentration of Riboflavin (Vitamin B2) in the unknown sample (μg/mL). (d) Calculate the coefficient of determination (R2).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 = 18.17 kgf/cm² for the modified mortar (m = 42) and y = 16.82 kgf/cm² for the unmodified mortar (n = 31). Let μ₁ and ₂ 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 0₁ = 1.6 and 0₂ = 1.3, test Ho: M₁ M₂ = 0 versus Ha: M₁ - H₂> 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 = (b) Compute the probability of a type Il error for the test of part (a) when µ₁ - H₂ = 1. (Round your answer to four decimal places.) (c) Suppose the investigator decided to use a level 0.05 test and wished B = 0.10 when M₁ M₂ = 1. If m = 42, what…
- 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 = 16.85 kgf/cm² for the unmodified mortar (n = 32). Let μ₁ and ₂ 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 0₁ = 1.6 and ₂ = 1.3, test Ho: M₁ M₂ = 0 versus Ha: M₁ M₂ > 0 at level 0.01. 1 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 = 4.74 X P-value = State the conclusion in the problem context. Ⓒ Reject Ho. The data suggests that the difference in average tension bond strengths exceeds 0. O Reject Ho. The data does not suggest that the difference in average tension bond strengths exceeds…Suppose that we are testing H0: μ = μ0 versus H1: μ ≠ μ0. Calculate the P-value for the following observed values of the test statistic: (a) Z0 = 1.95 (b) Z0 = -0.10Hypertension is a long-term medical condition in which the blood pressure in the arteries is persistently elevated. In elderly patients with hypertension, a high systolic pressure can cause a fainting episode, so X transporting them to an emergency clinic is imperative. Let be the systolic blood pressure of elderly patients with hypertension on arrival at the emergency clinic after a fainting episode. A random sample of such patients was considered. The measurements of each patient, in mm Hg, are given below. At the 5% level of significance, can we conclude from the data in this sample that the mean systolic blood pressure of all elderly hypertensive patients transported to an emergency clinic after a fainting episode is smaller than 175 mm Hg? patient 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 X 166 178 153 157 178 158 153 169 182 155 151 195 202 169 157 patient 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 X 173 161 155 154 178 200 168 196 180 191 167 170 161 191 166 patient 31 32 33 34 35…
