give the lower and upper bound The burning rates of two different solid-fuel propellants used in aircrew escape systems are being studied. It is known that bothpropellants have approximately the same standard deviation of burning rate; that is, a₁ = ₂3 cm/s. From a random sample ofsize n₁ = 20 and n₂ = 20, we obtain = 18.02 cm/s and ₂ = 24.31 cm/s.-Construct a 95% confidence interval on the difference in means μ₁ −μ₂.Round your answers to 3 decimal places (e.g. 98.765).

Answered: Image /qna-images/answer/8dceef98-dcc4-4a7e-8b1d-137882eb1fbe.jpg

Answered: The burning rates of two different…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Unfortunately, arsenic occurs naturally in some ground water. A mean arsenic level of u = 8 parts per billion (ppb) is considered safe for agricultural use. A well in Los Banos is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 37 tests gave a sample mean of = 7.3ppb arsenic. It is known that o = 1.9 ppb for this type of data. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use the classical approach. Use a = 0.01 What is the Decision (step 5) for this problem? There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 7.3 ppb. O There is sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. O There is…
Unfortunately, arsenic occurs naturally in some ground water. A mean arsenic level of µ = 8 parts per billion (ppb) is considered safe for agricultural use. A well in Los Banos is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 37 tests gave a sample mean of = 7.3ppb arsenic. It is known that o = 1.9ppb for this type of data. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use the classical approach. Use a = 0.01 What is the calculation (step 4) for this problem? Z = -2.34 z = 2.34 z = 2.24 z = -2.24
A biology student measured the ear lengths of an SRS of 10 Mountain cottontail rabbits, and an SRS of 10 Holland lop rabbits. The ear lengths for the two samples are listed in the two tables attached (see attached image). (a) Calculate a 95% confidence interval for the difference in mean ear lengths between Mountain cottontail rabbits and Holland lop rabbits. Make sure to define which is “µ1” and which is “µ2.” (You can complete the calculations either by hand or using R, but remember to show all your work.) (b) Do Holland lop rabbits have longer ears on average than Mountain cottontail rabbits? Carry out a test of significance to answer this question. Show your work at each step. Don’t forget to state the hypotheses at the start (making sure to define all parameters), and to include a conclusion in terms of the original problem. (You can complete the calculations either by hand or using R, but remember to show all your work.)
The melting points of two alloys used in formulating solder were investigated by melting 12 samples of each material. The sample mean and standard deviation for alloy 1 was x1 = 420°F and sf=6°F, and for alloy 2, they were x2= 426°F and s = 5°F. Do the sample data support the claim that both alloys have melting point with different mean? Use a = 0.05 and assume that both populations are normally distributed and their standard deviations are not equal. %3D at of estion 1. Determine the degrees of freedom (Round your answer to the nearest integer). 024 ONone of the choices are correct 020 022 26 2. What are the appropriate null and alternative hypotheses?
Unfortunately, arsenic occurs naturally in some ground watert. A mean arsenic level of u = 8.0 parts per billion (ppb) is considered safe for agricultural use. A well in Texas is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 36 tests gave a sample mean of x = 7.0 ppb arsenic, with s = 2.3 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. О но: и 3 8 рb; н,: и +8 ppb O Ho: H > 8 ppb; H: µ = 8 ppb O Ho: H = 8 ppb; H: µ > 8 ppb O Ho: H = 8 ppb; H,: µ < 8 ppb O Ho: H < 8 ppb; H,: µ = 8 ppb (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. O The standard normal, since the sample size is large and o is unknown. O The Student's t, since the sample size is large and o is unknown. O The standard normal, since the sample size is large and o is…
The mean score of a random sample of 35 Grade 11 students who took the first periodic test iscalculated to be 86. The population variance is known to be 0.10.
Unfortunately, arsenic occurs naturally in some ground watert. A mean arsenic level of u = 8.0 parts per billion (ppb) is considered safe for agricultural use. A well in Texas is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 41 tests gave a sample mean of x = 6.9 ppb arsenic, with s = 2.8 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use a = 0.05. (a) What is the level of significance? State the null and alternate hypotheses. О Но: и 8 pb О Но: и> 8 рpb; Hi: и %3D 8 рpb (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. O The Student's t, since the sample size is large and o is unknown. The standard normal, since the sample size is large and o is known. The Student's t, since the sample size is large and o is known. O The standard normal, since the sample size is large and o is unknown. What is the value of the sample test…
Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.8 and the sample standard deviation was 1.5. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use a = 0.05.) n USE SALT State the appropriate hypotheses. Ho:H = 48 Ha:µ > 48 = 48 Hạ: u + 48 Ho: H = 48 Hai u 48 H: µ = 48 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = What can you conclude? O The data…
Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.3 and the sample standard deviation was 1.4. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.) State the appropriate hypotheses. ⒸH₂: μ = 48 Ha: > 48 Ho:μ = 48 H₂H 48 H₂: μ = 48 Ho: μ< 48 H₂₁:μ = 48 USE SALT Ho: M = 48 H₂:48 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t= 7.68 X P-value =
Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.4 and the sample standard deviation was 1.5. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.) Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value =
Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.3. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.) USE SALT State the appropriate hypotheses. ⒸHO: μ> 48 H₂: μ = 48 Ho: μ 48 Ho: μ = 48 H₂: μ = 48 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = What can you conclude? O The data provides compelling…
The braking ability was compared for two car models. Random samples of two cars were selected. The first random sample of size 64 cars yield a mean of 36 and a standard deviation of 8. The second sample of size 64 yield a sample mean of 33 and a standard deviation of 8. Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models? Use Alpha= 0.01. Ho: µ1 – µ2 = 0 vs. Ha: µ1 – µ2 + 0 .p-value = 0.0017. Reject Ho Но: Д, — м2 — 0 vs. Ha:M1 — M2 + 0. p-value — 0.034. Ассеpt Ho O Ho:µ1 – Hz Но: И — 2 — 0 vs. Ha:M, — нz + 0. p-value —D 0.0017. Ассept Ho Ho: H1 – U2 = 0 vs. Ha: µ1 – µ2 + 0. p-value = 0.034. Reject Ho

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