Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yield of 86 and a sample standard deviation of 3. Fifteen batches were prepared usingcatalyst 2, and they resulted in an average yield of 89 with a standard deviation of 2 Assume that yield measurements are approximately normally distributed with the same standard deviationIs there evidence to support the claim that catalyst 2 produces higher mean yield than catalyst 1? Use -0.01.The appropriate decision is to reject the null hypothesisO TrueO False

Answered: Image /qna-images/answer/5bfb2829-62f8-446d-a813-8e712a08d1af.jpg

Answered: Two catalysts may be used in a batch…

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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Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yield of 85 and a sample standard deviation of 3. Fifteen batches were prepared using catalyst 2, and they resulted in an average yield of 89 with a standard deviation of 2. Assume that yield measurements are approximately normally distributed with the same standard deviation. (a) Is there evidence to support the claim that catalyst 2 produces higher mean yield than catalyst 1? Use a = 0.01. (b) Find a 99% confidence interval on the difference in mean yields that can be used to test the claim in part (a). Round your answer to two decimal places (e.g. 98.76). M1 - M2 < i
An engineer working for a tire manufacturer investigates the average life of a new material to be used in the manufacturing process. For this purpose, he builds a sample of 13 tires and tests them in a laboratory until the end of their useful life is reached. The data obtained from the sample are as follows: average of 60897.9 km and standard deviation of 2734.8 km. The engineer would like to demonstrate that the average service life of the tire with the new material exceeds 60,000 km. a) State the null and alternative hypothesis to be stated in this experiment. Test the stated hypotheses using an α = 0.05 What conclusions are reached? b) Find a 95% confidence interval for the average tire life with the new material. Interpret your answer. (RESPOND MANUALLY) (RESPOND MANUALLY)
Listed in the data table are amounts of strontium-90 (in millibrcquerels or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.10 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents. Find the test statistic, p value, and confidence interval
To compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 43 feet. Assume the population standard deviation is 4.6 feet. The mean braking distance for Make B is 46 feet. Assume the population standard deviation is 4.5 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. The critical value(s) is/are Find the standardized test statistic z for μ1−μ2.
A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 65 feet, with a population standard deviation of 13.6. The mean braking distance for SUVS equipped with tires made with compound 2 is 69 feet, with a population standard deviation of 8.5. Suppose that a sample of 55 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ₁ be the true mean braking distance corresponding to compound 1 and μ₂ be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance. Step 3 of 5 Find the p-value associated with the test statistic. Round your answer to four decimal places. Answer How to enter your answer (opens in…
Currently patrons at the library speak at an average of 61 decibels. Will this average increase after the installation of a new computer plug in station? After the plug in station was built, the librarian randomly recorded 57 people speaking at the library. Their average decibel level was 63.8 and their standard deviation was 9. What can be concluded at the the αα = 0.10 level of significance?
A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVS equipped with tires made with compound 1 is 68 feet, with a population standard deviation of 13.9. The mean braking distance for SUVs equipped with tires made with compound 2 is 73 feet, with a population standard deviation of 13.7. Suppose that a sample of 72 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVS equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ, be the true mean braking distance corresponding to compound 1 and μ₂ be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance. Step 1 of 5: State the null and alternative hypotheses for the test.
A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. The mean braking distance for SUVs equipped with tires made with compound 1 is 65 feet, with a population standard deviation of 13.6. The mean braking distance for SUVs equipped with tires made with compound 2 is 69 feet, with a population standard deviation of 8.5. Suppose that a sample of 55 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ₁ be the true mean braking distance corresponding to compound 1 and μ₂ be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance. Step 5 of 5: State the conclusion of the hypothesis test. Answer Tables Keypad Keyboard Shortcuts Previous Step Answers There is sufficient evidence…
Suppose there are two different vaccines for Covid, Vaccine X and Vaccine Y. An interesting question is which vaccine has a higher 6-month antibody effectiveness quotient (6AEQ). To examine this we randomly select 78 recipients of vaccine X and 93 recipients on vaccine Y. The vaccine X recipients had a mean 6AEQ of x = 151. The vaccine Y recipients had a mean 6AEQ of y = 148. It is recognized that the true standard deviation of 6AEQ for vaccine X recipients is 0x = 8.7 while it is recognized that the true standard deviation of 6AEQ for vaccine Y recipients is dy = 11.5. The true (unknown) mean 6AEQ for vaccine X recipients is μx, while the true (unknown) mean 6AEQ for vaccine Y recipients is y. 6AEQ measurements are known to be a normally distributed. In summary: Type Sample Size Sample Mean Standard Deviation Vaccine X 78 Vaccine Y 93 151 148 8.7 11.5 a) Calculate the variance of the random variable X which is the mean of the 6AEQ measurements of the 78 vaccine X recipients.…
Suppose there are two different vaccines for Covid, Vaccine X and Vaccine Y. An interesting question is which vaccine has a higher 6-month antibody effectiveness quotient (6AEQ). To examine this we randomly select 78 recipients of vaccine X and 93 recipients on vaccine Y. The vaccine X recipients had a mean 6AEQ of x = 151. The vaccine Y recipients had a mean 6AEQ of y = 148. It is recognized that the true standard deviation of 6AEQ for vaccine X recipients is o, = 9.7 while it is recognized that the true standard deviation of 6AEQ for vaccine Y recipients is o, = 10.5. The true (unknown) mean SAEQ for vaccine X recipients is ly, while the true (unknown) mean 6AEQ for vaccine Y recipients is 4,. 6AEQ measurements are known to be a normally distributed. summary: Sample Size Sample Mean Standard Deviation 9.7 Гуре Vaccine 78 151 Vaccine Y 93 148 10.5 What is the length of your 96% confidence interval for uy Hy ? ) If we used this data to test Hg: Hy - Hy =0 against the alternative H: uy…

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