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 forSUVs 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 foreach 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 whencompound 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.

In hypothesis testing, we start by stating the null hypothesis (H0) and the alternative hypothesis…

Answered: A researcher compares two compounds (1…

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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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 71 feet, with a population standard deviation of 11.0. The mean braking distance for SUVs equipped with tires made with compound 2 is 74 feet, with a population standard deviation of 12.3. Suppose that a sample of 70 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 2 of 5: Compute the value of the test statistic. Round your answer to two decimal places. Answer How to enter your answer (opens in new window)…
Suppose babies born after a gestation period of 32 to 35 weeks have a mean weight of 2500 grams and a standard deviation of 700 grams while babies born after a gestation period of 40 weeks have a mean weight of 2700 grams and a standard deviation of 340 grams. If a 34-week gestation period baby weighs 2275 grams and a 40-week gestation period baby weighs 2475 grams, find the corresponding z-scores. Which baby weighs less relative to the gestation period?
Based on sample data, newborn males have weights with a mean of3250.3g and a standard deviation of 726.3g. Newborn females have weights with a mean of 3061.3g and a standard deviation of 546.4g. Who has the weight that is more extreme relative to the group from which they came: a male who weighs 1600g or a female who weighs 1600g? Since the z score for the male is z= ?
The average size of a farm in one county is 162 acres with a standard deviation of 38 acres. The average size of a farm in a second county is 180 acres with a standard deviation of 40 acres. Assume the data were obtained from two samples that had 30 farms each. Can it be concluded that the average size of the farms in the two counties is different? (Let α = 0.05)
Suppose babies born after a gestation period of 32 to 35 weeks have a mean weight of 2600 grams and a standard deviation of 600 grams while babies born after a gestation period of 40 weeks have a mean weight of 2800 grams and a standard deviation of 390 grams. If a 32-week gestation period baby weighs 2150 grams and a 40-week gestation period baby weighs 2350 grams, find the corresponding z-scores. Which baby weighs less relative to the gestation period?
A survey found that the average hotel room rate in State I is RM88.22 and the average room in State II is RM80.61. Assume that the data were obtained from two samples of 50 hotels each. The population standard deviations were RM5.62 and RM4.83 respectively. At α = 0.05 , can it be concluded that there is a significant difference in the rates?
An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-temperature performance. A sample of 60 transistors of type A were tested and were found to have a mean lifetime of 1827 hours and a standard deviation of 174 hours. A sample of 180 transistors of type B were tested and were found to have a mean lifetime of 1658 hours and a standard deviation of 231 hours. Let ux represent the population mean for transistors of type A and µy represent the population mean for transistors of type B. Find a 95% confidence interval for the difference uy – µy . Round the answers to three decimal places. The 95% confidence interval is
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…
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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…
The circumference of a particular species of tree was thought to be different depending on certain geographical factors. An SRS of 60 trees is selected in Sequoyah National Forest, where it is found that = 6 meters with a standard deviation = 3 meters. A second SRS of size 40 is independently selected in Yellowstone National Forest, and it is found that = 4 meters with a standard deviation = 2 meters. Let μ1 and μ2 represent, respectively, the true mean circumferences of this particular species of tree. Assume the standard deviations of the two populations are equal. A 95% confidence interval for μ1 – μ2 is
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.…

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