Find a 99% upper confidence bound on the difference in mean yields µ1-µ2.
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A: Givensample size(n)=13Mean(x)=60897.9 standard deviation(s)=2734.8
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A: It is given that For catalyst 1: Sample mean, μ1 = 86 Sample standard deviation, s1 = 3 Sample…
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A: The p-value is 0.0174.
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Q: Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst…
A: given datax1 = 86n1 = 12s1 = 3n2 = 15x2 = 89s2 = 2
Q: concentration
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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 86 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
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- It was reported that last year the average price of gallons of gasoline in a city X was $3.15. This year a sample of 50 gas stations had an average price of $3.10 for a gallon. We assume that the population standard deviation of prices is $0.15. We are interested in determining whether this year's mean price is less than last year. Perform a hypothesis test at the level of significance α=0.05.A random sample of 76 eighth grade students' scores on a national mathematics assessment test has a mean score of 267. This test result prompts a state school administrator to declare that the mean score for the state's eighth graders on this exam is more than 260. Assume that the population standard deviation is 33. At α=0.11, is there enough evidence to support theadministrator's claim? Complete parts (a) through (e).Currently patrons at the library speak at an average of 61 decibels. Will this average decline 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 58 and their standard deviation was 10. What can be concluded at the the αα = 0.10 level of significance? For this study, we should use t-test for population mean The null and alternative hypotheses would be: H0:H0: Incorrect >, <, = or not equal to also provide the __?__ H1:H1: Incorrect >,<, = or not equal to also provide the __?__ The test statistic tc =(please show your answer to 3 decimal places.) The p-value = __?__ (Please show your answer to 4 decimal places.) e.Thus, the final conclusion is that .. a. The data suggest the populaton mean has significantly decreased from 61 at αα = 0.10, so there is statistically significant evidence…
- A manufacturer claims that the calling range (in feet) of its 900-MHz cordless telephone is greater than that of its leading competitor. A sample of 15 phones from the manufacturer had a mean range of 1280 feet with a standard deviation of 21 feet. A sample of 10 similar phones from its competitor had a mean range of 1230 feet with a standard deviation of 40 feet. Do the results support the manufacturer's claim? Let μ₁ be the true mean range of the manufacturer's cordless telephone and μ₂ be the true mean range of the competitor's cordless telephone. Use a significance level of a = 0.01 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places.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 < iAn 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)
- The average number of miles a person drives per day is 24. A researcher wishes to see if people over age 60 drive less than 24 miles per day. She selects a random sample of 40 drivers over the age of 60 and finds that the mean number of miles driven is 22.7. The population standard deviation is 2.9 miles. At =α0.01, is there sufficient evidence that those drivers over 60 years old drive less than 24 miles per day on average? Assume that the variable is normally distributed. Use the critical value method with tables. State the hypotheses and identify the claim with the correct hypothesis.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…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 recent study showed that the average number of sticks of gum a person chews in a week is 11. A college student believes that the guys in his dormitory chew a different amount of gum in a week. He conducts a study and samples 19 of the guys in his dorm and finds that on average they chew 13 sticks of gum in a week with a standard deviation of 1.6. Test the college student's claim at αα=0.01.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…
