what would be the value of the test statistic (assuming equal population standard deviations)?
Q: d) Calculate the maximum likelihood estimate for o2 using this data.
A: only D part
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A: p value is less than 0.05Explanation:
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A: Givensample size(n)=13Mean(x)=60897.9 standard deviation(s)=2734.8
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A: given datax1 = 86n1 = 12s1 = 3n2 = 15x2 = 89s2 = 2
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- The thickness of a plastic film (in mils) on a sub- strate material is thought to be influenced by the temperature at which the coating is applied. A completely randomized experiment is carried out. Eleven substrates are coated at 125°F, resulting in a sample mean coating thickness of = 103.5 and a sample standard deviation of s1 = 10.2. Another 13 substrates are coated at 150°F, for which , = 99.7 and s2 = 20.1 are observed. It was originally suspected that raising the process temperature would reduce mean coating thickness. Find a 99% confidence interval on the difference in mean coating thickness Hi - H2. Assume each population is normally distributed but that their variances are not equal. What is the lower bound? Two decimal placesListed in the data table are amounts of strontium-90 (in millibecquerels, 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 0.05 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. Click the icon to view the data table of strontium-90 amounts. a. The test statistic is (Round to two decimal places as needed.) b. The P-value is (Round to three decimal places as needed.)Listed in the data table are amounts of strontium-90 (in millibecquerels, 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 0.05 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. Click the icon to view the data table of strontium-90 amounts. What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 levels and population 2 consists of amounts from city #2. OA. Ho: H₁ H₂ H₁: Hy > H₂ OC. Ho: H₁ H₂ H₁: H₁ H₂ More Info City #1 106 86 121 118 101 20 104 213 130 290 100 277 145 Print City #2 117 63 100 85 87 107 110 111 143 133 101 209 Done ~ X OB. Ho: ₁₂ H₁: Hy > H₂ OD. Ho Hi#H H₁: H₁ H₂
- Listed in the data table are amounts of strontium-90 (in millibecquerels, 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 0.05 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. City_#1 City_#2108 11786 53121 100116 85101 83104 107213 110113 111290 142100 133266 101145 209 What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 levels and population 2 consists of amounts from city #2? The test statistic is? The P-value is? State the conclusion for the test?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)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…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.Assume that an internet provider monitors the amount of time that every customer accesses their home internet connection each day. The amount of time that all of their customers access their home internet connection is normally distributed. They access it for an average of 8.9 hours per day with a standard deviation of 2.1 hours. What proportion of samples of 100 will spend less than an average of 9 hours on their home internet per day? Your answer should have 4 decimal places.
- In a study of microplastics in our environment, it was determined that there are 5.25 trillion plastic pieces floating in the ocean, weighing more than 250,000 tons. In a recent expedition to examine whether increases in plastic production have led to changes in ocean microplastics, 16 samples were taken at various locations in the Pacific Ocean. It was determined that the average concentration was 9.5 particles per liter of ocean water with a sample standard deviation of 6. In one of the earliest research expeditions to study microplastics in the ocean conducted in 2009, it was determined that the average concentration of microplastics was 6 particles per liter (consider this the population or reference mean). What is the critical value for the two-tailed alternative hypothesis at alpha = 5%? (Hint: This is the value in the table that you will compare your test statistic to in order to determine the outcome of the hypothesis test). 2.131 1.746 2.120 1.753A 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…
