9.2 An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.
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- The breaking strengths of cables produced by a certain manufacturer have historically had a mean of 1775 pounds and a standard deviation of 60 pounds. The company believes that, due to an improvement in the manufacturing process, the mean breaking strength, μ, of the cables is now greater than 1775 pounds. To see if this is the case, 90 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1788 pounds. Can we support, at the 0.05level of significance, the claim that the population mean breaking strength of the newly-manufactured cables is greater than 1775 pounds? Assume that the population standard deviation has not changed. Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places, and round your responses as specified below. a. State the null hypothesis H0 and the alternative hyposthesis H1. b. Find the value of the test statistic. c. Find the p-value. d. Can we…For a certain knee surgery, a mean recovery time of 13 weeks is typical. With a new style of physical therapy, a researcher claims that the mean recovery time, μ, is less than 13 weeks. In a random sample of 32 knee surgery patients who practiced this new physical therapy, the mean recovery time is 12.8 weeks. Assume that the population standard deviation of recovery times is known to be 1.1 weeks. Is there enough evidence to support the claim that the mean recovery time of patients who practice the new style of physical therapy is less than 13 weeks? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis Ho and the alternative hypothesis H₁. H Ho: O H₁:0 OO 020 ローロ OO ? (b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test. • The value of the test statistic is given by x-μ √n • The p-value is the area under the curve to the left of the value of the test statistic. Standard Normal Distribution 04 Step 1:…A chain of restaurants has historically had a mean wait time of 5 minutes for its customers. Recently, the restaurant added several very popular dishes back to their menu. Due to this, the manager suspects the wait time, μ, has increased. He takes a random sample of 35 customers. The mean wait time for the sample is 5.1 minutes. Assume the population standard deviation for the wait times is known to be 1.1 minutes. Can the manager conclude that the mean wait time is now greater than 5 minutes? Perform a hypothesis test, using the 0.10 level of significance. (a) State the null hypothesis Ho and the alternative hypothesis H₁. Ho: O H₁:0 μ X 0° OO 20ם ローロ OO × 5 (b) Perform a hypothesis test. The test statistic has a normal distribution (so the test is a "Z-test"). Here is some other information to help you with your test. 0.10 is the value that cuts off an area of 0.10 in the right tail. *-μ ⚫ The test statistic has a normal distribution and the value is given by z=- σ √n Standard Normal…
- The lifetime of a certain brand of battery is known to have a standard deviation of 19.8 hours. Suppose that a random sample of 80 such batteries has a mean lifetime of 35.8 hours. Based on this sample, find a 95% confidence interval for the true mean lifetime of all batteries of this brand. Then give its lower limit and upper limit.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 isA potential buyer wants to decide which of the two brands of electric bulbs he should buy as he has to buy them in bulk. As a specimen, he buys 100 bulbs of each of the two brands-A and B. On using these bulbs, he finds that brand A has a mean life of 1,200 hours with a standard deviation of 50 hours and brand B has a mean life of 1,150 hours with a standard deviation of 40 hours. Do the two brands differ significantly in quality? Use a = 0.05.
- A researcher claims that the average life span of mice can be extended when the calories in their diet are reduced by approximately 40% from the time they are weaned. The restricted diets are enriched to normal levels by vitamins and protein. Suppose that a sample of 10 mice is fed a normal diet and has anaverage life span of 32.1 months with a standard deviation of 3.2 months, while a sample of 15 mice is fed the restricted diet and has an average life span of 37.6 months with a standard deviation of 2.8 months. Assume the distributions of life spans for the regular and restricted diets are approximately normal withequal variance.(a) State the appropriate hypotheses to test the researcher’s claim.(b) Conduct the test at the 0.05 level of significance. What do you conclude?A random sample of 32 showed that the mean shoe size for American men is 10.5 with a standard deviation of 1.12. Assuming normality, find the 95% confidence interval for the mean shoe size for American men. A random sample of 32 showed that the mean shoe size for American men is 10.5 with a standard deviation of 1.12. Assuming normality, find the 95% confidence interval for the mean shoe size for American men. (9.99, 11.01) (10.112, 10.888) (10.096, 10.904) (10.174, 10.826)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…A UCLA researcher claims that the average life span of mice can be extended by as much as 8 months when the calories in their diet are reduced by approximately 40% from the time they are weaned. The restricted diets are enriched to normal levels by vitamins and protein. Suppose that a random sample of 10 mice is fed a normal diet and has an average life span of 32.1 months with a standard deviation of 3.2 months, while a random sample of 15 mice is fed the restricted diet and has an average life span of 37.6 months with a standard deviation of 2.8 months. Test the hypothesis, at the 0.05 level of significance, that the average life span of mice on this restricted diet is increased by 8 months against the alternative that the increase is less than 8 months. Assume the distributions of life spans for the regular and restricted diets are approximately normal with equal variances. Do not use t-test statistic.