An economist wants to estimate the averageamount in savings account at banks in a givenregion. Arandom sample of 100 accounts gives a mean of357.60 and standard deviation of 1.40.(i) Given a 95% confidence interval for theaverage amount in any savings accounts at abank inthe Given region.(ii) 90% confidence interval for the averageamount in any savings account at a bank in thegiven region.(iii) On the basis of the above results, what wouldyou conclude about the relationship betweenconfidence level and confidence interval?

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Answered: An economist wants to estimate the…

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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From a random sample of 7 students in an introductory finance class that uses group-learning techniques, the mean examination score was found to be 76.64 and the sample standard deviation was 2.7. For an independent random sample of 8 students in another introductory finance class that does not use group-learning techniques, the sample mean and standard deviation of exam scores were 71.65 and 8.4, respectively. Estimate with 99% confidence the difference between the two population mean scores; do not assume equal population variances. A Click the icon to view the Student's t distribution table. The 99% confidence interval is from a lower limit of to an upper limit of (Round to two decimal places as needed.)
A new computer chip design(A) is being compared to the current design(B).Fifty independent chips of each design are tested for speed at performing a certain task. The new computer chip (A) performed the task in 481.2 ms, on average, with a sample standard deviation of 14.3 ms. The current computer chip (B) performed the task in 495.6 ms, on average, with a sample standard deviation of 16.4 ms. Can you conclude, at a 5% significance level, that chip A performs this task faster than chip B?
The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 19 minutes and a standard deviation of 3 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 3% of its customers, how long should it make the guaranteed time limit? Question Help v (a) The percent of customers that receive the service for half-price is %. (Round to two decimal places as needed.) (b) The guaranteed time limit is minutes. (Round up to the nearest integer as needed.) Enter your answer in each of the answer boxes. 1:20 PM P Type here to search AD4 d0E ENG 3/21/2021 DELL PgUp Delete PrtScr Insert F11 F12 F10 F9 F7 F8 F6 F4 F5 F2 F3 A Backspace Esc F1 &…
14.) A researcher would like to estimate the mean weight of all adult female American alligators in the Everglades. The mean weight of a random sample of 146 adult female American alligators from the Everglades was found to be 217 pounds. The standard deviation of the weights of all adult female American alligators in the Everglades is known to be 16 pounds.Using a 96% confidence level, determine the margin of error, EE, and a confidence interval for the mean weight of all adult female American alligators in the Everglades. Report the confidence interval using interval notation. Round solutions to two decimal places, if necessary.The margin of error is given by E= ___A 96% confidence interval is given by ___
Suppose two proposed computer mouse designs were compared by recording wrist extension in degrees for 23 people who each used both mouse types. The difference in wrist extension was calculated by subtracting extension for mouse type B from the wrist extension for mouse type A for each student. The mean difference was reported to be 8.71 degrees. Assume that it is reasonable to regard this sample of 23 people as representative of the population of computer users. (a) Suppose that the standard deviation of the differences was 10 degrees. Is there convincing evidence that the mean wrist extension for mouse type A is greater than for mouse type B? Use a 0.05 significance level. (Use μ = μA - MB.) Find the test statistic. (Round your answer to two decimal places.) t = Find the df. df = USE SALT Use technology to find the P-value. (Round your answer to four decimal places.) P-value = State your conclusion. O Fail to reject Ho. We have convincing evidence that the mean wrist extension for…
The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 19 minutes and a standard deviation of 2.5 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 4% of its customers, how long should it make the guaranteed time limit? Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2).
A random sample of 100 accounts from a branch of certain bank, has mean balance of Rs. 18,600 with S.D.= 1160. A random sample of 80 accounts from another branch of same bank has mean balance of Rs. 15,200; with S.D.= Rs. 920. Find 90% confidence interval of difference between Mean balance of all accounts in both branches.
The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17minutes and a standard deviation of 2 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 5% of its customers, how long should it make the guaranteed time limit?
The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 19 minutes and a standard deviation of 3.5 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 4% of its customers, how long should it make the guaranteed time limit?
The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 2.5 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for half-price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 3% of its customers, how long should it make the guaranteed time limit? %. (a) The percent of customers that receive the service for half-price is (Round to two decimal places as needed.)

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