Assessment Task: Statistical Intervals1. A random sample of size nl = 25, taken from a normal population with a standarddeviation ol = 5, has a mean 1 = 80. A second random sample of size n2 = 36, takenfrom a different normal population with a standard deviation o2 = 3, has a mean 2 =75. Find a 95% confidence interval for ul - u22. An electrical firm manufactures light bulbs that have a length of life that isapproximately normally distributed with a standard deviation of 40 hours. If asample of 30 bulbs has an average life of 780 hours, find a 98% confidence intervalfor the population mean of all bulbs produced by this firm.3. A machine produces metal pieces that are cylindrical in shape. A sample of piecesis taken, and the diameters are found to be 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 0.99,1.01, and 1.03 centimeters. The sample mean and standard deviation for the givendata are = 1.0056 and s = 0.0246.(a) Find a 99% confidence interval on the mean diameter.(b) Compute a 99% prediction interval on a measured diameter of a single metalpiece taken from the machine.(c) Find the 99% tolerance limits that will contain 95% of the metal pieces producedby this machine4. A random sample of 20 students yielded a mean of = 72 and a variance of s2 = 16for scores on a college placement test in mathematics. Assuming the scores to benormally distributed, construct a 98% confidence interval for a2.

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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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Find the critical value tcfor the confidence level c=0.80 and sample size n=28.
What critical value of t* should be used for a 95% confidence interval for the population mean based on a random sample of 21 observations? Find the t-table here. t* = 1.721 t* = 1.725 t* = 2.080 t* = 2.086
A researcher is testing a hypothesis of a single mean. The critical t value for a = .05 and a one-tailed test is 2.0639. The observed t value from sample data is 1.742. The decision made by the researcher based on this information is to the null hypothesis. reject not reject redefine change the alternate hypothesis into restate
If a Type I error is made, what conclusion will be drawn regarding the mean breaking strength?
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.
Currently patrons at the library speak at an average of 61 decibels. Will this average increase 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 63.8 and their standard deviation was 9. What can be concluded at the the αα = 0.10 level of significance?
Find the critical value tc for the confidence level c=0.99 and sample size n=19. tc=
A random sample of 23 bags of apples (marked as 10 pounds each) received by a large grocery chain tested out as having a mean of 9.2 pounds with a variance of 2.56 pounds. Test whether the true mean of all bags is under 10 pounds. Set up hypotheses. Perform the appropriate test by showing your formula. Interpret the results using a Type I (alpha) error of .05. Also provide the p value here. Also construct a 95% confidence interval around your sample mean (X bar) that should contain the true mean (mu). Also interpret this interval.
The braking ability was compared for two car models. Random samples of two cars were selected. The first random sample of size 64 cars yield a mean of 36 and a standard deviation of 8. The second sample of size 64 yield a sample mean of 33 and a standard deviation of 8. Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models? Use Alpha= 0.01. Ho: µ1 – µ2 = 0 vs. Ha: µ1 – µ2 + 0 .p-value = 0.0017. Reject Ho Но: Д, — м2 — 0 vs. Ha:M1 — M2 + 0. p-value — 0.034. Ассеpt Ho O Ho:µ1 – Hz Но: И — 2 — 0 vs. Ha:M, — нz + 0. p-value —D 0.0017. Ассept Ho Ho: H1 – U2 = 0 vs. Ha: µ1 – µ2 + 0. p-value = 0.034. Reject Ho
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…

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