Our research group is developing a chemically treated natural fiber for itstensile strength properties. Forty pieces were tested under similarconditions and the results showed an average tensile strength of 500.05MPa and a standard deviation of 80.55 MPa. What is the lower bound ofthe tensile strength (in MPa) at a one-sided 99% confidence level?O 464.30O 469.60O 466.90O 470.42

Given,sample size(n)=40sample mean(x¯)=500.05standard deviation(s)=80.55α=1-0.99=0.01α2=0.005degrees…

Answered: Our research group is developing a…

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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A group of scientists studied the adhesion of various biofilms to solid surfaces for possible use in environmental technologies. Adhesion assay is conducted by measuring absorbance at A590. If the scientists want the confidence interval to be no wider than 0.55 dyne-cm2 at 90% confidence interval, how many observations should they take? Assume that the standard deviation is known to be 0.66 dyne-cm2.
The sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 10 cans yields a sample standard deviation of s = 4.841 milligrams. Construct a 95% two-sided confidence interval for o. Round your answers to 2 decimal places.
Assume the height of a particular species of horse are normally distributed. A random sample of 7 horses has a mean height of 46.14 cm and a standard deviation of 1.19 cm. At a 0.05 level of significance, a test is undertaken to determine whether there is sufficient evidence to reject the claim of a mean height is higher than 45 cm. What statement can be made about the p-value?
Is a weight loss program based on exercise more effective than a program based on diet? The 44 overwelght people put on a strict one year exercise program lost an average of 30 pounds with a standard devilation of 10 pounds. The 50 overweight people put on a strict one year diet lost an average of 25 pounds with a standard deviation of 9 pounds. What can be concluded at the a = 0.01 level of signtficance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: Select an answerv Select an answerv Select an answerv (please enter a decimal) H: Select an answerv Select an answerv Select an answerv (Please enter a decimal) C. The test statistic ?v = (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ?v a f. Based on this, we should Select an answerv the null hypothesis. g. Thus, the final conclusion is that... O The results are statistically insignificant at a = 0.01,…
The number of chocolate chips in an 18oz bay of Chips Ahoy! chocolate chip cookies si approximately normally distributed with a mean of 1262 chips and standard deviation 118 chips according to a 1999 study. Assume the cookies have not changed since then. 1.) What proportion of 18oz bags of Chips Ahoy! contains more than 1200 chocolate chips? 2.)How many chips would be in a bag of Chips Ahoy! that is at the 30th percentile? 3.) What proportion of 18oz bags of Chips Ahoy! contains fewer than 1125 chocolate chips? 4.) What is the probability that a randomly selected 18oz bag of Chips Ahoy! contains between 1000 and 1400 chocolate chips, inclusive?
The life time (age) in hours of a random sample of one of the batteries produced in Jordan gave the following summary: of Sample size Sample Average Sample Standard deviation n = 9 X = 95 S = 3 A 98% confidence interval for the population standard deviation O is: a) (1.09 , 17.86) b) (1.09 , 6.61) c) (1.89 , 6.61) d) (3.58 , 43.7) e) (1.04, 4.23)
A health professional obtains records for 33 patients who have undergone a certain surgical procedure and finds the average length of stay is 3.23 days with a standard deviation of 1.58 days. Using a = evidence to suggest the actual average length of stay for this surgical procedure is 3 days? 0.10, is there What are the hypotheses? O Ho: µ = 3.23 vs H1: µ # 3.23 Ho:u = 3 vs H1 :µ#3 Họ : = %3D 3 vs H1: + 3 Ho: = 3.23 vs H1: + 3.23 %3D Calculate the test statistic using the correct letter. = - 0.84 t = 0.84 z = 0.84 = Z - 0.84 Determine the critical values. O+1.691 Q+1.694 ±1.307 +1.692 +1.282 土1.309 O+1.645
A health professional obtains records for 33 patients who have undergone a certain surgical procedure and finds the average length of stay is 5.79 days with a standard deviation of 1.37 days. Using a evidence to suggest the actual average length of stay for this surgical procedure is 5 days? What are the hypotheses? O Hoμ5 vs H₁ : μ #5 = Ho : x = 5 vs H₁ : x # 5 O H : |= 5.79 vs H1 : @#5.79 OHo5.79 vs H₁5.79 Calculate the test statistic using the correct letter. Oz= -3.31 Ot = 3.31 Ot= -3.31 Oz = 3.31 Determine the critical values. +2.576 +2.326 +2.449 ±2.739 ±2.733 ±2.728 +2.441 State your decision. O Accept Ho Fail to reject Ho Reject Ho days. = 0.01, is there The correct conclusion is that we Select an answer conclude that the average length of stay of all surgeries of this type Select an answer Select an answer What level of concern do you have for the validity of your results? O None = - our sample size is larger than 30 so there should be no problem with the conclusion Low - our…

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