20= -3.0320= -0.4286%0 = 3.03problem above isOto = 3.03 1. In the past, a chemical company produced 880 pounds of a certain type ofplastic per day. Now, using a newly developed and less expensive process, the meandaily yield of plastic for the first 50 days of production is 871 pounds; the standarddeviation is 21 pounds.Do the data provide sufficient evidence to indicate that the mean daily yield forthe new process is less than that of the old procedure? (Use ax=0.05)(b) Test statistic for the problem above is20 = -3.0320 = -0.428620 = 3.03to = 3.03OE(2#9KOL?+Rain...L

Answered: Image /qna-images/answer/4a259185-f5e8-4a3b-9192-a12d603b93a6.jpg

Answered: plastic per day. Now, using a newly…

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 generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 42 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.2 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) 1) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.)
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From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 41 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.2 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level? Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) Part 1) Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value Part 2) Alpha (Enter an exact number as an integer, fraction, or decimal.)α = Part 3) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper…
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