Answer to Question #215815 in Statistics and Probability for Dylan

Question #215815

The mean yield of a chemical process is being research by an engineer. From

previous experience with this process the standard deviation of yield is known

to be 3. He would like to be 99% confident that the estimate point should be

accurate within yield point with the value of one.

(i) Examine the error and then analyze how large a sample is necessary for

this research?

(ii) Suppose that an engineer reduced the sample size to 20. If it was found

that the sample mean is 10 and a standard deviation of 1.6, find a 99%

confidence interval for the mean yeild.


1
Expert's answer
2021-07-12T18:27:57-0400

(i) The critical value for α=0.01\alpha=0.01 is zc=z1α/2=2.5758.z_c=z_{1-\alpha/2}=2.5758.


SE=zc×σn=1SE=z_c\times\dfrac{\sigma}{\sqrt{n}}=1

n(zc×σ1)2n\geq(\dfrac{z_c\times\sigma}{1})^2

n(2.5758(3))2n\geq(2.5758(3))^2

n60n\geq60

(ii) The critical value for α=0.01\alpha=0.01 is zc=z1α/2=2.5758.z_c=z_{1-\alpha/2}=2.5758.

The corresponding confidence interval is computed as shown below:


CI=(xˉzc×σn,xˉ+zc×σn)CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})

=(202.5758×1.610,20+2.5758×1.610)=(20-2.5758\times\dfrac{1.6}{\sqrt{10}}, 20+2.5758\times\dfrac{1.6}{\sqrt{10}})

=(18.697,21.303)=(18.697, 21.303)

Therefore, based on the data provided, the 99% confidence interval for the population mean is 18.697<μ<21.303,18.697<\mu< 21.303, which indicates that we are 99% confident that the true population mean μ\mu is contained by the interval (18.697,21.303).(18.697, 21.303).



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