Consider the following sample of observations on coating thickness for low-visco
ID: 3261517 • Letter: C
Question
Consider the following sample of observations on coating thickness for low-viscosity paint. 0.840.88 0.88 1.05 1.09 1271.29 1.31 1.35 1.49 1.59 1.62 1.65 1.71 1.76 1.83 Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption). a) Calculate a point estimate of the mean value of coating thickness. (Round your answer to four decimal places.) State which estimator you used. O s/x (b) Calculate a point estimate of the median of the coating thickness distribution. (Round your answer to four decimal places.) State which estimator you used and which estimator you might have used instead. (Select all that apply.) (c) Calculate a point estimate of the value that separates the largest 10% of all values in the thickness distribution from the remaining 90%. [Hint: Express what you are trying to estimate in terms of and .] (Round your answer to our decimal places.)Explanation / Answer
a. Treat 16 observations as a random sample of coating thickness for all low viscosity paint. Therefore, to determine the point estimate of the population mean, compute the sample mean coating thickness of the 16 low viscosity paint.
The sample mean, xbar=sigma x/n, where, x denote the coating thickness, and n i snumber of observations.
=(0.84+0.88+...+1.83)/16=1.3510
The point estimate of mu is 1.3510 (ans)
The estimator that has been used is xbar.
b. Follow the instructions to compute median coating thickness.
1. Arrange data in ascending order.
0.84, 0.88, 0.88, 1.05, 1.09, 1.27, 1.29, 1.31, 1.35, 1.49, 1.59, 1.62, 1.65, 1.71, 1.76, 1.83
2. Find whether data set is even or odd. Here, n=16, therefore, data is even.
3. Since, data is even in number, therefore, median is the mean of two middle numbers, that is 8th and 9th number.
Therefore, x~=(1.31+1.35)/2=1.33 (ans)
The distribution of coating thickness is normal, therefore, assume the sample data follows normal distribution. Hence, one could have used xbar.
c. The point estimate of values that separate the largest 10% of all values in thickness ditribution from the remaining 90%, is the 90th percentile. That is in a normal curve, the x-value that puts a coating thickness in the top 10% is the same as the x-value for which 90% area is to the left. Look into the z table for entry closest to 0.90. It is 0.8997. This is the entry for z=1.28. Therefore, z=1.28 is the standardized value with area 0.90 to its' left.
Unstandardize the z value into its' original scale.
Compute standard deviation of coating thickness by substituting values in following formula.
s=sqrt[1/n-1 sigma (x-xbar)^2]
=sqrt[1/16-1 {(0.84-1.351)^2+(0.88-1.351)^2+...+(1.83-1.351)^2}]
=0.331
Now substitutue values in following equation to obtain the value that separate largest 10% of all values in thickness distribution from bottom 90%.
x=xbar+1.28*s
=1.351+1.28*0.331
=1.7747 (ans)
The estimator used is 90th percentile.
d. P(X<1.1)=P[Z<(1.1-1.351)/0.3201] [use, Z=(X-mu)/Sigma, where, mu is population mean, Sigma is population standard deviation, computation formula for Sigma=sqrt{1/n sigma (x-mu)^2}]
=P[Z<-0.78]
=0.2177 (ans)