Consider the following sample of observations on coating thickness for low-visco
ID: 3131503 • Letter: C
Question
Consider the following sample of observations on coating thickness for low-viscosity paint.
Assume that the distribution of coating thickness is normal (a normal probability plot strongly supports this assumption).
(a) Calculate a point estimate of the value that separates the largest 10% of all values in the thickness distribution from the remaining 90%. (Give answer accurate to 3 decimal places.)
State which estimator you used.
(b) Estimate P(X < 1.5), i.e., the proportion of all thickness values less than 1.5. (Give answer accurate to 4 decimal places.)
(c) What is the estimated standard error of the estimator that you used in part (b)? (Give answer accurate to 4 decimal places.)
Explanation / Answer
Normal Distribution
Solution:
Here, we have to find the point estimate of the value that separates the largest 10% of all values in the thickness distribution from the remaining 90%.
We are given
Mean = 1.4475
Standard deviation = 0.266045
Now, we have to find the z-value for upper 10% or lower 90% of the distribution.
Z = 1.281552
Now, the required estimate or score is given as below:
X = mean + Z*SD
X = 1.4475 + 1.281552*0.266045
X = 1.78845
Required answer = 1.788
State which estimator you used.
We will use 90th percentile.
Solution:
Here, we have to find P(X<1.5)
Z score is calculated as below:
Z = (X – mean) / SD
Z = (1.5 – 1.4475) / 0.266045
Z = 0.197335
P(X<1.5) = P (Z<0.197335) = 0.578217
Required probability or proportion = 0.5782
Solution:
Here, we have to find the standard error of the estimator.
The formula for the standard error is given as below:
Standard error = Standard deviation / sqrt (n)
Where n is the sample size.
We are given n = 16 and standard deviation = 0.266045
Now, plug these values in the formula and find the standard error for the estimator given as below:
Standard error = 0.266045 / sqrt (16) = 0.266045 / 4 = 0.066511
Required answer = 0.0665