Consider the tollowing sample ot observations on coating thickness tor low-visoo
ID: 3361465 • Letter: C
Question
Consider the tollowing sample ot observations on coating thickness tor low-visoosity paint 0.67 0.88 0.6 04 1.09 2 129 1.31 1.37 1.49 1.59 1.02 1.5. 1.11 1.7b 1.83 Assume that the distribution of coating thickness isal (a normal probability plot stronoly supports this assumption (a) Calculate a point eetimate o the mean value ot coating thickness. Round your answer to tour decimal places. Stata which stim tar you used. (b) Calculate point estimate of the median of the coating thickness distribution. (Round your answer to four dacimal places.) tate which estimator you used and which estimator you might have used instead. (Select all that apply.) e Calculate epuirnt estimete of the value that separetes the leryest 10% of al values in the thickriess distribution from the reiairing 90%. [Hnt: Exrss what you are trying to estirmate in terms of u and Round your ariswwer to four deal places.) State which estimator you used 10th percentila 90th percentile d Estimate P places.) 1.1 Le the pr portion of all thickness alues less than 1.1 Hint It you knew the values o s and you could calculate this probability. These values are not available, but they can be esti ated Round your ans to r ceci al (e What is the estimated standard errer of the estimator that you used in part (b) Round your answer to four decimal places)Explanation / Answer
Solution:- given that information 0.67,0.88,0.68,1.04,1.09,1.28,1.29,1.31,1.37,1.49,1.59,1.62,1.65,1.71,1.78,1.83
a) mean = sum of terms /no of terms
= 21.28/16
= 1.3300
=> option a.
b) median = 1.3400 => (1.31+1.37)/2 = 1.34
=> option a.
c) sd = 0.3728
=> Use the 90 percentile of the sample
= 1.3300 +(1.28*0.3728)
= 1.8072
d) P(X < 1.1) = P(Z < (1.1 - 1.33)/0.3728) = P( Z < -0.6170) = 0.2676
e) The estimated standard error is
SE = 0.3728/sqrt(16) = 0.0932