Question

Consider the following sample of observations on coating thickness for low-viscosity paint. 0.81 0.88 0.88 1.06 1.09 1.13 1.29 1.31 1.33 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.) 1.3394 State which estimator you used. OX Os Os/ ор (b) Calculate a point estimate of the median of the coating thickness distribution. (Round your answer to four decimal places.) 1.3222 x State which estimator you used and which estimator you might have used instead. (Select all that apply.) Os/ ✓ Os O X (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 u and o.] (Round your answer to four decimal places.) State which estimator you used. ох O 10th percentile Os O 90th percentile (d) Estimate Px < 1.3), i.e., the proportion of all thickness values less than 1.3. (Hint: If you knew the values of u and o, you could calculate this probability. These values are not available, but they can be estimated.] (Round your answer to four decimal places.) (e) What is the estimated standard error of the estimator that you used in part (b)? (Round your answer to four decimal places.)

Answer 1

exi from given Sample, n = 16 a) point estimate of mean value of coating thickness is X. ů = x 121.43 - 1-3 394 16 1.3394 Oy Here we estimator. x. to estimate mean value of coating thickness d bad use

mean b) point estimate of median # Normal distribution is symmeta letric -median = mode, inpoint estimate oord of median J.3394 used x to estimate median. it distribution is non-normal we might have used ☆ insted Here we we have to find, P. x-1 < x-che 0.90 Here - 5 11.3394 Erai-se ? n-1 - 1.705 49 15 0.33719 Р X-4 x-1.3394 0.90 0.33719 P[za 2-1.3394 =o.go 0.33 719 from 2-table, f[2 <1,2816] =0.go x-1.339.4 - 1.2816 0.33719 1,2816 X 0.33719 + 103394 x = 1.77154 i. The value of point estimate is 1177154 pe =

estimator used is go percentile. 7 z d). p. CXC103) by central limit, theorem, P[ x_4 1.3-1.3394 p[ze-0.1168 ] batan 0.9522 vodopier 99 p [xc113) = 0,4522 0,33719 04 2 12

e) estimate of standard error is = $s/\sqrt{n}$

= $0.33719/\sqrt{16}$

= 0.084297

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