Consider the following sample of observations on coating thickness for low-visco
ID: 3224737 • 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 mean value of coating thickness. (Round your answer to four decimal places.)
(b) Calculate a point estimate of the median of the coating thickness distribution. (Round your answer to four decimal places.)
(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 four decimal places.)
(d) Estimate P(X < 1.5), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of and , 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.)
Explanation / Answer
Answer:
The table for calculation is given as below:
No.
X
(X - mean)^2
1
0.81
0.287564063
2
0.88
0.217389063
3
0.88
0.217389063
4
1.05
0.087764062
5
1.09
0.065664062
6
1.11
0.055814062
7
1.29
0.003164062
8
1.31
0.001314062
9
1.47
0.015314063
10
1.49
0.020664063
11
1.59
0.059414063
12
1.62
0.074939063
13
1.65
0.092264063
14
1.71
0.132314063
15
1.76
0.171189063
16
1.83
0.234014063
Total
21.54
1.736175
Count
16
Var = 1.736175/15 = 0.115745
SD = sqrt(0.115745) = 0.340213
(a) Calculate a point estimate of the mean value of coating thickness. (Round your answer to four decimal places.)
Answer:
Point estimate for mean = Xbar = X/n = 21.54/16 = 1.3463
(b) Calculate a point estimate of the median of the coating thickness distribution. (Round your answer to four decimal places.)
Answer:
Point estimate for median = sample median = 1.3900
(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 four decimal places.)
Answer:
Estimate for µ = Sample mean = 1.3463
Estimate for = Sample SD = 0.3402
Point estimate of the value separating upper 10% area is X where
X = Mean + Z*SD
Z for upper 10% or lower 90% = 1.281552
X = 1.3463 + 1.281552*0.3402 = 1.7823
Required answer = 1.7823
(d) Estimate P(X < 1.5), i.e., the proportion of all thickness values less than 1.5. [Hint: If you knew the values of and , you could calculate this probability. These values are not available, but they can be estimated.] (Round your answer to four decimal places.)
Answer:
We have to find P(X<1.5)
Z = (X – mean) / SD
Z = (1.5 - 1.3463) / 0.3402
Z = 0.4518
P(X<1.5) = P(Z<0.4518) = 0.674291
Required probability = 0.6743
(e) What is the estimated standard error of the estimator that you used in part (b)? (Round your answer to four decimal places.)
Answer:
Standard error = SD/sqrt(n) = 0.3402/sqrt(16) = 0.3402/4 = 0.08505
Standard error = 0.0851
No.
X
(X - mean)^2
1
0.81
0.287564063
2
0.88
0.217389063
3
0.88
0.217389063
4
1.05
0.087764062
5
1.09
0.065664062
6
1.11
0.055814062
7
1.29
0.003164062
8
1.31
0.001314062
9
1.47
0.015314063
10
1.49
0.020664063
11
1.59
0.059414063
12
1.62
0.074939063
13
1.65
0.092264063
14
1.71
0.132314063
15
1.76
0.171189063
16
1.83
0.234014063
Total
21.54
1.736175
Count
16
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