Question

please be as clear as possible, take note of units and significant figures. thanks for the help

An engineer performed N= 20 tests to assess the load capacity of a new anchoring device. The measured load capacities are as follows (in kN): 11.0 20.0 24.0 13.0 16.0 24.0 18.0 7.0 20.0 22.0 17.0 18.0 28.0 27.0 20.0 15.0 12.0 11.0 31.0 28.0 After having assessed that the load capacity is normally distributed, the engineer wants to determine some statistics of the population. Q. 1 a) Determine mean, standard deviation and coefficient of variation of the population Then, the engineer randomly picks a sample of n = 9 tests out of the N = 20 (you will take the first 9 of those listed above). Q.2 b) Determine point estimates of mean, standard deviation and coefficient of variation c) The estimated standard error on the mean d) A 90% confidence interval on the sample mean, even though the sample is "small" The engineer, after having computed such values, asks for a second opinion to an "expert”, who provides a “subjective” estimate that the mean could be equal to 20 kN, with a coefficient of variation of 10%. Then the engineer decides to test, with a significance level a = 0.05 on a type I error, if the expert's hypothesis about the mean should be rejected or not: Ho: u = 20 kN, Hı: # 20 KN Q.3 e) Should the engineer reject the expert's hypothesis? If the engineer decides to consider the expert's opinion, then the sample statistics of Q.2 can be updated, from prior to posterior, considering both the sample statistics and the expert's opinion. Q.4 1) Find the updated point estimate of the mean, along with its posterior standard deviation and coefficient of variation. If the engineer decides not to consider the expert's opinion and keep the sample statistics of Q.2, then a possibility could be to work on improving the confidence interval in Q.2d. Therefore: Q. 5 a) How many more tests, in addition to the n = 9, should the engineer pick to make sure that the length of the confidence interval on the mean does not exceed 2 kN?

Answer 1

N = 20

Let x be the load capacities

x	$(x-\bar{x})^2$ $=(x-19.1)^2$
11.0	65.61
20.0	0.81
24.0	24.01
13.0	37.21
16.0	9.61
24.0	24.01
18.0	1.21
7.0	146.41
20.0	0.81
22.0	8.41
17.0	4.41
18.0	1.21
28.0	79.21
27.0	62.41
20.0	0.81
15.0	16.81
12.0	50.41
11.0	65.61
31.0	141.61
28.0	79.21
	$\sum (x-\bar{x})^2=819.8$

Q.1.

a)

Mean of the population:

$\mu=\frac{\sum x}{N}=\frac{382.0}{20}=19.1$

Mean of the population is 19.1

Standard deviation of the population:

σ Σ(1 – 1)2 N

$\sigma=\sqrt{\frac{819.8}{20}}$

$\sigma=\sqrt{40.99}$

$\sigma=6.402$ (Round to 3 decimal)

Standard deviation of the population is 6.402

Coefficient of variation (CV) :

CV = - * 100

$CV=\frac{6.402}{19.1}*100$

CV = 0,3352 * 100

CV = 33.52%

Coefficient of variation is 33.52%

Q.2

b)
n = sample size = 9

Consider first 9 values from above 20 values as a sample.

x	$(x-\bar{x})^2$ $=(x-17)^2$
11.0	36
20.0	9
24.0	49
13.0	16
16.0	1
24.0	49
18.0	1
7.0	100
20.0	9
$\sum x=153.0$	$\sum (x-\bar{x})^2=270$

Point estimate of mean:

$\bar{x}=\frac{\sum x}{n}=\frac{153.0}{9}=17$

Point estimate of mean is 17

Point estimate of Standard deviation:

$s=\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}$

$s=\sqrt{\frac{270}{9-1}}$

$s=\sqrt{\frac{270}{8}}$

$s=\sqrt{33.75}$

(Round to 3 decimal)

s= 5,809

Point estimate of Standard deviation is 5.809

Point estimate of Coefficient of variation (CV) :

$CV=\frac{s}{\bar{x}}*100$

$CV=\frac{5.809}{17}*100$

$CV=0.3417*100$

CV = 34.17%

Point estimate of Coefficient of variation is 34.17%

c)
Standard error of the mean:

$SE=\frac{s}{\sqrt{n}}$

$SE=\frac{5.809}{\sqrt{9}}$

$SE=\frac{5.809}{3}$

SE = 1.936    (Round to 3 decimal)

Standard error of the mean is 1.936

d)
90% confidence interval :

COnfidence level = c = 0.90

Sample size = n = 9

Sample mean = = 17

T

Population standard deviation = $\sigma$ = 6.402

Here Population standard deviation is known so we use z interval.

90% confidence interval is

$\bar{x}-z_c*\frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+z_c*\frac{\sigma}{\sqrt{n}}$

where zc is z critical value for (1+c)/2 = (1+0.90)/2 = 0.95

zc = 1.645 (From statistical table of z values, average of 1.64 and 1.65,(1.64+1.65)/2 = 1.645)

$17-1.645*\frac{6.402}{\sqrt{9}}<\mu<17+1.645*\frac{6.402}{\sqrt{9}}$

$17-1.645*2.134<\mu<17+1.645*2.134$

$17-3.51043<\mu<17+3.51043$

$13.490<\mu<20.510$ (Round to 3 decimal)

90% confidence interval is (13.490, 20.510)