Question

Please be as clear as possible.

Textbook - Applied Statistics and Probability for Engineers by Montgomery, 6th Edition

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

Final answers are highlighted in colour.

Q1.

a)Calculated using Excel.

Mean($\mu$)= AVERAGE(A1:A20)= 19.1

Standard Deviation($\sigma$)= STDEV.P(A1:A20)= 6.402

Coefficient of variation= $\frac{\sigma}{\mu}$ = 6.402/19.1= 0.3352= 33.52%

Q2.

b) Mean()= AVERAGE(C1:C9)= 17

X

Standard Deviation(S)= 5.809

Coefficient of variation= $\frac{\sigma}{\mu}$ = 5.809/17= 0.3417= 34.17%

c) Standard Error on the mean(SE)= =5.809/sqrt(9)= 5.809/3= 1.9365

S Vn

d)90% Confidence interval

= $\overline{X}\pm t^*SE$

t* for 90% confidence interval= 1.859547

CI= $\17\pm 1.859547 * 1.9365$ = (13.399, 20.601)

Q3.

e)

Expert's Hypothesis

H₀: $\mu = 20$

Ha: $\mu \neq 20$

Sample mean()=17, Standard Error(SE)=1.9365

X

t stat= $\frac{\overline{X}-\mu}{SE}$ = $\frac{17-20}{1.9365}$ = -1.549

p value for t= -1.549, degrees of freedom= 8 is 0.159974

As p value(0.159974) > level of significance(0.05), we cant reject the null hypothesis.

There is no sufficient evidence to say that the mean value is different than 20.

Q4.

f)

For the new sample,

Mean()= 21.11

X

Standard Deviation(S)= 7.59

Coefficient of variation= $\frac{\sigma}{\mu}$ = 7.59/21.11= 0.3595= 35.95%

Q5.

a)Length of the confidence interval should not exceed 2.

$t^* \frac{S}{\sqrt{n}} \leq 2$

If t* is assumed to be a constant.

$\frac{1.859547*5.809}{\sqrt{n}} \leq 2$

Solving for n gives, n>29

20 more tests should be done

But,

As degrees of freedom also changes with sample size(n), t* also varies. So this needs to be performed on Solver to get the value of n.

Solving the equation in solver gives n>= 25