Engineering Statistics Essay

Submitted By stelios1989
Words: 1697
Pages: 7

Engineering Statistics

Answer 1:

(a) Engineers wish to test for any changes in the compressive strength of concrete with new additives.

(i) The recommended test statistic for this experiment would be a two-tailed t-distribution hypothesis test.
In the short description of the experiment the population variance is unknown and has to be estimated from the sample variance. In addition, the sample is 27 concrete test blocks which makes it a small sample (n = 27 which is < 30) proving that the normal distribution would be unsuitable. Furthermore, the variables are continuous hence showing that the binomial distribution and Poisson should be ignored.
Finally, we are asked to assess if there is a significant difference in the compressive strength of the sample from the population mean therefore a two-tailed t-test is used and not an F-test (tool used to compare two samples).

(ii) Hypothesis to be carried out:

H0: μ=37.1kN
H1: μ≠37.1kN

The question has only provided us with the results of the 27 concrete test blocks (sample) and the mean compressive strength of concrete without the additives (population). Therefore, we are required to estimate the sample variances from data given on our own.

μ = 37.1kN X = 42.7kN

n = 27 concrete test blocks

α = 0.5% degrees of freedom = v = n – 1 = 26

X=∑Xin=1153.3627=42.7kN

VarX=∑Xi-X2n-1=37.66

SDX=VarX=39.11=6.14

t=X-μvarXn=42.7-37.137.6627=4.74

From the use of t-tables v = 26, α = 0.5%: t = 2.7787 < 4.74.
Our level of significance at 0.5% is much smaller than the calculated t-value, so it can be concluded that the null hypothesis, H0, is rejected. There is sufficient evidence indicating that there is no significant difference from the underlying mean concrete strength.

(iii) The recommended test statistic for this case would be a right-tailed hypothesis t-test.
In this case the sample is made up of 10 concrete test blocks which makes it a small sample (n < 20) and the Central Limit Theorem cannot be relied on. In addition, the variables are continuous hence allowing us to dismiss other tests such as the binomial distribution and Poisson.
The population variances are also unknown. It cannot be the F-test because we compare the sample to the population mean and not to another sample mean.
Finally, we are asked to assess if there is a significant increase in the underlying mean compressive strength of the sample so we use the right-tailed t-test.

Hypothesis to be carried out:

H0: μ = 37.1kN
H1: μ > 37.1kN

α = 5% degrees of freedom= n – 1 = 9

X=∑Xn=468.0810=46.81

VarX=∑Xi-X2n-1=32.53

SDX=VarX=32.53=5.70

t=X-μsn=5.38

From the use of t-tables v = 9, α = 5%: t = 1.8331 < 5.38.

Our level of significance at 5% is much smaller than the calculated t-value, so it can be concluded that the null hypothesis, H0, is rejected. There is sufficient evidence indicating that there isn’t significant increase in the underlying mean concrete strength.

(b) (i) A two sample t-test may be used to test the hypothesis that both samples come from the same population because of their small sample size (n<20). There are two samples hence comparing one to another would be sensible and the single sample t-test cannot be performed. Finally, both their variances are unknown. An assumption that was made is that the X (1 week) and Y (2 weeks) variables are normally and independently distributed with population means μ1week, μ2weeks.

(ii) Hypothesis to be tested:

H0: μx = μy
H1: μx ≠ μy α = 0.025 X = 38.284 Y = 42.096

X=∑Xn

Assuming equal population variances:

S2=∑X-X2+∑Y-Y2nx+ny-2=22.045

Samples come from the same population, hence, same population means (μx = μy = 0):

t=X-Y-μx-μyvarX-Y=1.82

varX-Y=s2nx+s2ny=22.04510+22.04510=4.41

Degrees of Freedom=nx+ny-2=18

From the use of t-tables we found that tCRIT = 2.1009 > 1.82 therefore we fail to reject the H0. There is not…