**I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jippo@campus.ie and I will add you to the mailing list.**

## This Week

We developed our applications of step and impulse functions to beam equations. Relevant notes p.22 – 50

In the tutorial we worked on the sample test.

## Next Week

We have a tutorial on Thursday. Please take this opportunity to nail down the differential equations. On the final exam there will be both a full question on beam equations and a full question on second order linear differential equations. Hence if you are comfortable with the material that is examinable in the test, feel free to move onto some of the exercises on beam equations.

## Test Date

**Tuesday 13 March at 6.45 p.m.**

## Timetable Changes

We are now going to schedule ourselves as follows:

**Week 6:** – & Tutorial

**Week 7:** Lecture/Test & –

**Week 8:** Lecture & Lecture

**Week 9:** Tutorial & Lecture

**Week 10:** Tutorial & Lecture

**Week 11:** – & Lecture

**Week 12:** Tutorial & Lecture

## Sample Test Answers

**Here I give you links to how I checked answers quickly using Wolfram Alpha. It takes Mathematica code (they are the same company) and will probably decipher your own stab at code also. Note that Wolfram Alpha gives us a lot more information than we need but that is the beauty of the thing really.**

Question 1 — note there is a small typo here it should be rather than just 16.

Question 2 — it’s not evaluating the constants using the boundary conditions for some reason… the answer is

Question 4 — the boundary conditions yield .

Question 5 — what is here and how do we write ?

Question 6 — again we need to realise that the notation here is different to ours. Applying the boundary conditions we get .

## 2 comments

Comments feed for this article

May 17, 2012 at 6:39 pm

Student 19Thanks for the reply today but I have a second problem with the question 1b part (i) winter 2008, would there be any chance that you could do the answer and further it to me,I know it is a big ask and your have other thinks to be doing but I do not understand the question PLEASE AND THANKS

May 17, 2012 at 7:02 pm

J.P. McCarthyIn the past it has been found that the mean life of components was 150 hours. The process of production was changed and a sample of 50 items yielded a mean value of 155 hours with a standard deviation of 12 hours. At the 5% level of significance, test

that the mean has changed and test that the mean has increased.

Solution:For the first problem we have: the mean is 150

: the mean is not 150

Set up a 95% ‘confidence interval’ for the sample means:

.

In this question , , and . All this yields a ‘confidence interval’:

.

As lies outside this ‘confidence interval’ we reject at the 5% level of significance.

ExplanationThe null hypothesis or the status quo is that the statistics are that the mean is 150. Based on this assumption, we construct an interval into which 95% of the sample means lie. This is a symmetric, two-sided interval so we have 2.5% in the ‘right tail’ (abnormally large sample means) and 2.5% in the ‘left tail’ (abnormally small sample means) (Cf. diagrams in Winter ’11 solutions). We take a sample mean (‘at random’) and if it falls outside this interval we say something is up… rather than taking the abnormal sample mean as a rare event we say that the statistics are in fact FALSE, and we reject the null hypothesis. If the sample mean lies inside this interval, it is not that we necessarily accept the null hypothesis, but we certainly cannot reject it in this framework. The 95% refers both to how rare our sample mean has to be for us to get suspicious as well as the probability of us making a Type I error (e.g. the probability of a sample mean falling outside this interval is 5% when the statistics are TRUE), rejecting the null hypothesis when it was correct.

For the second problem:

: the mean is 150

: the mean is greater than 150

Set up a 95% confidence interval of the form:

.

Here and are as before but here.

This yields a confidence interval:

.

As lies outside this ‘confidence interval’ we reject at the 5% level of significance.

ExplanationMuch the same except that we are only looking for abnormally large sample means that would suggest that the mean is higher.