Question

the codes should be done in matlab and please answer all questions.

Part B (Based off week 11 workshop content) An engineering ccmpony bas cedered steel to be used Soer a rail eustraction project. Before ing the steel they want to test it to eusure the materia properties from the supplier are correct A random sample of the ordered steel is tested for its Yield Strength The results are shown below 1. Caleulate the sample mean and sample vwrice by hand, then verify your answers with MATLAB'sean and var functions 2. The supplier claims that the average Yield Strength of their steel is 200 MPa. Conduct a hypothesis test to determine whether these b enough evidence to rject this claim using a siguificance lewd ofa 0,06 3. Determine a 96% coulidace interval kr tbe aver Yild Strength Does the conhdence interval sapport the result of your Inotis Use the tteat fusctioe in MATLAB to verily ywar aws to 2 d Q3 If we were to double the sample siae and obtain the same sample xnd sample variance, what ข๐uld happen to the 95% confere i trnal for Yiekl Strength? Explain why this happens

Answer 1

1. To find the sample mean, simply take the average of all the data given, this means that you should add them all up and divide by 8. So the mean is given by:

m = (197 + 203 + 194 + 196 + 196 + 197 + 198 + 206)/8 = 198.375

To find the variance simply take the differences of each data point from the mean, square them and divide by the number of data points we have. So the variance is given by:

s = [(197 - 198.375)² + (203 - 198.375)² + (194 - 198.375)² + (196 - 198.375)² + (196 - 198.375)² + (197 - 198.375)² + (198 - 198.375)² + (206 - 198.375)²]/8 = (1.8906 + 21.3906 + 19.1406 + 5.6406 + 5.6406 + 1.8906 + 0.1406 + 58.1406)/8 = 14.2343

To do this in matlab simply create a list as follows:

>> sample = [197, 203, 194, 196, 196, 197, 198, 206]

And then use mean(sample) and var(sample) respectively

2. The test goes as follows:

Null hypothesis:
μ 200
vs Alternative hypothesis:
μメ200

We need to perform a t test to prove or disprove the Null hypothesis. So we must find the standard error given by:

SE = s/sqrt(n) = 14.2343/sqrt(8) = 5.0325

The t-statistic is then given by

t_c = (m - μ) / SE = (198.375 - 200)/5.0325 = -0.3229​​​​​​​

The degrees of freedom of this statistic are given by n-1 = 7 using a t table we see that the t-statistic for two tailed hypotheses with 95% confidence and 7 degrees of freedom is 2.365. This means that our calculated t_cshould be greater than 2.365 or less than -2.365 to reject the null hypothesis, since this is not the case we can conclude that with 95% confidence the mean is 200.

3. The confidence interval is simply given by the formula:

772土2 0,025 S E

Where z_0.025means the value on the z table where we find P(z > Z) = 0.025. So the confidence interval is given by

198.375 ± 1.96(5.0325)

Clearly 200 is included in this interval so it supports the results of the previous question.

4. To do this question, using the list we defined earlier, run the command:

>> ttest(sample, 200)

This tests whether the mean of the sample data is 200. It returns 0 if we keep the null hypothesis (mean is 200) and 1 if we reject the null hypothesis (mean is not 200).

5. Recall that the confidence interval is given by . And SE depends on the number of observations. If we were to double the sample size and keep the same sample mean and sample variance we would be left with

772土2 0,025 S E

SE = 14.2343/sqrt(16) = 3.5585

And a 95% confidence interval of

198.375 ± 1.96(3.5585)

Meaning that the interval got smaller.