We skipped a week!

There was nothing noteworthy during class number 11. We simply reviewed exams and homework. I suspect this was very useful for most of you, though. Here is a list of the stuff I have included in this set of notes.

1. Eyob asked that I put the solutions to problems 4.43, 4.45, and 4.46 on this web page. They follow this list immediately.
2. Please go back to the previous set of notes and look at the analysis of variance found there. We will use analysis of variance this coming class period when we begin looking at regression and modeling.
3. On friday I spent a short time showing how to use the Chi Squared statistic in non-parametric tests. Go to the library sometime soon and have a gander at the July-August, 1999, issue of the Journal of Advertising Research in the article entitled Use of Rhetorical Devices in Advertising and note the use of Chi-squared tests.
4. As far as projects are concerned, each of you are expected to give an oral, in-class update on your project(s) this coming friday. Be prepared.

Problem 4.43

Here the problem is to look at the screening of material though a set of sieves to find the mean particle size. The table on page 206, which I will not reproduce here, shows results for the past 25 samples. The sample variance is calculated by subtracting the sample mean from each observation, squaring, adding all the results together, and dividing by 24 (i.e. 25 minus one). The result is s²=13.67. Now the first order of analysis is to test the hypothesis concerning the true variance of the process. The engineers claim that the true process variance is 144. Therefore the nominal hypothesis is...

H0:s2=144

and the alternative is 

Ha:s2!=144

This is a two-tailed test. Our test statistic is built using the Chi-squared deviate as follows.

X2=(n-1)s2/s2

X2=2.278 in our case.

At the 5% level of significance the lower critical value for Chi squared is 12.40 and the upper is 39.36. You can check these from the columns for 2.5% and 97.5%, respectively on the table on pages 259 and 260. Because our observed value of 2.278 is below this acceptance region, we have ample evidence at the 5% level to reject the nominal hypothesis.

The 95% confidence interval for the true process variance is based on our observed value of s² because this is an estimator of the true variance. Refer back to the 2.5% and 97.5% values of X². The 95% confidence interval is [ 24*13.67/39.36, 24*13.67/12.40] = [8.33,26.46].

Finally, in order to perform this analysis we had to assume that the data were independent deviates drawn from a normal distribution. You must graph the data to see how closely they approximate this.

Problem 4.45

This problem relies on your results for problem 4.7 earlier in the class. In problem 4.7 you tested a nominal hypothesis

H0:m=5.6

against the alternative  

Ha:m!=5.6

The test was, therefore, a two tailed test, and we shall calculate a two-tailed p value as a result.

The observed sample mean is 5.8, the target mean is 5.6, and there are three measurement in the sample. Thus our test statistic (Z) must reflect a variability that compares samples of size three to one another. In other words the standard deviation is to be replaced with the sample standard error by dividing the standard deviation by the square root of 3.

Z = Y-m/s/Sqrt(3)

since s=0.7

then Z=0.49

We now wish to find the probability of observing this value of Z or a larger value in absolute magnitude. The probability of a larger value is p = 1-CDF(0.49) straight from the table on page 458. The probability of Z<-0.49 is the same. So our final answer is p= 2(1-CDF(0.49))=0.617

Problem 4.46

This problem relies on your results for problem 4.8 earlier in the class. In problem 4.8 you tested a nominal hypothesis

H0:m=100

against the alternative  

Ha:m > 100

The test was, therefore, a one tailed test, and we shall calculate a one-tailed p value as a result.

The observed sample mean is 101.4, the target mean is 100, and there are four measurements in the sample. Thus our test statistic (Z) must reflect a variability that compares samples of size four to one another. In other words the standard deviation is to be replaced with the sample standard error by dividing the standard deviation by the square root of 4.

Z = Y-m/s/2

since s=8.0

then Z=0.35

We now wish to find the probability of observing this value of Z or a larger value. Magnitude doesn't matter in this one-tailed calculation. The probability of a larger value is p = 1-CDF(0.35) straight from the table on page 458, which is 1-.637 or 363.

Link forward to the next set of class notes for Friday, November 19, 1999