Question

The response time (in seconds) was determined for three different types of circuits that could be used in an automatic valve shutdown mechanism. The results are provided in the table below:

Circuit type
1	9	12	10	8	15
2	20	21	23	17	30
3	6	5	8	16	7

Construct a set of orthogonal contrasts, assuming that at the outset of the experiment you suspected the response time of circuit type 2 to be different from the other two. ii. If you wished to minimize the response time, which circuit type would you select? iii. Analyse the residuals from this experiment and comment on whether the assumptions of analysis of variance are satisfied.

Answer 1

i) Now, orthogonal contrasts are the comparison of mean between groups where the sum of coefficients should be zero. It is given that we should assume that circuit type 2's response time is different than 1 and 3. So we can group 1 and 3 together. Providing a coefficient of 1When we check for orthogonal contrasts, our null hypotheses are

H₀=

H₁=

where is the sum of values of each circuit type. Putting values in, we get

54-2*111+42=-126.

We can conclude that Type 2 is different than 1 and 3 since the sum did not come to zero. We can analyze further.

To calculate sum of square contrast, we can use

(-126)²/5*6. This gives us 529.2

If we need to calculate the F value, we will first need mean squared error. Using SAS or excel, we will get it at 16.9. The value then, is= 529.2/16.9=31.3.

ii) We will use either 1 or 3, as they are similar and their response times are lower too.

iii)

We will need to use Levene's test to see if the assumption of equal variances are valid. You can use Proc GLM in SAS for the test. I am putting the output below.

Since the p-value is .6145, this assumption is valid. For the normal variance assumption, one needs to plot the Normal Probability curve in SAS. You can use probplot for the same.If the residuals in the plot are within +/-2, we can assume that there are no outliers and that assumption is also valid.