Question

1.       A mechanic is restoring an antique car. She checked the compression of each cylinder twice. She checked all six cylinders in sequence and then replicated the effort. The standard pressure for a cylinder is 100 PSI. The attached table has shows the results of her measurements. She has asked you to help analyze the data.

While she performed three tests she asked you to first analyze the data as one pooled set (that is, without regard to the fact that she ran two tests). Use what you think would be the appropriate test and explain your results.

She then asked you to consider the two different tests and to explain if there is any difference

Pressure (PSI)
Test 1	Test 2	Test 3
65	70	68
67	70	69
68	72	70
70	73	72
71	74	73
73	75	74

Answer 1

1.)

Considering data as one pooled set, we can check whether average pressure from tests is equal to standard pressure of cylinder or it is unequal.

So, Our Hypotheses becomes:

Null: Average pressure of cylinder from test is equal to standard pressure

Alternate: Average pressure of cylinder from test is not equal to standard pressure

We will be checing our hypotheses at 5% significance level which means = 0.05

We will use one sample t test to compare the value of average pressure with standard pressure

Pooled data is:

65
67
68
70
71
73
70
70
72
73
74
75
68
69
70
72
73
74

Formula is:

t = (x - m) / (s / )

Now x = 100

Mean for pooled data is, m = 70.77

Standard deviation, s = 2.71

Number of Observations, n = 18

Putting in above formula we get, t = 45.71

Checking t distribution table, we get probability as, p < 0.00001

Since, p < 0.05, we can reject Null Hypotheses and conclude that Average Pressure is not equal to standard pressure

2.)

Comparing the mean of all 3 Test at once. We will perform one way ANOVA to compare the mean of all 3 Tests at once.

We will calculate F statistic and p value to compare the means

Please find the tables below:

Data:

Test 1	Test 2	Test 3
65	70	68
67	70	69
68	72	70
70	73	72
71	74	73
73	75	74

Summary of Data
	Treatments
	1	2	3	4	5	Total
N	6	6	6			18
∑X	414	434	426			1274
Mean	69	72.3333	71			70.7778
∑X2	28608	31414	30274			90296
Std.Dev.	2.8983	2.0656	2.3664			2.7128

Result Details
Source	SS	df	MS
Between-treatments	33.7778	2	16.8889	F = 2.77372
Within-treatments	91.3333	15	6.0889
Total	125.1111	17

Checking table to calulate p-value we get, p = .094406

We will check at 5% significance level

Since p > 0.05, result is not significant meaning that average of Pressure for 6 cylinders under 3 different tests is same at 5% siginificance level.

If we want to check for only 2 Tests at a time, we can perform Paired t Test to compare that