Question

Following is a problem description. For all hypothesis tests, you MUST show all 7 steps we have been using and use the P-VALUE METHOD to make your decision. For confidence intervals, there are not specific steps, but there is a specific Excel tool for each interval. They should not be done by-hand for this set, nor should you simply use Excel formulas to use it as a calculator. Treat each question as a separate problem -- we use the same data set but are answering different “research questions”.

Many parts of cars are mechanically tested to be certain that they do not fail prematurely. In an experiment to determine which one of two types of metal alloy produces superior door hinges, 40 of each type were tested until they failed. Car manufacturers consider any hinge that does not survive 1 million openings and closings to be a failure. The number of openings and closings as observed and recorded in the accompanying table (to the closest 0.1 million). A statistician has determined that the number of openings and closings is normally distributed.

NOTE: use ONLY the P-value method for hypothesis tests. If you include both rules in step 4 or include both in your decision step, I will have to conclude that you do not yet understand the p-value rule.

Number of Openings and Closings

Alloy 1				Alloy 2
1.5	1.5	0.9	1.3	1.4	0.9	1.3	0.8
1.8	1.6	1.3	1.5	1.3	1.3	0.9	1.4
1.6	1.2	1.2	1.8	0.7	1.2	1.1	0.9
1.3	0.9	1.5	1.6	1.2	0.8	1.2	1.1
1.2	1.3	1.4	1.4	0.8	0.7	1.1	1.4
1.1	1.5	1.1	1.5	1.1	1.4	0.8	0.8
1.3	0.8	0.8	1.1	1.3	1.1	1.5	0.9
1.1	1.6	1.6	1.3	1.4	1.2	1.3	1.6
0.9	1.4	1.7	0.9	0.6	0.9	1.8	1.4
1.1	1.3	1.9	1.3	1.5	0.8	1.6	1.3

Estimate with 90% confidence the variance of the number of openings and closings with the hinges made from Alloy 2. Interpret the interval.

Answer 1

step -1

The F hypothesis test is defined as:

Ho: S₁^2 = S2^2

Ha: S₁^2 ≠ S2^2

step -2

F =  S2^2 /S₁^2

S₁^2 = variance of Alloy 1 = 0.07897

S₁^2 = variance of Alloy 2 = 0.08613

F = 1.090667342

step -3

degree of freedom:

numerator = (40-1) =39

denominator = (40-1) = 39

step -4

confidence level is 90 %

step -5

p value = 0.393951

step- 6

step -6

The result is not significant at p < .10. because p value < 0.10

means there is no significant difference in opening and closing