Question

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

The quality control manager is not only concerned about the openings and closings of the hinges but is also concerned about the proportion of hinges that fail. Can we infer at the 10% significance level that the proportion of hinges made with Alloy 2 that fail exceeds 18%?

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

p = 39.3951 % we Can infer at the 10% significance level that the proportion of hinges made with Alloy 2 that exceeds 18% or not fail.