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

Do the data provide enough evidence to allow us to infer at the 5% significance level that hinges made with Alloy 1 last longer than hinges made with Alloy 2?

Answer 1

Given Level Of Significance = 5%

So, = 0.05

We need to check if hinges made with Alloy 1 last longer than hinges made with Alloy 2

Let 1 = Mean for Alloy 1

2 = Mean for Alloy 2

Hypotheses are:

Ho: 1 < 2

Ha: 1 > 2

For this we need to perform t-test to compare means for data sets

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

1 = 1.323, 2 = 1.145

sample standard deviation

s1 = 0.2833, s2 = 0.2935

n1 = n2 = 40

So, t-statistic = 2.76

degree of freedom, df = n1 + n2 - 2 = 78

Hence p-value = 0.0036

Since p-value < 0.05, we can reject the Null Hypotheses.

Hence we conclude that hinges made with Alloy 1 last longer than hinges made with Alloy 2