we have been using and use the
we have been using and use the P-VALUE METHOD tomake your decision. For confidence intervals, there are notspecific steps, but there is a specific Excel tool for eachinterval. They should not be done by-hand for this set, nor shouldyou simply use Excel formulas to use it as a calculator. Treat eachquestion as a separate problem — we use the same data set but areanswering different “research questions”.
Many parts of cars are mechanically tested to be certain thatthey do not fail prematurely. In an experiment to determine whichone of two types of metal alloy produces superior door hinges, 40of each type were tested until they failed. Car manufacturersconsider any hinge that does not survive 1 million openings andclosings to be a failure. The number of openings and closingsas observed and recorded in the accompanying table (to the closest0.1 million). A statistician has determined that the number ofopenings and closings is normally distributed.
NOTE: use ONLY the P-value method for hypothesistests. If you include both rules in step 4 or include both in yourdecision step, I will have to conclude that you do not yetunderstand 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 the5% significance level that hinges made with Alloy 1 last longerthan hinges made with Alloy 2?
Answer:
Given Level Of Significance = 5%
So, 
= 0.05
We need to check if hinges made with Alloy 1 last longer thanhinges made with Alloy 2
Let 
1 = Mean for Alloy1
2 = Mean for Alloy2
Hypotheses are:
Ho: 1 < 2
Ha: 1 > 2
For this we need to perform t-test to compare means for datasets
| 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 thanhinges made with Alloy 2
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