New bumper (12 repair incidents): 9, 2, 5, 12, 5, 4, 7, 5, 11, 3, 7, 1
comprise group (9 repair incidents): 7, 5, 7, 4, 18, 4, 8, 14, 13
Solution
a) Dot plot of the data is devoted below
[pic]
[pic]
b) The null and alternative hypothesis are
[pic] and [pic].
c) Assuming equal variance, at ? = 0.05, from Students t distribution with [pic] = 12 + 9 2 = 19 degrees of freedom, the critical value of the test is obtained as 1.729.
Thus, the decision rule is
Reject H0 if [pic] < 1.729.
The sketch of the do away withion region is devoted below.

[pic]
d) The test statistic is given by,
[pic],
where [pic].
From the given data,
[pic] = 5.92, [pic]= 8.89, [pic] = 3.42, [pic] = 4.96.
Therefore,
[pic]= 17.148
Thus,
[pic] = 1.63.
e) Since t = 1.63 > 1.729, we disregard to reject the null hypothesis.
f) The p-value of the test is given by,
P[ t < -1.63] = 0.06
The decision rule using p-value is reject the null hypothesis Ho if p-value < 0.05.
Since the p-value > 0.05, we fail to reject the null hypothesis. This is a close decision.
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