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if anyone could please help with this it would be great. (Happy)
The mechanic at a manufacturing plant has made the claim that his machine will make, on average, no more than four defective parts per hour. Over a period of 16 hours, the machine makes an average of 4.6 defective parts per hour, with a standard deviation of .8 parts per hour. Test the mechanic's claim.
thanks again to whomever can help.
It is hoped that you can help. What have you done?
Did you calculate the test statistic?
Did you identify the Critical Region?
What distribution are you supposing for the parameter you are estimating?
*sigh* another one of these...
The mechanic at a manufacturing plant has made the claim that his machine will make, on average, no more than four defective parts per hour. Over a period of 16 hours, the machine makes an average of 4.6 defective parts per hour, with a standard deviation of .8 parts per hour. Test the mechanic's claim.
H_0: u ≤ 4
H_A: u > 4
x = 4
u = 4.6
s = 0.8
n = 16
Personally, now, I would use a student's t-test to find the p-value, since n<30. However, I will do the set-up using the z-test statistic. If you need to change it, you are free to do so. The t-test follows almost the same pattern and guidelines.
z = x-u / s(sqrt n)
z = 4-4.6 / (0.8*sqrt16) = -0.6 / (0.8*4) = -0.6 / 3.2 = -0.1875
The one-tailed p-value is 1 - 0.5743657 = 0.4256343. This is absurdly high and is well above the standard significance-level of 0.05. We fail to reject the null hypothesis.
Good luck! If you can clarify this question any more, please do so.
-Andy