In 64 randomly selected hours of production the mean and the

In 64 randomly selected hours of production, the mean and the standard deviation of the number of acceptable pieces produced by a automatic stamping machine are x = 1,040 and s = 180. At the 0.02 level of significance, does this, enable us to reject the null hypothesis H0 mu = 1000 against the alternative hypothesis H1 : umu Less then 1000. Calculate and interpret the P-value.

Solution

H0 :m = 1000 ag. H1 : m > 1000...

sample size = 64....
sample mean = 1040..
sample s.d = 180...
sample std. error = 180 / sqrt ( 64) = 22.5

test statistic = ( 1040 - 1000) / 22.5
= 1.777778..

d.f = 64 -1 = 63.....

this follows a t-distributin with df =63..

p-value = 0.04013324

p-value is 0.04 i.e, the probability of rejecting the null hypothesis when it is true is 0.04!

our level of significance = 0.02..
which is less than the p-value...

so, we accept the null hypothesis!
i.e, we can\'t reject the null hypothesis!