An investigation into two kinds of photocopying equipment showed that 75 failure

Question

An investigation into two kinds of photocopying equipment showed that 75 failures of the first kind of equipment took on the average 83.2 minutes to repair with a standard deviation of 19.3 minutes, while 75 failures of the second kind of equipment took on the average 90.8 minutes to repair with a standard deviation of 21.4 minutes.

(a)        Test the null hypothesis 1 – 2 = 0 (the hypothesis that on average it takes an equal amount of time to repair either kind of equipment) against the alternative hypothesis 1 – 2 0 at a level of significance = 0.05.

(b)        Using 19.3 and 21.4 as estimates of 1 and 2, find the probability of failing to reject the null hypothesis 1 – 2 = 0 with the criterion of part (a) when actually 1 – 2 = -12.

Explanation / Answer

a.
Given that,
mean(x)=83.2
standard deviation , 1 =19.3
number(n1)=75
y(mean)=90.8
standard deviation, 2 =21.4
number(n2)=75
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=83.2-90.8/sqrt((372.49/75)+(457.96/75))
zo =-2.28
| zo | =2.28
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =2.284 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.28 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: -2.28
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

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