An experiment is conducted at a large bank to determine weather a new computer p

Question

An experiment is conducted at a large bank to determine weather a new computer program will speed up the processing of credit card billing. The mean time to process billing using the current program is 12.3 minutes with a standard deviation of 3.5 minutes. The new program optimized the current algorithm and there is reason to believe that the average processing time could not be worse than it is now.The new program is tested with 100 billings and yielded a sample mean of 10.9 minutes. Assume that the standard deviation for the processing time with the new program is the same as the current one.

Does the new program signficantly reduce the time of processing? State the null hypothesis

Explanation / Answer

Given that,
population mean(u)=12.3
standard deviation, =3.5
sample mean, x =10.9
number (n)=100
null, Ho: =12.3
alternate, H1: <12.3
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 10.9-12.3/(3.5/sqrt(100)
zo = -4
| zo | = 4
critical value
the value of |z | at los 5% is 1.645
we got |zo| =4 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -4 ) = 0.00003
hence value of p0.05 > 0.00003, here we reject Ho
ANSWERS
---------------
null, Ho: >12.3
alternate, H1: <12.3
test statistic: -4
critical value: -1.645
decision: reject Ho
p-value: 0.00003

signficantly reduce the time of processing

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