A plant in the manufacturing sector is concerned about its sulfur dioxide emissi

Question

A plant in the manufacturing sector is concerned about its sulfur dioxide emissions. The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25. Recently, a “cleaner” technology has been adopted. In such a scenario, the CEO would like to investigate whether there has been a significant change in the emissions and has hired you for advice. In other words, the CEO wants to know if the mean level of emissions is different from 1000. What decision do you arrive at for the 1% level of significance?

a. Reject H0

b. Accept H0

c. Fail to reject H0

d. None of these

What decision do you arrive at for the 7% level of significance?

a. Reject H0

b. Accept H0

c. Fail to reject H0

d. None of these

What decision do you arrive at for the 12% level of significance?

a. Reject H0

b. Accept H0

c. Fail to reject H0

d. None of these

Explanation / Answer

a.
Given that,
population mean(u)=1000
standard deviation, =25
sample mean, x =1000
number (n)=7
null, Ho: =1000
alternate, H1: !=1000
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1000-1000/(25/sqrt(7)
zo = 0
| zo | = 0
critical value
the value of |z | at los 1% is 2.576
we got |zo| =0 & | z | = 2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 0 ) = 1
hence value of p0.01 < 1, here we do not reject Ho
ANSWERS
---------------
null, Ho: =1000
alternate, H1: !=1000
test statistic: 0
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 1
option :c. Fail to reject H0

b.
Given that,
population mean(u)=1000
standard deviation, =25
sample mean, x =1000
number (n)=7
null, Ho: =1000
alternate, H1: !=1000
level of significance, = 0.07
from standard normal table, two tailed z /2 =1.812
since our test is two-tailed
reject Ho, if zo < -1.812 OR if zo > 1.812
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1000-1000/(25/sqrt(7)
zo = 0
| zo | = 0
critical value
the value of |z | at los 7% is 1.812
we got |zo| =0 & | z | = 1.812
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 0 ) = 1
hence value of p0.07 < 1, here we do not reject Ho
ANSWERS
---------------
null, Ho: =1000
alternate, H1: !=1000
test statistic: 0
critical value: -1.812 , 1.812
decision: do not reject Ho
p-value: 1
option :c. Fail to reject H0

c.
Given that,
population mean(u)=1000
standard deviation, =25
sample mean, x =1000
number (n)=7
null, Ho: =1000
alternate, H1: !=1000
level of significance, = 0.12
from standard normal table, two tailed z /2 =1.555
since our test is two-tailed
reject Ho, if zo < -1.555 OR if zo > 1.555
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1000-1000/(25/sqrt(7)
zo = 0
| zo | = 0
critical value
the value of |z | at los 12% is 1.555
we got |zo| =0 & | z | = 1.555
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 0 ) = 1
hence value of p0.12 < 1, here we do not reject Ho
ANSWERS
---------------
null, Ho: =1000
alternate, H1: !=1000
test statistic: 0
critical value: -1.555 , 1.555
decision: do not reject Ho
p-value: 1
option :c. Fail to reject H0

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