Please can you help me with this by only using statistical tables no excel etc .
ID: 3325878 • Letter: P
Question
Please can you help me with this by only using statistical tables no excel etc ... thanks a lot
A certain product has been produced by machine A. Machine B has been procured recently for producing the same product. A random sample of 41 units produced by machine A has a standard deviation of 0.1 mm, while a random sample of 61 units produced by machine B has a standard deviation of 0.07 mm, for a certain critical dimension of the product. Suppose both populations are normally distributed. (i) Is there sufficient evidence to conclude that the units produced by machine B has a smaller population variance than that produced by machine A for the critical (7 marks) dimension? Use a 0.01 level of significance for the test. (ii) Suppose the population standard deviation of the critical dimension for units produced by machine B is 0.082 mm. What is the probability that the standard deviation of that dimension for a random sample of 61 units produced by machine (4 marks) B is less than 0.07 mm?Explanation / Answer
i.
Given that,
sample 1
s1^2=0.01, n1 =41
sample 2
s2^2 =0.0049, n2 =61
null, Ho: ^2 = ^2
alternate, H1: ^2 > ^2
level of significance, = 0.01
from standard normal table,right tailed f /2 =1.936
since our test is right-tailed
we use test statistic fo = s1^1/ s2^2 =0.01/0.0049 = 2.041
| fo | =2.041
critical value
the value of |f | at los 0.01 with d.f f(n1-1,n2-1)=f(40,60) is 1.936
we got |fo| =2.041 & | f | =1.936
make decision
hence value of | fo | > | f | and here we reject Ho
ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 > ^2
test statistic: 2.041
critical value: 1.936
decision: reject Ho
we have enough evidence to support the claim that machine B is small population variance than machine A
ii.
Given that,
population standard deviation ()=0.07
sample standard deviation (s) =0.082
sample size (n) = 61
we calculate,
population variance (^2) =0.0049
sample variance (s^2)=0.006724
null, Ho: =0.07
alternate, H1 : <0.07
level of significance, = 0.01
from standard normal table,left tailed ^2 /2 =88.379
since our test is left-tailed
reject Ho, if ^2 o < -88.379
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(61 - 1 ) * 0.006724 / 0.0049 = 60*0.006724/0.0049 = 82.335
| ^2 cal | =82.335
critical value
the value of |^2 | at los 0.01 with d.f (n-1)=60 is 88.379
we got | ^2| =82.335 & | ^2 | =88.379
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.0295
ANSWERS
---------------
null, Ho: =0.07
alternate, H1 : <0.07
test statistic: 82.335
critical value: -88.379
p-value:0.0295
decision: do not reject Ho
we do not have enough evidence to support the claim that machine B is less than 0.07mm