In a test of the reliability of products produced by two mac

In a test of the reliability of products produced by two machines, machine 1 produced 10 defective parts in a run of 200, while machine 2 produced 15 defective parts in a run of 260. Test the claim that proportion of defectives produced by machine 1 produces is less than that of machine 2. Use 5% level of significance. Find a 95% upper confidence bound for the difference in proportion of defectives between machine 1 and machine 2. Use the corresponding lower confidence interval to test the hypothesis found. Please show how answers were found.

Solution

Q1.
Null, Ho: p1 = p2
Alternate,proportion of defectives produced by machine 1
produces is less than that of machine 2 H1: p1 < p2
Test Statistic
Sample 1 : X1 =10, n1 =200, P1= X1/n1=0.05
Sample 2 : X2 =15, n2 =260, P2= X2/n2=0.058
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.054
Q^ Value For Proportion= 1-P^=0.946
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.05-0.058)/Sqrt((0.054*0.946(1/200+1/260))
Zo =-0.361
| Zo | =0.361
Critical Value
The Value of |Z | at LOS 0.05% is 1.645
We got |Zo| =0.361 & | Z | =1.645
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Left Tail - Ha : ( P < -0.3608 ) = 0.35914
Hence Value of P0.05 < 0.35914,Here We Do not Reject Ho

We don\'t have evidence to indicate that proportion of defectives produced by machine 1
produces is less than that of machine 2

Q2.
Find a 95% upper confidence bound for the difference in proportion of defectives between machine 1 and machine 2?
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=10
No.Of Observed (n1)=200
P1= X1/n1=0.05
Proportion 2
No. of chances(X2)=15
No.Of Observed (n2)=260
P2= X2/n2=0.058
C.I = (0.05-0.058) ±Z a/2 * Sqrt( (0.05*0.95/200) + (0.058*0.942/260) )
=(0.05-0.058) ± 1.96* Sqrt(0)
=-0.008-0.041,-0.008+0.041
=[-0.049,0.034]