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For a comparison of the rates of defectives produced by two assembly lines, independent random samples of 100 items were selected from each line. Line A yielded 18 defectives in the sample, and line B yielded 12 defectives. Find a 98% confidence interval for the true difference in proportions of defectives for the two lines. Is there evidence here to suggest that one line produces a higher proportion of defectives than the other? A health care professional wishes to estimate the average birth weight of infants. Assume the birth weight is normally distributed with standard deviation sigma = 8 ounces. How large a sample must be obtained if she desires to be 90% confident that the true mean is within 2 ounces of the sample mean? she desires to be 98% confident that the true mean is within 1 ounces of the sample mean?

Solution

16.
a.CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
Probability Succuses( X1 )=18
No.Of Observed (n1)=100
P1= X1/n1=0.18
Proportion 2
Probability Succuses(X2)=12
No.Of Observed (n2)=100
P2= X2/n2=0.12
C.I = (0.18-0.12) ±Z a/2 * Sqrt( (0.18*0.82/100) + (0.12*0.88/100) )
=(0.18-0.12) ± 2.33* Sqrt(0.003)
=0.06-0.117,0.06+0.117
=[-0.057,0.177]

b.
We have value zero in Lower & Upper value interval. Yes we have evidence