1) Two different types of polishing solution are being evaluated for possible us

Question

1) Two different types of polishing solution are being evaluated for possible use in a tumble-polish operation for manufacturing intraocular lenses used in the human eye following cataract surgery. Three hundred lenses were tumble-polished using the first polishing solution, and of this number 250 had no polishing-induced defects. Another 300 lenses were tumble-polished using the second polishing solution, and 200 lenses were satisfactory on completion.

a) Is there any reason to believe that two polishing solutions differ? Perform hypothesis testing at significance level 0.05. [Hint: Rather than calculating the p-value, it is easier to compare your statistic to 1.96 to make the conclusion.]

b) Construct a 95% two-sided confidence interval for the true population difference? How is the conclusion compared to a)?

please answer both questions, thanks!

Explanation / Answer

PART A.
Given that,
sample one, x1 =250, n1 =300, p1= x1/n1=0.833
sample two, x2 =200, n2 =300, p2= x2/n2=0.667
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.833-0.667)/sqrt((0.75*0.25(1/300+1/300))
zo =4.714
| zo | =4.714
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =4.714 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 4.714 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 4.714
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0

we have a reason to believe that two polishing solutions differ

PART B.
TRADITIONAL METHOD
given that,
sample one, x1 =250, n1 =300, p1= x1/n1=0.8333
sample two, x2 =200, n2 =300, p2= x2/n2=0.6667
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.8333*0.1667/300) +(0.6667 * 0.3333/300))
=0.0347
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0347
=0.068
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.8333-0.6667) ±0.068]
= [ 0.0987 , 0.2347]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =250, n1 =300, p1= x1/n1=0.8333
sample two, x2 =200, n2 =300, p2= x2/n2=0.6667
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.8333-0.6667) ± 1.96 * 0.0347]
= [ 0.0987 , 0.2347 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 0.0987 , 0.2347] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2

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