Styles Cipboard Alignment crit 102 parts made from alloy 1 and 46 parts made fro
ID: 3341277 • Letter: S
Question
Styles Cipboard Alignment crit 102 parts made from alloy 1 and 46 parts made from alloy 2 were subjected to stress tests. 18 parts from alloy 1 and 14 parts from alloy 2 did not pass the test. Can we reject the hypothesis that the proportion of nonpassing parts from alloy 1 is at least as large as the proportion of nonpassing parts from alloy 2 at a-0.025? Assume independent samples. Question 1For the hypothesis stated above (in terms of alloyl- alloy2 Part A What is the decision rule? Fill in only one of the following statements f the hypothesis is one tailed Reject Ho if the hypothesis is two tailed: Reject Ho ifcrit t stat Part B What is the test statistic? Question 2 Find the 99% confidence interval (in terms of alloy2-alloy 1). Left Endpoint- Right Endpoint - Click to Grade Your Work Daily ProblemExplanation / Answer
PART A.
Given that,
sample one, x1 =18, n1 =102, p1= x1/n1=0.176
sample two, x2 =14, n2 =46, p2= x2/n2=0.304
null, Ho: p1 >= p2
alternate, H1: p1 < p2
level of significance, = 0.025
from standard normal table,left tailed z /2 =1.96
since our test is left-tailed
reject Ho, if zo < -1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.176-0.304)/sqrt((0.216*0.784(1/102+1/46))
zo =-1.749
| zo | =1.749
critical value
the value of |z | at los 0.025% is 1.96
we got |zo| =1.749 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -1.749 ) = 0.04014
hence value of p0.025 < 0.04014,here we do not reject Ho
ANSWERS
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null, Ho: p1 >= p2
alternate, H1: p1 < p2
test statistic: -1.749
critical value: -1.96
reject Ho, if zo < -1.96
decision: do not reject Ho
p-value: 0.04014
proportion rate of alloy 1 is as large as alloy 2
PART B.
given that,
sample one, x1 =18, n1 =102, p1= x1/n1=0.1765
sample two, x2 =14, n2 =46, p2= x2/n2=0.3043
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 = [ (0.1765-0.3043) ± 2.58 * 0.0776]
= [ -0.3282 , 0.0724 ]