ANSWER 1,2, AND 3 COMPLETELY AND FULLY! DO NOT COPY Two suppliers manufacture a

Question

ANSWER 1,2, AND 3 COMPLETELY AND FULLY! DO NOT COPY

Two suppliers manufacture a plastic gear used in a laser printer. The impact of these gears measured in foot-pounds is an important characteristic of interest. A random sample of 10 gears from Supplier 1 results in x 1 = 296 and s1 = 12. Another random sample of 25 gears from Supplier 2 results in x 2 = 315 and s2 = 18. Assume the populations are independent and normally distributed.

a.) What is the parameter of interest? What assumptions are made?

b.) Write the null and alternative hypostheses.

c.) Calculate the test statistic.

d.) Determine the reject region.

e.) Make a decision and write a thorough interpretation in context of the problem.

a.) What is the parameter of interest? What assumptions are made? (Note: Use information from Part 1)

b.) Write the null and alternative hypostheses.

c.) Calculate the test statistic.

d.) Determine the reject region.

e.) Make a decision and write a thorough interpretation in context of the problem.

3.) Construct a one-sided upper confidence interval for mean impact strength and explain how the interval can be used to answer #2.

Explanation / Answer

PART A.
Given that,
mean(x)=296
standard deviation , s.d1=12
number(n1)=10
y(mean)=315
standard deviation, s.d2 =18
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =296-315/sqrt((144/10)+(324/25))
to =-3.6324
| to | =3.6324
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 3.63242 & | t | = 2.262
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.6324 ) = 0.005
hence value of p0.05 > 0.005,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.6324
critical value: -2.262 , 2.262
decision: reject Ho
p-value: 0.005

sufficient evidence to conclude that the variance of the impact strength
is different for the two suppliers

PART B.
Given that,
mean(x)=296
standard deviation , s.d1=12
number(n1)=10
y(mean)=315
standard deviation, s.d2 =18
number(n2)=25
null, Ho: u1 = u2
alternate, claim that supplier 2 provides gears with higher mean impact strength, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.833
since our test is left-tailed
reject Ho, if to < -1.833
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =296-315/sqrt((144/10)+(324/25))
to =-3.6324
| to | =3.6324
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 1.833
we got |to| = 3.63242 & | t | = 1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -3.6324 ) = 0.00273
hence value of p0.05 > 0.00273,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -3.6324
critical value: -1.833
decision: reject Ho
p-value: 0.00273
claim that supplier 2 provides gears with higher mean impact strength

PART C.
given that,
mean(x)=296
standard deviation , s.d1=12
number(n1)=10
y(mean)=315
standard deviation, s.d2 =18
number(n2)=25
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((144/10)+(324/25))
= 5.231
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table,right tailed and
value of |t | with min (n1-1, n2-1) i.e 9 d.f is 1.833
margin of error = 1.833 * 5.231
= 9.588
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (296-315) ± 9.588 ]
= [-28.588 , -9.412]

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