| Two-Sample Hypothesis A national manufacturer of ball bearings is experimentin
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Question
| Two-Sample Hypothesis A national manufacturer of ball bearings is experimenting with two different processes for producing precision ball bearings. It is important that the diameters be as close as possible to an industry standard. The output from each process is sampled and the average error from the industry standard is measured in millimeters. The results of the samples are presented next. Process A Process B Sample mean 2.0 Standard deviation 1.0 12 3.0 0.5 14 Sample size The researcher is interested in determining whether there is evidence that the two processes yield different average errors. The population standard deviations are unknown but are assumed equal. Answer the following Assume you are trying to conclude if the two processes yield different average errors a. Determine if the appropriate hypothesis test should be an upper tail, lower tail, or 2-tail test b. Determine the value of the test statistic. C. Determine the critical z or t value using a level of significance of 5% d. Determine the p-value. e. State your conclusion: Reject Ho or Do not Reject Ho? (Using a significance level of 5%). f. What is the probability of making a Type I Error?Explanation / Answer
a.
two samples are given so we are using t test for for difference of mean
Given that,
mean(x)=2
standard deviation , s.d1=1
number(n1)=12
y(mean)=3
standard deviation, s.d2 =0.5
number(n2)=14
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.201
since our test is two-tailed
reject Ho, if to < -2.201 OR if to > 2.201
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2-3/sqrt((1/12)+(0.25/14))
to =-3.144
| to | =3.144
critical value
the value of |t | with min (n1-1, n2-1) i.e 11 d.f is 2.201
we got |to| = 3.14362 & | t | = 2.201
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.1436 ) = 0.009
hence value of p0.05 > 0.009,here we reject Ho
ANSWERS
---------------
b.
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.144
c.
critical value: -2.201 , 2.201
d.
p-value: 0.009
e.
decision: reject Ho
we have enough evidence to support the claim
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