A bearing used in an automotive application is required to have a nominal inside

Question

A bearing used in an automotive application is required to have a nominal inside diameter of 1.5 inches. A random sample of 25 bearings is selected and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard devistion " 0.01 inch. (a)Test the hypotheses How= 1.5 versus H1, 15 using -001. The true mean hole diameter significantly different from 1.5 in, at -0.01 (b) What is the P-value for the test in part (a) P-value Round your answer to two decimal places (e.q. 98.76). (c) Compute the power of the test if the true mean diameter is 1.495 inches. Power of the test Round your answer to four deciral places (e.g. 98.7654). (d) what sample size would be required to detect a true mean dameter as low as 1.495 inches if we wanted the power of the test to be at least 0.90 bearings

Explanation / Answer

PART A.
Given that,
population mean(u)=1.5
standard deviation, =0.01
sample mean, x =1.4975
number (n)=25
null, Ho: =1.5
alternate, H1: !=1.5
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1.4975-1.5/(0.01/sqrt(25)
zo = -1.25
| zo | = 1.25
critical value
the value of |z | at los 1% is 2.576
we got |zo| =1.25 & | z | = 2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.25 ) = 0.21
hence value of p0.01 < 0.21, here we do not reject Ho
ANSWERS
---------------
null, Ho: =1.5 , alternate, H1: !=1.5
test statistic: -1.25
critical value: -2.576 , 2.576
decision: do not reject Ho

PART B.
p-value: 0.21

PART C.
Given that,
Standard deviation, =0.01
Sample Mean, X =1.4975
Null, H0: =1.5
Alternate, H1: !=1.5
Level of significance, = 0.01
From Standard normal table, Z /2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-1.5)/0.01/(n) < -2.5758 OR if (x-1.5)/0.01/(n) > 2.5758
Reject Ho if x < 1.5-0.025758/(n) OR if x > 1.5-0.025758/(n)
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Suppose the size of the sample is n = 25 then the critical region
becomes,
Reject Ho if x < 1.5-0.025758/(25) OR if x > 1.5+0.025758/(25)
Reject Ho if x < 1.4948484 OR if x > 1.5051516
Implies, don't reject Ho if 1.4948484 x 1.5051516
Suppose the true mean is 1.495
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(1.4948484 x 1.5051516 | 1 = 1.495)
= P(1.4948484-1.495/0.01/(25) x - / /n 1.5051516-1.495/0.01/(25)
= P(-0.0758 Z 5.0758 )
= P( Z 5.0758) - P( Z -0.0758)
= 1 - 0.4698 [ Using Z Table ]
= 0.5302
For n =25 the probability of Type II error is 0.5302

power of the test = 1 - 0.5302 = 0.4698

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