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 deviation sigma = 0.01 inch. (a) Test the hypotheses H_0: mu = 1-5 versus H_1: mu notequalto 1.5 using alpha = 0.01. The true mean hole diameter significantly different from 1.5 in. at alpha = 0.01. (b) What is the P-value for the test in part (a)? P-value = (c) Compute the power of the test if the true mean diameter is 1.495 inches. Power of the test = (d) What sample size would be required to detect a true mean diameter as low as 1.495 inches if we wanted the power of the test to be at least 0.93? bearings

Explanation / Answer

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 ) = 0.211
hence value of p0.01 < 0.211, 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
p-value: 0.211

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