Question

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 14 of the plates have blistered. If it is really the case that 15% of all plates blister under these circumstances and a sample size of 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected (to 4 decimal places). (The probability of making a type 2 error when the true value of p = 0.15)

Answer 1

Given that,
possibile chances (x)=14
sample size(n)=100
success rate ( p )= x/n = 0.14
success probability,( po )=0.15
failure probability,( qo) = 0.85
null, Ho:p=0.15
alternate, H1: p!=0.15
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.14-0.15/(sqrt(0.1275)/100)
zo =-0.28
| zo | =0.28
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =0.28 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.28006 ) = 0.77943
hence value of p0.05 < 0.7794,here we do not reject Ho
ANSWERS
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a.
null, Ho:p=0.15
alternate, H1: p!=0.15
test statistic: -0.28
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.77943
we do not have enough evidence to support the claim that when the true value of p = 0.15
Type 2 error is possible because when its fails to reject the null hypothesis.