3. The life in hours of a battery is known to be approximately normally distribu

Question

3. The life in hours of a battery is known to be approximately normally distributed, with standard deviation of 1.25 hours. A random sample of 10 batteries has a mean life of 40.5 hours. (a) Is there evidence to support the claim that battery life exceeds 40 hours? Use -0.05 (b) What is the p-value for the test in part (a)? (c) What is the -error for this test if the true mean is 42 hours? What is the power of the test? (d) What sample size would be required to ensure that does not exceed 0.10 if the true mean life is 44 hours?

Explanation / Answer

3.

Given that,
population mean(u)=40
sample mean, x =40.5
standard deviation, s =1.25
number (n)=10
null, Ho: =40
alternate, H1: !=40
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.26
since our test is two-tailed
reject Ho, if to < -2.26 OR if to > 2.26
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =40.5-40/(1.25/sqrt(10))
to =1.265
| to | =1.265
critical value
the value of |t | with n-1 = 9 d.f is 2.26
we got |to| =1.265 & | t | =2.26
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.2649 ) = 0.2377
hence value of p0.05 < 0.2377,here we do not reject Ho
ANSWERS
---------------
null, Ho: =40
alternate, H1: !=40
test statistic: 1.265
critical value: -2.26 , 2.26
a.
decision: do not reject Ho
we do not have enough evidence to support the claim that battery life exceeds 40 hours
b.
p-value: 0.2377
c.
Given that,
Standard deviation, =1.25
Sample Mean, X =40.5
Null, H0: =40
Alternate, H1: !=40
Level of significance, = 0.05
From Standard normal table, Z /2 =1.96
Since our test is two-tailed
Reject Ho, if Zo < -1.96 OR if Zo > 1.96
Reject Ho if (x-40)/1.25/(n) < -1.96 OR if (x-40)/1.25/(n) > 1.96
Reject Ho if x < 40-2.45/(n) OR if x > 40-2.45/(n)
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Suppose the size of the sample is n = 10 then the critical region
becomes,
Reject Ho if x < 40-2.45/(10) OR if x > 40+2.45/(10)
Reject Ho if x < 39.22524 OR if x > 40.77476
Implies, don't reject Ho if 39.22524 x 40.77476
Suppose the true mean is 42
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(39.22524 x 40.77476 | 1 = 42)
= P(39.22524-42/1.25/(10) x - / /n 40.77476-42/1.25/(10)
= P(-7.01965 Z -3.09964 )
= P( Z -3.09964) - P( Z -7.01965)
= 0.001 - 0 [ Using Z Table ]
= 0.001
For n =10 the probability of Type II error is 0.001
power of the test = 1-beta =1-0.001 = 0.999

d.
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(39.22524 x 40.77476 | 1 = 44)
= P(39.22524-44/1.25/(n) x - / /n 40.77476-44/1.25/(n)

beta value can not exceeds 0.10 so that we can find sample size

0.10 = 39.22524-44/1.25/(n < = X-u//n <= 40.77476-44/1.25/(n) = 0.10
0.10 = 39.22524-44/1.25/(n
(n) = 39.22524-44/1.25/0.10
(n) = -38.198
n= 1459.093 or

40.77476-44/1.25/(n) = 0.10

(n) = 40.77476-44/1.25/0.10

(n) = -25.08192
n= 665.739

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