Question

A battery company claims that its batteries last an average of 100 hours under normal use. After several complaints that the batteries do not last this long, an independent testing laboratory decided to test the company’s claim with a random sample of 42 batteries. The data from the 42 batteries appeared to be unimodal and symmetric with a mean 97 hours and a standard deviation of 12 hours. What sample size would allow us to increase our confidence level to 95% while reducing the margin of error to only 3 hours? Round up to the nearest whole number.

Answer 1

a.

Given that,
population mean(u)=100
sample mean, x =97
standard deviation, s =12
number (n)=42
null, Ho: μ=100
alternate, H1: μ<100
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.683
since our test is left-tailed
reject Ho, if to < -1.683
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =97-100/(12/sqrt(42))
to =-1.62
| to | =1.62
critical value
the value of |t α| with n-1 = 41 d.f is 1.683
we got |to| =1.62 & | t α | =1.683
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.6202 ) = 0.05643
hence value of p0.05 < 0.05643,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=100
alternate, H1: μ<100
test statistic: -1.62
critical value: -1.683
decision: do not reject Ho
p-value: 0.05643
we do not have enough evidence to support the claim that its batteries last an average of 100 hours under normal use
b.
given data,
margin of error =3 hours
confidence level is 95%
sample size = n
standard deviation =12 hours
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 12
ME =3
n = ( 1.96*12/3) ^2
= (23.52/3 ) ^2
= 61.47 ~ 62