Many smartphones, especially those of the LTE-enabled persuasion, have earned a

Question

Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad battery life. Battery life between charges for the Motorola Droid Razr Max averages 20 hours when the primary use is talk time and 7 hours when the primary use is Internet applications (The Wall Street Journal, March 7, 2012). Since the mean hours for talk time usage is greater than the mean hours for Internet usage, the question was raised as to whether the variance in hours of usage is also greater when the primary use is talk time. Sample data showing battery hours of use for the two applications follows.

b. What are the standard deviations of battery hours of use for the two samples? Round your answers to two decimal places.

c. Conduct the hypothesis test and compute the p-value. Round your answers to three decimal places.

Using a .05 level of significance, what is your conclusion?

Talk Time Internet 35.8 14.0 22.2 12.5 4.0 16.4 32.6 1.9 8.5 9.9 42.5 5.4 8.0 1.0 3.8 15.2 30.0 4.0 12.8 4.7 10.3 35.5

Explanation / Answer

b.

s.d of talk time = 14.1336

s.d of internet = 5.7698

c.

Given that,

sample 1
s1^2=199.7586, n1 =12
sample 2
s2^2 =33.2906, n2 =10
null, Ho: ^2 = ^2
alternate,variance in hours of usage is also greater when the
primary use is talk time H1: ^2 < ^2
level of significance, = 0.05
from standard normal table,left tailed f /2 =3.102
since our test is left-tailed
reject Ho, if F o < -3.102
we use test statistic fo = s1^1/ s2^2 =199.7586/33.2906 = 6
| fo | =6
critical value
the value of |f | at los 0.05 with d.f f(n1-1,n2-1)=f(11,9) is 3.102
we got |fo| =6 & | f | =3.102
make decision
hence value of | fo | > | f | and here we reject Ho
ANSWERS
---------------
null, Ho: ^2 > ^2
alternate, H1: ^2 < ^2
test statistic: 6
critical value: -3.102
decision: reject Ho

we have evidence to support variance in hours of usage is also greater when the
primary use is talk time

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