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

ID: 3218004 • Letter: M

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.

Primary Use: Talking

35.8 22.2 4.0 32.6 8.5 42.5 8.0 3.8 30.0 12.8 10.3 35.5

Primary Use: Internet

14.0 12.5 16.4 1.9 9.9 5.4 1.0 15.2 4.0 4.7

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

s1 = ___

s2 = ___

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

The value of F is ___

p-value is ____

Primary Use: Talking 35.8 22.2 4.0 32.6 8.5 42.5 8.0 3.8 30.0 12.8 10.3 35.5 Primary Use: Internet 9.9 14.0 12.5 16.4 1.9 5.4 4.7 1.0 15.2 4.0

Explanation / Answer

Given that,
sample 1
s1^2=199.75864, n1 =12
sample 2
s2^2 =33.29059, n2 =10
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
level of significance, = 0.01
from standard normal table, two tailed f /2 =6.314
since our test is two-tailed
reject Ho, if F o < -6.314 OR if F o > 6.314
we use test statistic fo = s1^1/ s2^2 =199.75864/33.29059 = 6
| fo | =6
critical value
the value of |f | at los 0.01 with d.f f(n1-1,n2-1)=f(11,9) is 6.314
we got |fo| =6 & | f | =6.314
make decision
hence value of |fo | < | f | and here we do not reject Ho

ANSWERS
---------------
null, Ho: ^2 = ^2
alternate, H1: ^2 != ^2
test statistic: 6
critical value: -6.314 , 6.314
decision: do not reject Ho

b. What are the standard deviations of battery hours of use for the two samples? Round your answers to two decimal places.
s1 = 14.1336
s2 = 5.7698
c. Conduct the hypothesis test and compute the p-value. Round your answers to three decimal places.
The value of F is 6.314

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