Question

salaries are equal. At a.05 level of significance, Whl Isy Many smartphones, especially those of the LTE-enabled persuasion, have been criticized for exceptionally poor 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 pri- mary 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. file EB BatteryTime Primary Use: Talking 35.8 22.2 4.0 32.6 8.5 42.5 3.8 30.0 12.8 10.3 35.5 8.0 Primary Use: Internet 14.0 12.5 16.4 1.99.9 5.4 1.0 15.2 4.04.7 Formulate hypotheses about the two population variances that can be used to determine if the population variance in battery hours of use is greater for the talk time application What are the standard deviations of battery hours of use for the two samples? a. b. c. Conduct the hypothesis test and compute the p-value. Using a.05 level of significance. what is your conclusion?

Answer 1

a)

The hypothesis is defined as,

$Null\ hypothesis,\ H_0: \sigma_1^2=\sigma_2^2$

$Alternate\ hypothesis,\ H_a: \sigma_1^2>\sigma_2^2$

where $\sigma_1^2$ is the variance of phone use for talktime and $\sigma_2^2$ is the variance of phone use for internet

b)

The standard deviation of for talk time data is obtained using the formula,

$\sigma_1=\sqrt{\frac{\sum \left ( X-\mu \right )^2}{n-1}}$

X	$\left ( X-\mu \right )^2$
35.8	234.09
22.2	2.89
4	272.25
32.6	146.41
8.5	144
42.5	484
8	156.25
3.8	278.89
30	90.25
12.8	59.29
10.3	104.04
35.5	225

$\mu=20.5$
$\sum \left ( X-\mu \right )^2=2197.36$

$\frac{\sum \left ( X-\mu \right )^2}{n-1}=\frac{2197.36}{11}=199.76$

$\sqrt{\frac{\sum \left ( X-\mu \right )^2}{n-1}}=14.13365$

The standard deviation of internet data is,

X	(X-\mu)^2
14	30.25
12.5	16
16.4	62.41
1.9	43.56
9.9	1.96
5.4	9.61
1	56.25
15.2	44.89
4	20.25
4.7	14.44

$\mu=8.5$

$\sum \left ( X-\mu \right )^2=299.62$

$\frac{\sum \left ( X-\mu \right )^2}{n-1}=33.2911$

$Standard\ deviation=\sqrt{\frac{\sum \left ( X-\mu \right )^2}{n-1}}=5.7698$

c)

The f distribution is used to test the hypothesis, the F statistic is defined as,

$F=\frac{\sigma_1^2}{\sigma_2^2}$

$F=\frac{\sigma_1^2}{\sigma_2^2}=\frac{199.76}{33.2911}=6.0004$

The P-value is calculated using f distribition table for F = 6.004 , degree of freedom for numerator = 11 and degree of freedom for denominator = 9 and significance level = 0.005

$P-value=0.005994$

P-value < 0.05 at 5% significance level. The null hypothesis is rejected. There is a significant difference in variances