A phone manufacturer wants to compete in the touch screen phone market. Manageme

Question

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 147 units of the leading product provides a mean battery life of 5 hours and 52 minutes with a standard deviation of 38 minutes. A similar analysis of 94 units of the new product results in a mean battery life of 8 hours and 14 minutes and a standard deviation of 19 minutes. It is not reasonable to assume that the population variances of the two products are equal. Sample 1 is from the population of new phones and Sample 2 is from the population of old phones. All times are converted into minutes. Let new products and leading products represent population 1 and population 2, respectively. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product. H_0: mu_1 - mu_2 lessthanorequalto 120; H_A: mu_1 - mu_2 > 120 H_0: mu_1 - mu_2 Greaterthanorequalto 120; H_A: mu_1 - mu_2

Explanation / Answer

sample 1 denotes the new phones and sample 2 denotes the old phones.

let X1 denotes the battery life in months for new phones and X2 denotes the same for the leading phones

we have a sample of size n1=94 from X1 with sample mean=X1bar=8 hours and 14 minutes=494 minutes and sample standard deviation=s1=19 minutes

we have a sample of size n2=147 from X2 with sample mean X2bar=5 hours 52 minutes=352 minutes and sample standard deviation=s2=38

assumption is that X1~N(u1,sigma12) and X2~N(u2,sigma22) all the parameters are unknown.

a) we want to test whether the new product has mean battery life more than two hours(120 minutes) longer than the leading phones.

hence the null hypothesis is H0: u1-u2<=120 and the alternative hypothesis is H1: u1-u2>120 [answer]

b)-1) to test this we have the test statistic as

T=((X1bar-X2bar)-120)/sqrt[s12/n1+s22/n2] which under H0 follows a t distribution with df=(s12/n1+s22/n2)2/[(s12/n1)2/(n1-1)+(s22/n2)2/(n2-1)]

now X1bar=494 n1=94 s1=19 X2bar=352 n2=147 s2=38

so value of the test statistic is

t=((494-352)-120)/sqrt[192/94+382/147]= 5.95 [answer]

b-2) now the df of the t distribution is

(s12/n1+s22/n2)2/[(s12/n1)2/(n1-1)+(s22/n2)2/(n2-1)]=(192/94+382/147)2/[(192/94)2/(94-1)+(382/147)2/(147-1)]=227.8111

since the alternative hypothesis is a right tailed hence at 5% level of significance H0 will be rejected iff t>t0.05;227.8111

where t0.05;227.8111 is the upper 5% point of a t distribution with df 227.8111

using R we find t0.05;227.8111=1.65157

hence t=5.95>1.65157

hence H0 is rejected

and hence the conclusion is

reject H0: the battery life of the new product is more than two hours longer than the leading product [answer]

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