The following table shows the results of two random samples that measured the av

Question

The following table shows the results of two random samples that measured the average number of minutes per charge for AA Lithium-ion (Li-ion) rechargeable batteries versus Nickel- Metal Hydride (NiMH) rechargeable batteries. Li-ion NiMH 96.1 82.8 Sample mean Sample standard deviation6.6 11.7 Sample size Perform a hypothesis test using -0.05 to determine if the average number of minutes per charge differs between these two battery types Ho: 1 = 2, H1: 1 2. Assume the population variances for the number of minutes per charge are equal. (Pooled: Yes) 15 10

Explanation / Answer

Given that,
mean(x)=96.1
standard deviation , s.d1=6.6
number(n1)=15
y(mean)=82.8
standard deviation, s.d2 =11.7
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.07
since our test is two-tailed
reject Ho, if to < -2.07 OR if to > 2.07
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (14*43.56 + 9*136.89) / (25- 2 )
s^2 = 80.0804
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=96.1-82.8/sqrt((80.0804( 1 /15+ 1/10 ))
to=13.3/3.6533
to=3.6405
| to | =3.6405
critical value
the value of |t | with (n1+n2-2) i.e 23 d.f is 2.07
we got |to| = 3.6405 & | t | = 2.07
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 3.6405 ) = 0.0013
hence value of p0.05 > 0.0013,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 3.6405
critical value: -2.07 , 2.07
decision: reject Ho
p-value: 0.0013

Option B

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