A telephone company claims that the mean duration of all long-distance phone cal

Question

A telephone company claims that the mean duration of all long-distance phone calls made by its residential customers is 10 minutes. A random sample of 100 long-distance calls made by its residential customers taken from the records of this company showed that the mean duration of calls for this sample is 9.20 minutes. The population standard deviation is known to be 3.80 minutes

a. Find the p-value for the test that the mean duration of all longdistance calls made by residential customers of this company is different from 10 minutes. If = .02, based on this p-value, would you reject the null hypothesis? Explain. What if = .05?

b. Test the hypothesis of part a using the critical-value approach and = .02. Does your conclusion change if = .05?

Explanation / Answer

PART A.
WITH 0.02 LOS
Given that,
population mean(u)=10
standard deviation, =3.8
sample mean, x =9.2
number (n)=100
null, Ho: =10
alternate, H1: !=10
level of significance, = 0.02
from standard normal table, two tailed z /2 =2.326
since our test is two-tailed
reject Ho, if zo < -2.326 OR if zo > 2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 9.2-10/(3.8/sqrt(100)
zo = -2.10526
| zo | = 2.10526
critical value
the value of |z | at los 2% is 2.326
we got |zo| =2.10526 & | z | = 2.326
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -2.10526 ) = 0.03527
hence value of p0.02 < 0.03527, here we do not reject Ho
ANSWERS
---------------
null, Ho: =10
alternate, H1: !=10
test statistic: -2.10526
critical value: -2.326 , 2.326
decision: do not reject Ho
p-value: 0.03527

WITH 0.05 LOS

Given that,
population mean(u)=10
standard deviation, =3.8
sample mean, x =9.2
number (n)=100
null, Ho: =10
alternate, H1: !=10
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 9.2-10/(3.8/sqrt(100)
zo = -2.10526
| zo | = 2.10526
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.10526 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -2.10526 ) = 0.03527
hence value of p0.05 > 0.03527, here we reject Ho
ANSWERS
---------------
null, Ho: =10
alternate, H1: !=10
test statistic: -2.10526
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.03527

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