00:01
Hello students in this question.
00:03
In 2015, the average duration of long distance telephone calls from a certain town was 3 .9 minutes.
00:12
A telephone company wants to perform a test to determine whether this average duration of long distance call has changed.
00:25
50 calls that is n equal to 50 originating from the town was randomly selected and had followed.
00:34
Summary that is summation x equal to 205 and summation x minus x bar whole square equal to 56 .43 first we need to find sample mean that is x bar that is equal to summation x i divided by n that is 2 .05 divided by 50 that is 2 .05 divided by 50 that is is equal to 4 .5 and next we want sample standard deviation that is s equal to square root of summation x i minus x bar whole square divided by n minus 1 that is square root of 56 .43 divided by 49.
01:30
That is equal to 1 .0731.
01:38
And next it is asked, using p value approach at alpha equal to 1 % level of significance, test whether the mean duration of long distance calls from the town has increased.
01:52
For that, h0 is new equal to 3 .9 versus the alternative hypothesis is new is greater than 3 .9.
02:06
Here we assume that sample is given from normally distributed population with unknown standard deviation so we use t statistic that is x bar minus mu divided by s by root n that is 4 .1 minus 3 .9 divided by 1 .0731 divided by root 50 and that is equal to 1 .3179.
02:43
And here p value is 0 .0968...