Question

Test at the 5% significance level whether the mean length of telephone conversations equals eight minutes against the alternative that it doesn't.

mean of sample data = 9
variance of sample data = 6.25

please help :(

Answer 1

do you know the size of the sample?

Answer 2

yes it's 9

Answer 3

i believe the test statistic can be found using:
\[t* = \frac{9-8}{\sqrt{6.25/n}}\]

Answer 4

plug in 9 for n

Answer 5

that's correct but why do you do 9-8 and not 8-9? isn't it the observed value minus the mean?
and also what do you do with this t value?

Answer 6

degrees of freedom = n-1 = 8.............
then you have to use a t-distribution table to find critical value............

yes the observed value is 9, the testing value is 8

Answer 7

the value gained from the sample is 9 which is greater than our hypothesized value of 8 so it must be "9-8" to get a positive test-statistic

Answer 8

the critical value is 2.306 i believe.

so because 1.2 is less then 2.306 i accept the null hypothesis?

Answer 9

that is correct.
A high test-statistic corresponds to low p-value, --> Reject....................................
Low test-statistic corresponds to high p-value --> Accept