CNBC presented statistics on the average number of programmi

CNBC presented statistics on the average number of programming minutes in a half-hour sitcom. The following data are representative of their findings. Number of programming minutes in a half-hour sitcom (23.22, 23.09, 22.17, 21.98, 20.56, 22.11, 20.32, 24.96, 24.65, 23.53, 21.59, 24.35)

Can you conclude the average number of programming minutes during a half-hour television sitcom is different from 21 minutes at ?=0.1?

What is the Null and the Alternative?

What are the critical Values? (+,-, +/-)

Reject the Null if what?

What is the test statistic?

What is the Decision

What is the Conclusion

What is the P-Value

Solution

What is the Null and the Alternative?

Let mu be the population mean

Null hypothesis: mu=21

Alternative hypothesis: mu not equal to 21

----------------------------------------------------------------------------------------------------

What are the critical Values? (+,-, +/-)

It is a two-tailed test.

The degree of freedom =n-1=12-1=11

Given a=0.1, the critical values are t(0.05, df=11) =1.796 or -1.796 (from student t table)

----------------------------------------------------------------------------------------------------

Reject the Null if what?

Reject the null if t<-1.796 or t>1.796

----------------------------------------------------------------------------------------------------

What is the test statistic?

sample mean=22.71083

sample standard deviation =1.518555

t=(xbar-mu)/(s/vn)

=(22.71083-21)/(1.518555/sqrt(12))

=3.90

----------------------------------------------------------------------------------------------------

What is the Decision

Since t=3.9 is larger than 1.796, we reject the null hypothesis.

----------------------------------------------------------------------------------------------------

What is the Conclusion

So we can conclude that the average number of programming minutes during a half-hour television sitcom is different from 21 minutes

----------------------------------------------------------------------------------------------------

What is the P-Value

The p-value= 2*P(t with df=11 >3.9) =0.0025 (from student t table)