Question

Media Metrix, Inc., monitors internet users in seven countries: Australia, Great Britain, France, Canada, Germany. Japan and the United States. According to recent measurement figures, US users occupy the first place in Internet use with an average of 13 hours per month (The Washington Post, August 4, 2000). Suppose that in a follow-up study involving 145 Canadian Internet users, the sample mean was 10.8 hours per month and the sample standard deviation was 9.2 hours.

a) Formulate the null and alternative hypotheses that will be used to determine if the sample data support the conclusion that Canadian internet users have a population mean lower than the US average of 13 hours per month.

b) With a = 0.01, what is the critical value for the test statistic? Express the rejection rule.

c) What is the value of the test statistic?

d) What is your conclusion?

Answer 1

a)

Null hypothesis

              
              
Alternative hypothesis  

b)

Sample size =n=   145      
              
Degree of freedom = n-1 =144  

   

t critical value is =-2.353 ..............by using t table.

Decision rule: Reject H0 if t calculated value < -2.353

c)
              
We have given here,
Population mean for given sample=13 hours
Sample mean=10.8 hours      
Sample standard deviation s=9.2 hours  

              
              
              
t test statistic formula is              


=-2.88      

d) t calculated value = -2.88 < t critical value -2.353  

Therefore, reject H0 at

Therefore,We have sufficient evidence at to say that , Canadian internet users have a population mean lower than the US average of 13 hours per month.