Question

A survey taken several years ago found that the average time a person spent reading the local daily newspaper was 10.8 minutes. The standard deviation of the population was 3 minutes. To see whether the average time had changed since the newspaper's format was revised, the newspaper editor surveyed 36 individuals. The average time that the 36 people spent reading the paper was 12.2 minutes.  

(a) At α = 0.02, is there a change in the average time an individual spends reading the newspapers?  

Test by using confidence interval for the mean. Round confidence interval to nearest thousandth.

Show step1-step3.  

(b). Do the results agree?  Explain why.  

Answer 1

population std dev ,    σ =    3.0000
Sample Size ,   n =    36
Sample Mean,    x̅ =   12.2000

Level of Significance ,    α =    0.02          

z value=   z α/2=   2.326   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   3/√36=   0.5000          
margin of error, E=Z*SE =   2.3263   *   0.5000   =   1.163
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    12.20   -   1.1632   =   11.0368
Interval Upper Limit = x̅ + E =    12.20   -   1.1632   =   13.3632
98%   confidence interval is (   11.037 < µ <   13.363 )

b)

As 10.8 is not fall into above interval we can conclude that there is a change in the average time an individual spends reading the newspapers

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!