Question

1) Nielsen Company - which publishes the Nielsen ratings for television shows - claims that the average American watches 34.5 hours of television per week.

a) State the appropriate null and alternate hypothesis to determine if the average number of hours of TV watched by an American per week is different than the Nielsen Claim.

b) To test the Nielsen claim, you randomly select 10 Americans and record the number of hours of television watched per week. The data appear in the TV worksheet of the Hypothesis Tests HW data workbook on Moodle. Use JMP to test the hypothesis of part a).

c) Using  = 0.05, draw a conclusion for the hypothesis test. Make sure you state your conclusion in the context of the problem.

d) If the true average number of hours of television watched per week is 40, what is the probability that you would conclude the alternative hypothesis of part a) with a sample of n = 10 at  = 0.05, i.e., determine the power of the test.

e) If the true average number of hours of television watched per week is 40, what sample size should you take so that you would conclude the alternative hypothesis of part a) at  = 0.05 with probability of at least 0.90.

Hours
30
64
49
40
21
58
9
43
39
43

Answer 1

a)

null hypothesis:the average number of hours of TV watched by an American per week is 34.5 hours of television per week

alternate hypothesis:the average number of hours of TV watched by an American per week is different from 34.5 hours of television per week

b)

Ho :   µ =   0  
Ha :   µ ╪   0  
          
Level of Significance ,    α =    0.05  
sample std dev ,    s =    16.3992  
Sample Size ,   n =    10  
Sample Mean,    x̅ =   39.6  
          
degree of freedom=   DF=n-1=   9  
          
Standard Error , SE =   s/√n =   5.1859  
          
t-test statistic=   (x̅ - µ )/SE =    7.6361  
          
critical t value, t*   =   2.2622   [Excel formula =t.inv(α/no. of tails,df) ]
          
p-Value   =   0.0000  
Conclusion:     p-value<α, Reject null hypothesis

c)

so, there is enough evidence that the average number of hours of TV watched by an American per week is different from 34.5 hours of television per week   

d)

true mean ,    µ =    40                          
                                  
hypothesis mean,   µo =    34.5                          
significance level,   α =    0.05                          
sample size,   n =   10                          
std dev,   σ =    16.39918697                          
                                  
δ=   µ - µo =    5.5                          
                                  
std error of mean,   σx = σ/√n =    5.1859                          
                                  
Zα/2   = ±   1.960   (two tailed test)                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -1.960   and   1.960
these Z-critical value corresponds to some X critical values ( X critical), such that                                  
                                  
-1.960   ≤(x̄ - µo)/σx≤   1.960                          
24.336   ≤ x̄ ≤   44.664                          
                                  
now, type II error is ,ß =        P (   24.336   ≤ x̄ ≤   44.664   )          
       Z =    (x̄-true mean)/σx                      
       Z1 =   -3.021                      
       Z2 =    0.899                      
                                  
   so, P(   -3.021   ≤ Z ≤   0.899   ) = P ( Z ≤   0.899   ) - P ( Z ≤   -3.021   )
                                  
       =   0.816   -   0.001   =   0.8145      
                                  
power =    1 - ß =   0.1855

e)

True mean,   µ =    40
hypothesis mean,   µo =    34.5
      
Level of Significance ,    α =    0.05
std dev =    σ =    16.39918697
power =    1-ß =    0.9
      
δ=   µ - µo =    5.5
      
Z (α/2)=       1.9600
      
Z (ß) =        1.2816
      
sample size needed =    n = ( σ [ Z(ß)+Z(α/2) ] / δ )² =    93.4148
      
so, sample size =        94.000