A sample of 400 high school students showed that they spend an average of 70 min

Question

A sample of 400 high school students showed that they spend an average of 70 minutes a day watching television with a standard deviation of 14 minutes. Another sample of 500 college students showed that they spend an average of 55 minutes a day watching television with a standard deviation of 12 minutes. Construct a 99% confidence interval for the difference between the mean times spent watching television by all high school students and all college students. Test at the 2.5% significance level if the mean time spent watching television per day by high school students is higher than the mean time spent watching television by college students.

Explanation / Answer

a)

Calculating the means of each group,              
              
X1 =    70          
X2 =    55          
              
Calculating the standard deviations of each group,              
              
s1 =    14          
s2 =    12          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    400          
n2 = sample size of group 2 =    500          

Also, sD =    0.882043083          
              
For the   0.99   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.005          
z(alpha/2) =    2.575829304          
          
Thus,
  
lower bound = [X1 - X2] - z(alpha/2) * sD =    12.72800758          
upper bound = [X1 - X2] + z(alpha/2) * sD =    17.27199242          
              
Thus, the confidence interval is              
              
(   12.72800758   ,   17.27199242   ) [ANSWER]

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   <=   0  
Ha:   u1 - u2   >   0  

At level of significance =    0.025          

As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    70          
X2 =    55          
              
Calculating the standard deviations of each group,              
              
s1 =    14          
s2 =    12          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    400          
n2 = sample size of group 2 =    500          

Also, sD =    0.882043083          
              
Thus, the z statistic will be              
              
z = [X1 - X2 - uD]/sD =    17.00597203          
              
where uD = hypothesized difference =    0          
              
Now, the critical value for z is              
              
zcrit = 1.96
              
Also, using p values, as this is right tailed,              
              
p = 3.70824E-65
  
              
As z > 1.96, and P < 0.025,    WE REJECT THE NULL HYPOTHESIS.          

There is significant evidence at 0.025 level that the mean time spent watching television per day by high school students is higher than that of college students. [CONCLUSION]

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