Question

A researcher wants to test the claim that the mean weight of sugar in cola drinks is more today than it was in the past. They tested 49 cans of soda produced last year and found they had a mean weight of sugar of 13.5 g with a standard deviation of 1.2 grams. They also tested 25 cans produced 30 years ago and found they had a mean weight of sugar of 10.1 grams with a standard deviation of 1.6 grams.

(a) Use a 0.05 significance level to test the claim that there is more sugar in today’s cola drinks than there was in cola produced in the past.

(b) Construct the confidence interval appropriate for (a).

Answer 1

a)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   13.500                  
standard deviation of sample 1,   s1 =    1.2000                  
size of sample 1,    n1=   49                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   10.100                  
standard deviation of sample 2,   s2 =    1.6000                  
size of sample 2,    n2=   25                  
                          
difference in sample means =    x̅1-x̅2 =    13.5000   -   10.1   =   3.40  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.3466                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.3310                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   3.4000   -   0   ) /    0.33   =   10.2729
                          
Degree of freedom, DF=   n1+n2-2 =    72                  

p-value =        0.0000   [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis                      

there is more sugar in today’s cola drinks than there was in cola produced in the past

b)

Degree of freedom, DF=   n1+n2-2 =    72              
t-critical value =    t α/2 =    1.9935   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.3466              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.3310              
margin of error, E = t*SE =    1.9935   *   0.33   =   0.66  
                      
difference of means =    x̅1-x̅2 =    13.5000   -   10.100   =   3.4000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    3.4000   -   0.6598   =   2.7402
Interval Upper Limit=   (x̅1-x̅2) + E =    3.4000   +   0.6598   =   4.0598