Question

We want to compare sugar contents ow two cracker brands. 17 crackers are chosen from the first brand and it is observed that the sample has an average sugar content of x1= 2.73mg, with a sample standart deviation of s1=0.1mg, 9 crackers are chosen from the second brand and it is observed that this sample has and average sugar content of x2=2.53 mg, with a sample standart deviation of s2=0.1mg. We want to determine whether the average sugar content of the first cracker brand exceeds that of the order brand by 0.1 miligram (mg).

We want to test the null hypothesis p1-p2=0.1 against the alternative hypothesis p1-p2 not equal to 0.1 at the 0.05 level of significance under the assumption that sugar contents of both brands share the same variance.

a) what is the test statistic you would use? calculate it.

b) make the test by using critical value approach. Discuss whether H0 should bu rejected or not.

c) make the test by using P-value. Discuss whether H0 should be rejected or not.

d) make the test bu using 100(1-a)% confidence interval approach. Discuss whether H0 should be rejected or not.

e) suppose the true avg1-avg2=0.18. What is the minimum sample size necessary to have the probability of falsely accepting H0 is at most 0.2. Assume that the population variance is equal to the sample variances.

Answer 1

Solution :

a)

Ho :  p1-p2 =   0.1                  
Ha :  p1-p2 ╪   0.1                  
                          
Level of Significance ,    α = 0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,   x1= 2.73                  
standard deviation of sample 1,   s1 = 0.10                  
size of sample 1,    n1=17
                          
Sample #2   ---->   sample 2                  
mean of sample 2, x2= 2.53            
standard deviation of sample 2,   s2 = 0.10                  
size of sample 2,    n2=9                  
                          
difference in sample means =    x̅1-x̅2 = 2.73  -  2.53   = 0.2
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) = 0.10   
std error , SE =    Sp*√(1/n1+1/n2) = 0.0412   
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = ( 0.2 -   0.1 ) / 0.04   =   2.4

b)

Degree of freedom, DF=   n1+n2-2 = 17+9-2 = 24   
t-critical value , t = 1.711
Decision:   | t-stat | > | critical value |,

So, Reject Ho                      

c)

p-value = 0.02451   
Conclusion :    

P-value <α , Reject null hypothesis                      

d)

Degree of freedom, DF=  n1+n2-2 =    23              
t-critical value = 1.711   
                      
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) = 0.10  
                      
std error , SE = Sp*√(1/n1+1/n2) = 0.0412   
margin of error, E = t*SE =    1.711   *  0.0412 = 0.0705
                      
difference of means =    x̅1-x̅2 = 2.7100 -   2.510   =  0.2000
confidence interval is                       
Interval Lower Limit =   (x̅1-x̅2) - E = 0.2000 -   0.0705   =   0.1295
Interval Upper Limit =   (x̅1-x̅2) + E = 0.2000  +   0.0705   =   0.2705

As confidence interval does not contain 0.1 , reject the hypothesis.