Question

1. Recently there has been a marketing push promoting the new wave of electronic cigarettes in the media. Suppose a new study claims that they have determined that e-cigarette smokers still have a much higher risk for esophageal cancer than non-smokers. Assume the following table of summary statistics describes the SRSs of those in the e-cigarette group versus non-smokers.

Study Group	Average # of cases of esophageal cancer	Standard deviation	Sample Size
E-cigarette Smokers	21.4	1.8	749
Non-Smokers	2.7	0.8	898

1. What is the appropriate t-test to use here?
2. Carry out the appropriate t test to determine if there is a significant difference in the number of esophageal cancer cases in e-cigarette smokers versus nonsmokers. Use an alpha level of 0.05. Be sure to include each of the four steps of hypothesis testing.
3. Construct a 95% confidence interval for the difference in number of esophageal cancer cases between the groups. Interpret your interval correctly.

Answer 1

A) We can use 2 indepenedent sample t test

B)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   21.40                  
standard deviation of sample 1,   s1 =    1.80                  
size of sample 1,    n1=   749                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   2.70                  
standard deviation of sample 2,   s2 =    0.80                  
size of sample 2,    n2=   898                  
                          
difference in sample means =    x̅1-x̅2 =    21.4000   -   2.7   =   18.70  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.3499                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.0668                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   18.7000   -   0   ) /    0.07   =   279.943
                          
Degree of freedom, DF=   n1+n2-2 =    1645                  
  

p-value =        0.000000   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis                      
                          
There is enough evidence that there is a significant difference in the number of esophageal cancer cases in e-cigarette smokers versus nonsmokers

c)

Degree of freedom, DF=   n1+n2-2 =    1645              
t-critical value =    t α/2 =    1.9614   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.3499              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.0668              
margin of error, E = t*SE =    1.9614   *   0.0668   =   0.1310  
                      
difference of means =    x̅1-x̅2 =    21.4000   -   2.700   =   18.7000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    18.7000   -   0.1310   =   18.569
Interval Upper Limit=   (x̅1-x̅2) + E =    18.7000   +   0.1310   =   18.831

THANKS

revert back for doubt

please upvote