Question

A new study suggests that women with asthma exposed to air pollution, even before conception, significantly increase their risk of delivering a premature baby. The researcher reported gestational ages for two groups of women with asthma – one group of 44 women exposed to air pollution (mean = 35 weeks with a standard deviation of 1 week) and another group of 48 women not exposed to air pollution (mean = 39 weeks with a standard deviation of 1 week). Which one of the following statements is INCORRECT? (Set alpha level at 0.05; Pooled standard deviation = 1)

The obtained test statistic is approximately -19.17.

The critical test statistic is 1.960.

The null hypothesis is that gestational age is not different between women with asthma who are exposed to air pollution and those who are not exposed to air pollution.

The decision should be to reject the null hypothesis.

The ratio of variances is 1.

The degrees of freedom is 91.

need help answering question

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   A                  
mean of sample 1,    x̅1=   35.0000                  
standard deviation of sample 1,   s1 =    1.0000                  
size of sample 1,    n1=   44                  
                          
Sample #2   ---->   B                  
mean of sample 2,    x̅2=   39.000                  
standard deviation of sample 2,   s2 =    1.0000                  
size of sample 2,    n2=   48                  
                          
difference in sample means =    x̅1-x̅2 =    35.0000   -   39.000   =   -4.0000  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.00                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.2087                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -4.0000   -   0   ) /    0.2087   = -19.17
                          
Degree of freedom, DF=   n1+n2-2 = 90                 
t-critical value , t* = ±1.96 (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                      

Conclusion:     p-value <α , Reject null hypothesis  

so, every thing is correct excel degree of freedom, because df is 90

so, answer is degree of freedom