Answer:
a) For this case the value of the significanceis  and
 and  , we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:
, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

If the calculated statistic  we can reject the null hypothesis at 5% of significance
 we can reject the null hypothesis at 5% of significance
b) Where  
  
c) 
    
d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed. 
Step-by-step explanation:
Data given and notation    
 represent the number of people with the characteristic 1
 represent the number of people with the characteristic 1 
 represent the number of people with the characteristic 2
 represent the number of people with the characteristic 2  
 sample 1 selected
 sample 1 selected  
 sample 2 selected
 sample 2 selected  
 represent the proportion estimated for the sample 1
 represent the proportion estimated for the sample 1  
 represent the proportion estimated for the sample 2
 represent the proportion estimated for the sample 2  
 represent the pooled estimate of p
 represent the pooled estimate of p
z would represent the statistic (variable of interest)    
 represent the value for the test (variable of interest)
 represent the value for the test (variable of interest)  
 significance level given
 significance level given  
Concepts and formulas to use    
We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    
Null hypothesis: 
    
Alternative hypothesis: 
    
We need to apply a z test to compare proportions, and the statistic is given by:    
 (1)
   (1)  
a.State the decision rule.
For this case the value of the significanceis  and
 and  , we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:
, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

If the calculated statistic  we can reject the null hypothesis at 5% of significance
 we can reject the null hypothesis at 5% of significance
b. Compute the pooled proportion.
Where  
  
c. Compute the value of the test statistic.                                                                                              
z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    
Replacing in formula (1) the values obtained we got this:    
 
    
d. What is your decision regarding the null hypothesis?
Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.