Question

A researcher conducts a study of white and black attitudes toward the police in his state.

The percentage of a random sample of white respondents (N = 300) who say they have a favorable attitude toward the police is 61%. The percentage of a random sample of black respondents (N = 250) who say they have a favorable attitude toward the police is 47%.

• Is there a real (statistically significant) difference between the percentage of Whites and African Americans who have a positive attitude toward the police in the larger population, or are these results likely to have occurred by chance?
• Construct a 95% confidence interval for the proportion of Whites in the population who have a favorable attitude toward the police.
• Construct a 95% confidence interval for the proportion of African Americans in the population who have a favorable attitude toward the police.

Please show work. I want to learn how to execute the question. Class does not use any statistical softwares or excel.

Answer 1

a)

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->   white          
first sample size,     n1=   300          
proportion success of sample 1 , p̂1=      0.6100  

number of successes, sample 1 =     x1= n1*p̂1 = 183          
       
                  
sample #2   ----->   black          
second sample size,     n2 =    250          
proportion success of sample 1 , p̂ 2 =    0.4700          
number of successes, sample 2 = n2* p̂ 2 = 117.5          
                  
difference in sample proportions, p̂1 - p̂2 =     0.6100   -   0.4700   =   0.1400
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.5464          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0426          
Z-statistic = (p̂1 - p̂2)/SE = (   0.140   /   0.0426   ) =   3.284
                  
p-value =        0.00102 [excel formula =2*NORMSDIST(z) or can be find from z table]      
decision :    p-value<α,Reject null hypothesis               
                  
Conclusion:   There is enough evidence to conclude that there a real (statistically significant) difference between the percentage of Whites and African Americans who have a positive attitude toward the police in the larger population

b)

Level of Significance,   α =    0.05          
Sample Size,   n =    300          
                  
Sample Proportion ,    p̂ = x/n =    0.610          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0282          
margin of error , E = Z*SE =    1.960   *   0.0282   =   0.0552
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.610   -   0.0552   =   0.5548
Interval Upper Limit = p̂ + E =   0.610   +   0.0552   =   0.6652
                  
95%   confidence interval is (   0.5548   < p <    0.6652   )

c)

Level of Significance,   α =    0.05          
Sample Size,   n =    250          
                  
Sample Proportion ,    p̂ = x/n =    0.470          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0316          
margin of error , E = Z*SE =    1.960   *   0.0316   =   0.0619
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.470   -   0.0619   =   0.4081
Interval Upper Limit = p̂ + E =   0.470   +   0.0619   =   0.5319
                  
95%   confidence interval is (   0.4081   < p <    0.5319   )

please revert for doubts..