Question

Let us return to the content in Exercise 49. The data in Table 7.1 indicates the results of a GSS survey which asked white citizens who have either never been married or are married whether they own or are buying versus rent their home. Let us use confidence intervals to compare people who own or are buying a home among those that are married versus those who pay rent among those that are married.
Calculate pˆ for the group of homeowners among those that are married.
For a confidence level of ell = .97, determine the z-score for which 97% of normally
distributed data falls within z deviations of the mean. Review Example 7.1.
Use Equation 7.1 to find the standard error and then use Equation 7.2 to determine the
confidence interval for p.
Review the section notes to carefully explain what the interval tells us.
Repeat the above steps for the group of renters among that that are married.
Review the section notes to carefully explain what the interval tells us.
Now compare these two intervals. Do the intervals overlap or not? What association do we have or not have between marriage and homeownership due to whether or not the intervals overlap?
	Married	Never Married	Total
				Owns or is Buying	9,178	1,785	10,963
Pays Rent	1,867	2,282	4,149
				Total	11,045	4,067	15,112

Answer 1

Number of Items of Interest,   x =   9178      
Sample Size,   n =    10963      
Sample Proportion ,    p̂ = x/n =    0.8372      
              
Level of Significance,   α =    0.03      
z -value =   Zα/2 =    2.170   [excel formula =NORMSINV(α/2)]  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0035      
margin of error , E = Z*SE =    2.17*0.0035=   0.0077      
              
Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.8372-0.0077=   0.8295      
Interval Upper Limit = p̂ + E =   0.8372+0.0077=   0.8448      
              
97% Confidence interval is (   0.830   < p <    0.845   )
======================

Number of Items of Interest,   x =   11045      
Sample Size,   n =    15112      
Sample Proportion ,    p̂ = x/n =    0.7309      
              
Level of Significance,   α =    0.03      
z -value =   Zα/2 =    2.170   [excel formula =NORMSINV(α/2)]  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0036      
margin of error , E = Z*SE =    2.17*0.0036=   0.0078      
              
Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.7309-0.0078=   0.7230      
Interval Upper Limit = p̂ + E =   0.7309+0.0078=   0.7387      
              
97% Confidence interval is (   0.723   < p <    0.739   )

two intervals do not overlap each other, hence, there is significant difference in proportion of  homeowners that are married and renters that are married