In: Statistics and Probability
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 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.
Calculate pˆ for the group of renters that are married.
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 |
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