Question

The National Health Interview Survey surveyed 1,480Americans and asked them how confident they felt that they could get health-related advice or information if they needed it. Of those surveyed, 23.51% said they felt completely confident that they could get health-related advice or information.

6. Construct a 99% confidence interval for the percentage of Americans who say they could get health-related advice or information if they needed it.

7. Construct a 90% confidence interval for the percentage of Americans who say they could get health-related advice or information if they needed it.

8. Compare the relative width of these two confidence intervals. Why do they differ?

9. Interpret one of these confidence intervals.

Answer 1

6)

Level of Significance,   α =    0.01          
Sample Size,   n =    1480          
                  
Sample Proportion ,    p̂ = x/n =    0.2351   
z -value =   Zα/2 =    2.576   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0110          
margin of error , E = Z*SE =    2.576   *   0.0110   =   0.0284
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.2351 -   0.0284   =   0.2067
Interval Upper Limit = p̂ + E =   0.2351 +   0.0284   =   0.2635
                  
99%   confidence interval is (   0.207   < p <    0.263   )

7)

Level of Significance,   α =    0.10          
Sample Size,   n =    1480          
                  
Sample Proportion ,    p̂ = x/n =    0.2351   
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0110          
margin of error , E = Z*SE =    1.645   *   0.0110   =   0.0181
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.2351 -   0.0181   =   0.2170
Interval Upper Limit = p̂ + E =   0.2351 +   0.0181   =   0.2532
                  
90%   confidence interval is (   0.217   < p <    0.253   )

9)

99% confidence interval is wider than that of 90%

because for 99% CI, critical value is larger

9)

there is 99% confidence that true  percentage of Americans who say they could get health-related advice or information if they needed it lies with confidence interval (   0.207   < p <    0.263   )