Question

# The National Health Interview Survey surveyed 1,480Americans and asked them how confident they felt that they...

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.

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   )