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 )**

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