Question

80. An NHANES report gives data for 654 women aged 20 to 29 years. The mean BMI of these 654 women was 26.8. On the basis of the sample, estimate the mean BMI in the population of all 20.6 million women in the age group. Let us assume that the sample is from a normal population with standard deviation of 7.5.

    1. a) Construct a 95% confidence interval for mean BMI of all women

    2. b) Construct both a 90% and 99% confidence interval for the mean BMI in this population?

    3. c) How does increasing the confidence level change the margin of error of a confidence interval when the

       sample size and population standard deviation remain the same?

    4. d) Suppose that the survey above had a sample size of just 100 women. What is the 95% confidence interval

       for this situation? Is it different from the initial case?

    5. e) Construct a 95% confidence interval for samples with 400 and 1600 women. How are these confidence

       intervals different?

Answer 1

x̅ = 26.8, σ = 7.5, n = 654

a) 95% Confidence interval :

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.96

Margin of error , E = z-crit*σ/√n = 1.96*7.5/√654 = 0.5748

95% Confidence interval :

Lower Bound = x̅ - E = 26.8 - 0.5748 = 26.2252

Upper Bound = x̅ + E = 26.8 + 0.5748 = 27.3748

b)

90% Confidence interval :

At α = 0.1 two tailed critical value, z_c = ABS(NORM.S.INV(0.1/2)) = 1.645

Margin of error , E = z-crit*σ/√n = 1.6449*7.5/√654 = 0.4824

Lower Bound = x̅ - E = 26.8 - 0.4824 = 26.3176

Upper Bound = x̅ + E = 26.8 + 0.4824 = 27.2824

99% Confidence interval :

At α = 0.01 two tailed critical value, z_c = ABS(NORM.S.INV(0.01/2)) = 2.576

Margin of error , E = z-crit*σ/√n = 2.5758*7.5/√654 = 0.7554

Lower Bound = x̅ - E = 26.8 - 0.7554 = 26.0446

Upper Bound = x̅ + E = 26.8 + 0.7554 = 27.5554

c)

The margin of error increases as the confidence level increases when the sample size and population standard deviation remain the same

d)

95% Confidence interval :

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.96

Margin of error , E = z-crit*σ/√n = 1.96*7.5/√100 = 1.47

Lower Bound = x̅ - E = 26.8 - 1.47 = 25.33

Upper Bound = x̅ + E = 26.8 + 1.47 = 28.27

Yes it is different from the initial case.

e)

For n = 400

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.96

Margin of error , E = z-crit*σ/√n = 1.96*7.5/√400 = 0.735

95% Confidence interval :

Lower Bound = x̅ - E = 26.8 - 0.735 = 26.065

Upper Bound = x̅ + E = 26.8 + 0.735 = 27.535

For n = 1600:

At α = 0.05 two tailed critical value, z_c = ABS(NORM.S.INV(0.05/2)) = 1.96

Margin of error , E = z-crit*σ/√n = 1.96*7.5/√1600 = 0.3675

95% Confidence interval :

Lower Bound = x̅ - E = 26.8 - 0.3675 = 26.4325

Upper Bound = x̅ + E = 26.8 + 0.3675 = 27.1675

As the sample size increases the length of confidence interval decreases.