Question

The Center for Disease Control takes periodic randomized surveys to track health of Americans. In a survey of 4881 adults in 2007-2008, 1674 were obese (body mass index BMI ≥ 30). In a survey of 5181 adults in 2011-2012, 1808 were obese. Complete parts A, B, and C by hand.

a)   Estimate with 95% confidence the change in the population proportion from the first survey to the second survey that were obese and interpret the results.

b)   The standard error for this difference is quite small. What is the primary reason that causes the SE to be so small?

c) Conduct a hypothesis test for a difference between the two proportions at a 5% level of significance. Include hypotheses, p-value, decision and conclusion.

d)   Use Minitab to complete the above exercise. Upload your results or attach output to assignment.

Answer 1

Result:

The Center for Disease Control takes periodic randomized surveys to track health of Americans. In a survey of 4881 adults in 2007-2008, 1674 were obese (body mass index BMI ≥ 30). In a survey of 5181 adults in 2011-2012, 1808 were obese. Complete parts A, B, and C by hand.

a) Estimate with 95% confidence the change in the population proportion from the first survey to the second survey that were obese and interpret the results.

CI = (p1-p2) ± z*se

Z value at 95% =1.96

p1=1674/4881 = 0.343

p2=1808/5181 = 0.349

p1-p2=-0.006

Standard error of p1-p2

=0.0095

95% CI = -0.006 ± 1.96*0.0095

= (-0.0246, 0.0126)

b) The standard error for this difference is quite small. What is the primary reason that causes the SE to be so small?

Since sample size is very large, standard error for this difference is small.

c) Conduct a hypothesis test for a difference between the two proportions at a 5% level of significance. Include hypotheses, p-value, decision and conclusion.

Ho: P₁ = P₂   H₁: P₁ ≠ P₂

p1=1674/4881 = 0.343

p2=1808/5181 = 0.349

P = (x1+x2)/(n1+n2)

=(1674+1808)/(4881+5181) =0.3461

= -0.6328

Table value of z at 0.05 level = 1.96

Rejection Region: Reject Ho if z < -1.96 or z > 1.96

Calculated z = -0.6328 not in the rejection region

The null hypothesis is not rejected.

P value =0.5269 which is > 0.05 level of significance.

We conclude that there is no difference between the two proportions.

d)   Use Minitab to complete the above exercise. Upload your results or attach output to assignment.

Test and CI for Two Proportions

Method

p₁: proportion where Sample 1 = Event

p₂: proportion where Sample 2 = Event

Difference: p₁ - p₂

Descriptive Statistics

Sample	N	Event	Sample p
Sample 1	4881	1674	0.342963
Sample 2	5181	1808	0.348967

Estimation for Difference

Difference	95% CI for Difference
-0.0060049	(-0.024600, 0.012591)

CI based on normal approximation

Test

Null hypothesis	H₀: p₁ - p₂ = 0
Alternative hypothesis	H₁: p₁ - p₂ ≠ 0

Method	Z-Value	P-Value
Normal approximation	-0.63	0.527
Fisher's exact		0.529