1. Nationally, 37.8% of the people in the United States are obese. Determine if

Question

1. Nationally, 37.8% of the people in the United States are obese. Determine if the proportion of people who are obese in the Framingham population differs from the rest of the country by calculating the proportion and two-sided 95% confidence interval for the sample. (10 points)

2. Around 10.7% of the people in the United States are diabetic. Is the proportion of people who are diabetic in Framingham sample different from this national value? Determine this by calculating the proportion for the Framingham sample and its two-sided 95% confidence interval. (10 points)

3. Physicians believe that obesity and diabetes are related. Using the sample of subjects from Framingham, determine if being obese and having diabetes are independent. Use a significance level of =0.05.

a. State your null and alternative hypotheses (3 points)

b. Calculate the appropriate test statistic. (9 points)

c. Explain the results in a way that the physicians will be able to understand it. (3 points)

4. The physicians also believe that people who are obese are more likely to die earlier than those who are not. Again use the data in the Framingham sample to test this theory by comparing the mean number of days until death for the two groups. Use a significance level of =0.05.

a. State your null and alternative hypotheses (3 points)

b. Calculate the appropriate test statistic. (9 points)

c. Explain the results in a way that the physicians will be able to understand it. (3 points)

Explanation / Answer

Solution:

1) z_c = 1.96 at 95% confidence
proportion = p = 0.378
standard error = s.e. = [p(1-p)/n] = (0.2351/n)
margin of error = E = s.e.× z_c = (0.2351/n) × 1.96
95% confidence interval for this proportion = ( 0.378 - (0.2351/n) × 1.96 , 0.378 + (0.2351/n) × 1.96 )

= ( 0.378 - (0.2351/4434) × 1.96 , 0.378 + (0.2351/4434) × 1.96 )

= (0.1152, 0.1438)

2)

Test Hypothesis:
H0 : p = 0.107
Ha : p 0.107
Here n = 4434, x = 121 and p0 = 0.107

np0 = 4434(0.107) = 121
n(1-p0) = 4434(1-0.107) = 4313
p-hat = x/n = 121/4434 = 0.0273

Standard error SE = sqrt(p0(1-p0)/n)
= sqrt(0.107(1-0.107)/4434)
= 0.004642
Then, the 95% confidence interval for the proportion of diabetic people can be defined as,
= p ± Z *SE
= (0.0273) ± (1.96)(0.004642)
= (0.0183, 0.0364)

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