Question

Question: do these data represent compelling evidence that the proportion of unemployed 18-29 year-olds that have diabetes is di↵erent from the proportion of employed 18-29 year-olds that have diabetes?

[Counts]	Unemployed	Employed	Total
Diabetes	146	717	863
No Diabetes	5709	47057	57766
Total	5855	47774	58629

        (a) In the notation we are using in this course, what is the value of n2?

(b) In the notation we are using in this course, what is the value of pˆ1?

       (c) In the notation we are using in this course, what is the value of pˆ2?

      (d) Verify that the conditions are met for you to test H0 using a confidence interval approach.

     (e) Test H0 at the 0.001 significance level, using a confidence interval approach. Conclude with a clear

verdict as to whether you reject the null.

    (f) Making reference to your work so far this problem, comment on the distinction between a difference being “significant” and a difference being “large.”

Answer 1

a) n2 = 47774

b)

n1 = 5855

x1 = 146

p̂1 = x1/n1 = 0.0249

c)

n2 = 47774

x2 = 717

p̂2 = x2/n2 = 0.0150

d) Assumption:

The data is random and independent of each other.

The sample size must be sufficiently large.

When the sample size, n, should be no more than 10% of the population.

e)

99.9% Confidence interval for the difference:

At α = 0.001, two tailed critical value, z_c = NORM.S.INV(0.001/2) = 3.291

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ] = (0.0249 - 0.015) - 3.291*√[(0.0249*0.9751/5855) + (0.015*0.985/47774)] = 0.0030

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ] = (0.0249 - 0.015) + 3.291*√[(0.0249*0.9751/5855) + (0.015*0.985/47774)] = 0.0169

0.003 < p1 -p2 < 0.0169

As the confidence interval do not contain 0, so we reject the null hypothesis.