Question

A medical group studies the ailments of adults with diabetes. Of 8750 adults who are diabetic, 2100 have high cholesterol. Of 12350 adults who are not diabetic, 1482 have high cholesterol. At 99% confidence can you show the proportions of adults with high cholesterol differs depending on being diabetic?

Answer 1

Solution:

Confidence interval for difference between two population proportions:

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Solution:

We are given

P1 = 2100/8750 = 0.24

N1 = 8750

P2 = 1482/12350 = 0.12

N2 = 12350

(P1 – P2) = 0.24 – 0.12 = 0.12

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence interval = (P1 – P2) ± Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Confidence interval = 0.12 ± 2.5758*sqrt[(0.24*(1 – 0.24)/8750) + (0.12*(1 – 0.12)/12350)]

Confidence interval = 0.12 ± 2.5758*0.0054

Confidence interval = 0.12 ± 0.0140

Lower limit = 0.12 - 0.0140 = 0.1060

Upper limit = 0.12 + 0.0140 = 0.1340

Confidence interval = (0.1060, 0.1340)

The proportions of adults with high cholesterol differ depending on being diabetic, because above confidence interval does not contain the value 0.