Question

Diabetes In a simple random sample of 1200 people from the First Nations population in Canada, the proportion with diabetes was found to be 0.175 (or 17.5%, 210 out of 1200).

a. What is the estimated standard deviation for the sample percentage of First Nations people who have diabetes?

b. Find the margin of error, using a 95% confidence level, for estimating this proportion.

c. Report the 95% confidence interval for the proportion of all First Nation people in Canada who have diabetes.

d. Statistics Canada reported in 2009 that 6.8%, or 0.068, of non-First Nations Canadians, have diabetes. Does the confidence interval you found in part c support or refute the claim that the proportion of First Nations people in Canada who have diabetes is higher than the proportion of non–First Nations people? Explain.

Answer 1

Given : = 0.175 , n =1200

a) Estimated standard deviation =  =   

Estimated standard deviation = 0.011

b) Margin of error (E) : z *

Confidence level = 0.95 ,  Therefore α = 1 - 0.95 = 0.05 and (α/2) = 0.025

zα/2 = 1.96 ---- ( from the z score table )

Margin of error (E) = 1.96*0.011

Margin of error (E) = 0.0215

c)

Lower bound = - E = 0.175 - 0.0215 = 0.1535

Upper bound = + E = 0.175 + 0.0215 = 0.1965

95% confidence interval for the proportion of all First Nation people in Canada who have diabetes is ( 0.1535 , 0.1965 )

d) We are given population proportion P = 0.068 , it is not fall in the interval ( 0.1535 , 0.1965 ) and it is fall below the lower limit 0.1535

Therefore it supports the claim that the proportion of First Nations people in Canada who have diabetes is higher than the proportion of non–First Nations people.