Question

When reviewing health records, a sample of size 280 indicates that 46% of Americans over the age of 45 suffer from type II diabetes. Use the 95% confidence level to:

25. Estimate the margin of error. Thousandths

26. Estimate the true proportion. Thousandths

27. If you are conducting a new study using the 95% confidence level and no prior sample proportion information is known, estimate the sample size needed to be within a margin of error of ±3%

Answer 1

Solution:

Given,

n = 280 ....... Sample size

= 0.46

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [ 0.46 *(1 - 0.46 )/280 ]

= 0.058

Now the confidence interval is given by

( - E)   ( + E)

( 0.46 - 0.058 )   ( 0.46 + 0.058 )

0.402   0.518  

Required 95% Confidence Interval is ( 0.402 , 0.518 )

Solution:

Given,

E = 0.03

c = 95% = 0.95

When is not given then we use,

or p = 0.5

1 - 1- p = 1 - 0.5 = 0.5

Now,

= 1 - c = 1 - 0.95 = 0.05

/2 = 0.025

= 1.96 (using z table)

The sample size for estimating the proportion is given by

n =

= (1.96)2 * 0.5 * 0.5 / (0.032)

= 1067.1111

= 1068 ..(round to the next whole number)

Answer : Required Sample size is n = 1068