Question

We wish to estimate the proportion of all Americas that are planning to buy a tablet computer within the next year.

a) How many Americas need to be sampled if we want to estimate the proportion to within 5% with 99% confidence?

b) Assume we sampled the number of Americans found in (a) and 40% say that they will be buying a tablet computer in the next year. Construct and interpret a 90% confidence interval for the proportion of Americas who will make this purchase next year

Answer 1

Solution:

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 5% = 0.05

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.02

Z_/2 = Z_0.02 = 2.054

Sample size = n = ((Z _{/ 2}) / E)² * * (1 - )

= (2.054 / 0.05)² * 0.5 * 0.5

= 421.89

= 422

n = sample size = 422

b ) n = 422

= 40% = 0.400

1 - = 1 - 0.40 = 0.600

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.400 * 0.600) / 422)

= 0.039

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.400 - 0.039 < p < 0.400 + 0.039

0.361 < p < 0.439

The 90% confidence interval for the population proportion p is : ( 0.361 , 0.439)