Question

Problem 2: The product manager of a large smart home security device company would like to conduct a survey to find the proportion p of consumers were happy with a new surveillance camera they bought.
(a) How many samples are needed to estimate p with 2% margin of error and 90% confidence?
(b) When 1000 consumes were surveyed, 400 consumers responded that they were satisfied with their security cameras. Find 95% confidence interval for p.

Answer 1

Solution:

a ) Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 2% = 0.04

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

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

= (1.645 / 0.02)² * 0.5 * 0.5

= 1691.26

= 1691

n = sample size = 1691

b ) Given that,

n = 1000

x = 400

= x / n = 225 /1000 = 0.400

1 - = 1 - 0.400 = 0.600

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (((0.400 * 0.600) /1000)

= 0.024

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.400 - 0.030 < p < 0.400 + 0.030

0.370 < p < 0.430