Question

Suppose a survey was done this year to find out what percentage of all Americans own a bread machine. Out of their random sample of 1,000 Americans, 300 own a bread machine. The margin of error of 95% confidence interval for this survey was plus or minus 3%.

95% confidence interval will be

		0 to 6%
		3% to 6%
		27% to 33%
		None of the above

Answer 1

Solution :

Given that,

n = 1000

x = 300

= x / n = 300 / 1000 = 0.300

1 - = 1 - 0.300 = 0.700

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 * (((0.300 * 0.700) / 1000)

= 0.028

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.300 - 0.028 < p < 0.300 + 0.028

0.272 < p < 0.328

( 27 % t o 33% )

Option  ( 27 % t o 33% ) is correct.