Question

In a poll 1040 of 2260 respondents said they would prefer to live somewhere else. Confirm the distribution for p-hat is approximately normal and find a 95% confidence interval for the proportion of people who prefer to live somewhere else.

Answer 1

Sample proportion = 1040 / 2260 = 0.460

n * = 2260 * 0.460 = 1040 >= 10

n ( 1 - ) = 2260 * ( 1 - 0.460) = 1220.4 >= 10

Since n >= 10 and n(1-) >=10 , distribution of is approximately normal.

95% confidence interval for p is

- Z * sqrt ( (1-) / n) < p < + Z * sqrt ( (1-) / n)

0.460 - 1.96 * sqrt( 0.46 * ( 1 - 0.46) / 2260) < p < 0.460 + 1.96 * sqrt( 0.46 * ( 1 - 0.46) / 2260)

0.439 < p < 0.481

95% CI is ( 0.439 , 0.481 )