Question

Use the one-proportion z-interval procedure to find the required confidence interval.

A researcher wishes to estimate the proportion of adults in the city of Darby who are vegetarian. In a random sample of 1323 adults from this city, the proportion that are vegetarian is 0.068. Find a 90% confidence interval for the proportion of all adults in the city of Darby that are vegetarians.

Answer 1

Solution :

n = 1323

=0.068

1 - = 1 - 0.068 = 0.932

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 * (((0.068 * 0.932) / 1323)

=0.011

A 90 % confidence interval for population proportion p is

- E < P < + E

0.068 - 0.011 < p < 0.068 + 0.011

0.057 < p < 0.079