Question

From a random sample of first year college students, 132 out of 200 identified "being well-off financially" as an important personal goal. We are interested in the percentage of all first year college students who have this personal goal. Construct a 90% confidence interval.

		(58.2%, 73.8%)
		(59.4%, 72.6%)
		(61.7%, 70.3%)
		(60.5%, 71.5%)

Answer 1

Solution :

Given that,

n = 200

x = 132

Point estimate = sample proportion = = x / n = 200 / 132 = 0.660

1 - = 1 - 0.660 = 0.340

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

= 1.645 ((0.660*(0.340) /200 )

= 0.055

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.660 - 0.055 < p < 0.660 + 0.055

0.605 < p < 0.715

( 60.5% , 71.5% )

The 90% confidence interval for the population proportion p is : ( 60.5% , 71.5% )

Correct option is :- ( 60.5% , 71.5% )