A poll reported in The Sacramento Bee (Nov. 19, 2003, p. A20) that 818 of the...

A poll reported in The Sacramento Bee (Nov. 19, 2003, p. A20) that 818 of the 1515 respondents agreed that homosexual couples could be good parents. The report also stated that,

Source: Pew Research Center for the People and the Press survey of U.S. adults, Oct 15-19;.

The true proportion of the entire adult population at that time who would that agree that homosexual couples could be good parents was unknown (p).

What is the sample proportion (p)?
Calculate the margin of error for a 95% confidence interval.
Find the 95% confidence interval.

Interpret the 95% confidence interval.

Solutions

Expert Solution

Solution :

Given that,

n = 1515

x = 818

Point estimate = sample proportion = = x / n = 818 / 1515 = 0.540

1 - = 1 - 0.540 = 0.460

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 ( $\sqrt$ ((0.540 * 0.460) / 1515 )

= 0.013

A 95% confidence interval for population proportion p is ,

± E

= 0.540 ± 0.013

= ( 0.527, 0.553 )

We are 95% confident that the true proportion of the entire adult population at that time who would that agree that homosexual couples could be good parents between 0.527 and 0.553.

orchestra answered 1 year ago

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