Question

In: Statistics and Probability

Suppose a survey of 500 people age 18 to 34 indicated that 32.2% of them live...

Suppose a survey of 500 people age 18 to 34 indicated that 32.2% of them live

with one or both of their parents. Calculate and interpret a confidence interval estimate for the true proportion

of all people age 18 to 34 who live with one or both parents. Use a 94% confidence level.

Expert Solution

Solution :

Given that,

n = 500

Point estimate = sample proportion = = 0.322

1 - = 1-0.322=0.678

At 94% confidence level

= 1 - 94%

= 1 - 0.94 =0.06

/2 = 0.03

Z/2 = Z_{0.03 = 1.881} ( Using z table )

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

=1.881 ( $\sqrt$ ((0.322*0.678) /500 )

E = 0.0393

A 94% confidence interval for population proportion p is ,

- E < p < + E

0.322 -0.0393< p < 0.322 +0.0393

0.2827< p < 0.3613

The 94% confidence interval for the population proportion p is : 0.2827< p < 0.3613

orchestra answered 1 year ago

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