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According to a report in USA Today, more and more parents are helping their young adult children purchase their first home. Suppose eight persons in a random sample of 40 young adults who recently purchased a home in Kentucky received help from their parents. You have been asked to construct a 95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents. What is the margin of error for a 95% confidence interval for the population proportion

Respuesta :

ogorwyne

Answer:

the margin of error

= 1.96 x 0.0632

= 0.124

Step-by-step explanation:

this question has the sample size, n = 40

8 people have received help from their parents from this sample.

8/40 = 0.2

which is the sample proportion

z = 1 - 0.2

= 0.8

to calculate standard error

√pz/n

= √0.2 x 0.8/40

= √0.16/40

= 0.0632

at 95% confidence level

z(alpha/2) = 1.96

therefore the margin of error

= 1.96 x 0.0632

= 0.124

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