In a random sample of 800 men aged 25 to 35 years, 24% said they live...

In a random sample of 800 men aged 25 to 35 years, 24% said they live with one or both parents. Construct a 95% confidence interval for the proportion of men aged 25 to 35 that live with their parents. Make sure to include every step and an interpretation.

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Expert Solution

Solution :

The 95% confidence interval for population proportion is given as follows :

Where, p̂ is sample proportion, q̂ = 1 - p̂, n is sample size and Z(0.05/2) is critical z-value to construct 95% confidence interval.

Sample proportion of men aged 25 to 35 years, who live with one or both parents is given by,

n = 800

Using Z-table we get, Z(0.05/2) = 1.96

Hence, 95% confidence interval for the proportion of men aged 25 to 35 that live with their parents is,

The 95% confidence interval for the proportion of men aged 25 to 35 that live with their parents is (0.2104, 0.2696).

Interpretation : We are 95% confident that the proportion of men aged 25 to 35 that live with their parents, lies between 0.2104 and 0.2696.

In other words, we are 95% confident that the percentage of men aged 25 to 35 that live with their parents, lies between 21.04% and 26.96%.

Please rate the answer. Thank you.

orchestra answered 1 year ago

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