business statistics

Question

Out of 100 people sampled, 49 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

_________ < p < ________

Answer 1

Hi Salma ,

To construct a 95% confidence interval for the true population proportion of people with kids, you can use the formula for the confidence interval for a proportion:

Confidence Interval = p̂ ± Z * √(p̂ * (1 - p̂) / n)

Where:

- p̂ is the sample proportion.

- Z is the critical value for a 95% confidence level. For a 95% confidence level, Z ≈ 1.96.

- n is the sample size.

Given:

- Sample proportion (p̂) = 49/100 = 0.49

- Sample size (n) = 100

Now, calculate the confidence interval:

Confidence Interval = 0.49 ± 1.96 * √(0.49 * (1 - 0.49) / 100)

Confidence Interval = 0.49 ± 1.96 * √(0.49 * 0.51 / 100)

Confidence Interval = 0.49 ± 1.96 * √(0.02499)

Confidence Interval = 0.49 ± 1.96 * 0.158113883

Now, calculate the upper and lower bounds of the confidence interval:

Lower Bound = 0.49 - 1.96 * 0.158113883 ≈ 0.49 - 0.310337788 ≈ 0.179662212

Upper Bound = 0.49 + 1.96 * 0.158113883 ≈ 0.49 + 0.310337788 ≈ 0.689662212

So, the 95% confidence interval for the true population proportion of people with kids is approximately 0.1797 to 0.6897 (rounded to three decimal places).

The confidence interval can be written as:

0.1797 < p < 0.6897

I hope this will help. I am happy to tutor you on any other questions you may have; please feel free to send me a message!