If n=530 and p' (p-prime) = 0.78, construct a 90% confidence interval. Give your answers to three decimals. ________< p < ________

Answer 1

Hi Salma,

To construct a 90% confidence interval for a population proportion (p), you can use the formula for the confidence interval:

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

Where:

- p̂ (p-hat) is the sample proportion.

- Z is the critical value for a 90% confidence level.

- n is the sample size.

Given:

- Sample proportion (p̂) = 0.78

- Sample size (n) = 530

To find the critical value Z for a 90% confidence level, you can refer to a standard normal distribution table or use a calculator. For a 90% confidence level, Z is approximately 1.645.

Now, calculate the confidence interval:

Confidence Interval = 0.78 ± 1.645 * √(0.78 * (1 - 0.78) / 530)

Confidence Interval = 0.78 ± 1.645 * √(0.78 * 0.22 / 530)

Confidence Interval = 0.78 ± 1.645 * √(0.1644 / 530)

Confidence Interval = 0.78 ± 1.645 * √(0.00031057)

Confidence Interval = 0.78 ± 1.645 * 0.017625

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

Lower Bound = 0.78 - 1.645 * 0.017625 ≈ 0.78 - 0.029011 ≈ 0.750989

Upper Bound = 0.78 + 1.645 * 0.017625 ≈ 0.78 + 0.029011 ≈ 0.809011

So, the 90% confidence interval for the population proportion (p) is approximately 0.751 to 0.809 when rounded to three decimal places.

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!

Answer 2

Salma,

You posted this question earlier without the notation specification. The previous confidence interval, converted to that notation, is:

0.750 < p < 0.810