Construct a 90​% confidence interval of the population proportion using the given information. x=120, n=200

Question

Construct a

90​%

confidence interval of the population proportion using the given information.

x=120, n=200

Answer 1

To construct a 90% confidence interval for the population proportion, we can use the formula:

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

Where:

p̂ is the sample proportion (x/n),

Z is the Z-value corresponding to the desired confidence level (90% confidence level corresponds to a Z-value of approximately 1.645),

n is the sample size.

Given x = 120 and n = 200, we can calculate p̂:

p̂ = x/n = 120/200 = 0.6

Now, substituting the values into the formula:

Confidence Interval = 0.6 ± 1.645 * √[(0.6(1 - 0.6)) / 200]

Calculating the value inside the square root:

(0.6(1 - 0.6)) / 200 = 0.24 / 200 = 0.0012

Calculating the square root:

√(0.0012) ≈ 0.0346

Calculating the confidence interval:

Confidence Interval = 0.6 ± 1.645 * 0.0346

Confidence Interval ≈ 0.6 ± 0.0569

Thus, the 90% confidence interval for the population proportion is approximately 0.5431 to 0.6569.

Answer 2

Here is the step by step solutions for calculating a confidence interval for proportions. Good luck!