Question

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 379 with 299 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Answer 1

n = sample size = 379

x = success = 299

Sample proportion formula

We find the Z_c Critical Using z table

C = level of confidence = 0.90 ( We converted 90% in to decimal )

We calculate the area

We look for the area 0.9500 inside the body of the table

We get the area 0.9505 and 0.9495 Both are close to 0.9500

we get the two z score as 1.64 and 1.65

We take the average of of z scores to get the critical value

we calculate margin of error

Confidence interval formula

Round the confidence interval up to 3 decimal place

Final answer :

The required confidence interval for the population proportion P in the form of tri linear inequality is

Some more information Using technology

If you have Ti - 84 or Ti -83 calculate you can calculate this confidence interval directly

Press " STAT"then select " Tests " then select 1 Prop Z Int

x =299

n=379

C-level = 0.90

You get the same answer

I hope this will help you :)