Question

Construct the 99% confidence interval estimate of the population proportion p if the sample size is n=900 and the number of successes in the sample is x=333. Use the Agrest-Coull method.

Answer 1

Answer:

An 99% confidence interval  of the given proportion

(0.355 , 0.385)

Step-by-step explanation:

Given sample size n= 900

the number of successes in the sample is x=333

The proportion P = $\frac{x}{n} = \frac{333}{900} = 0.37$

            Q = 1-P =1 - 0.37 = 0.63

Confidence interval:-

99% of confidence interval zα = 2.93

$(P - z_{\alpha } \sqrt{\frac{PQ}{n} } , P + z_{\alpha } \sqrt{\frac{PQ}{n} })$

$(0.37 - 2.93 \sqrt{\frac{0.37(0.63}{900} } ,0.37 +2.93 \sqrt{\frac{0.37(0.63}{900} } })$

(0.37 - 0.015 , 0.37 + 0.015)

(0.355 , 0.385)

Conclusion:-

An 99% Confidence interval (0.355 , 0.385)