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shusha [124]
3 years ago
5

Construct the 99% confidence interval estimate of the population proportion p if the sample size is n=900 and the number of succ

esses in the sample is x=333. Use the Agrest-Coull method.
Mathematics
1 answer:
oee [108]3 years ago
7 0

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

<u>Confidence interval</u>:-

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)

<u>Conclusion</u>:-

<u>An 99% Confidence interval (0.355 , 0.385)</u>

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