Question

Construct a 99% confidence interval of the population proportion using the given information. x = 40, n=200 Click here to view the table of critical values. The lower bound is The upper bound is (Round to three decimal places as needed.)

Answer 1

Solution:

We need to construct the 99% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:

Favorable Cases X =	40
Sample Size N =	200

The sample proportion is computed as follows, based on the sample size N=200 and the number of favorable cases X=40

p^​=X​/N=40/200​=0.2

The critical value for α=0.01 is zc​=z1−α/2​=2.576. The corresponding confidence interval is computed as shown below:

p(1-P) p(1-) CI(Proportion) P -%c - + V n2 n 0.2 - 2.576 x 0.2(1 -0.2) 0.2 +2.576 x 200 0.2(1 -0.2) 200 (0.127, 0.273)

Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.127<p<0.273, which indicates that we are 99% confident that the true population proportion pp is contained by the interval (0.127,0.273).

Lower Bound= 0.127

Upper Bound= 0.273

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