Question

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 203 with 16% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

C.I. =

Explain step by step

Answer 1

In order to Construct a confidence Interval for population proportion the first step is that we need to check the following

(i)np≥5

(ii)n(1-p)≥5

Under these condition the sampling distribution of the sample proportion is approximately normal with mean p and standard deviation

Here n=203 and p=16/100

(i)203*16/100=32.48≥5

(ii)203*84/100=170.52≥5

sampling distribution of the sample proportion is approximately normal

then the general form of confidence interval is

point estimatemargin of error *standard error of the estimate

Here the point estimate is the sample proportion and the standard error is given by

and the margin of error is the critical value which is   

then the Confidence interval is given by

*

in other words

now

=16/100=0.16

=2.326

*