Question

Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. nbsp 90 % confidence; the sample size is 3200 comma of which 15 % are successes 90% confidence; the sample size is 3200, of which 15% are successes The margin of error Eequals=nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Answer: The margin of error E = 0.0104

Explanation:

The formula to find the margin of error that corresponds to the given statistics and confidence level for population proportion is given by :-

`E=z*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}` , where

n= Sample size

`\hat{p}` = Sample proportion

z* = critical value.

As per given , we have

n= 3200

`\hat{p}=0.15`

Confidence level : 90%

The critical z-value for 90% confidence is z* = =1.645[By z-table]

Substitute all values in the formula , we get

`E=(1.645)\sqrt{(0.15(1-0.15))/(3200)}`

`E=(1.645)√(0.00003984375)`

`E=(1.645)(0.00631219058648)=0.0103835535148\approx0.0104`

Hence, the margin of error E = 0.0104