According to Runzheimer International, in a survey of relocation administrators

Question

According to Runzheimer International, in a survey of relocation administrators 63% of all workers who rejected relocation offers did so for family considerations. Suppose this figure was obtained by using a simple random sample of the files of 672 workers who had rejected relocation offers. Use this information to construct 98% confidence interval to estimate the population proportion of workers who reject relocatoin offers for family considerations.

Note: I already have the answer, but I am trying to figure out the steps to the solution so please describe in detail. Thank you.

Explanation / Answer

Here p=0.63 and n=672

Now z value for 98% CI is 2.33

So margin of error=z*sqrt(p(1-p)/n)=2.33*sqrt(0.63*0.37/672)=0.043

So CI=p+/-Margin of error=0.63+/-0.043=[0.587,0.673]

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