The FBI wants to determine the effectiveness of their 10 Most Wanted list. To do so, they need to find out the fraction of people who appear on the list that are actually caught. In an earlier study, the population proportion was estimated to be 0.26. How large a sample would be required in order to estimate the fraction of people who are captured after appearing on the 10 Most Wanted list at the 98% confidence level with an error of at most 0.04?

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yogeshkumar49685 yogeshkumar49685 · Answer 1 · 2022-07-20T14:31:59+02:00

The sample size required to estimate the fraction of people who are captured after appearing on the 10 most wanted list is 228.

Given standard deviation of 0.26 ,margin of error of 0.04, and confidence interval of 98%.

We have to determine the sample size required to estimate the fraction of people who are captured after appearing on the 10 ost wanted list.

We can find the sample size with the help of margin of error.

Margin of error is the difference between the calculated values and real values.

Margin of error=z*σ/[tex]\sqrt{n}[/tex]

where σ is standard deviation,

n is sample size and z is critial z value.

We have to find the z value for 98% confidence level from z table.

z value=2.326.

Put the values in the formula of margin of error.

0.04=2.326*0.26/[tex]\sqrt{n}[/tex]

[tex]\sqrt{n}[/tex]=2.326*0.26/0.04

[tex]\sqrt{n}[/tex]=15.119

squaring both sides we get

n=228.58

By rounding we get

n=228.

Hence the sample size needed is 228.

Learn more about margin of error at https://brainly.com/question/10218601

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