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Land's Beginning is a company that sells its merchandise through the mail. It is considering buying a list of addresses from a magazine. The magazine claims that at least 25% of its subscribers have high incomes (they define this to be household income in excess of $100,000). Land's Beginning would like to estimate the proportion of high-income people on the list. Checking income is very difficult and expensive but another company offers this service. Land's Beginning will pay to find incomes for an SRS of people on the magazine's list. They would like the margin of error of the 95% confidence interval for the proportion to be 0.05 or less. Use the guessed value p = 0.25 to find the required sample size.

Answer :

Solution :

It is given that we use CI = 95%

Therefore, the value of z = 1.96 as the [tex]$P(-1.96 <z<1.96)=0.95$[/tex]

Also, here it is given that E = 0.05 and the value of p = 0.25

Thus from the formula of E, we can find n

[tex]$E= z \times \sqrt{\frac{pq}{n}}[/tex]

[tex]$n= \left(\frac{z}{E}\right)^2 \times p \times q$[/tex]

[tex]$n= \left(\frac{1.96}{0.05}\right)^2 \times 0.25 \times 0.75$[/tex]

  = 288.12

  = 289

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