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The Labor Bureau wants to estimate, at a 90% confidence level, the proportion of all households that receive welfare. A preliminary sample showed that 17.5% of households in this sample receive welfare. The sample size that would limit the margin of error to be within 0.025 of the population proportion is:_________.

Respuesta :

Answer:

Sample size should be atleast 625

Step-by-step explanation:

Given that the  Labor Bureau wants to estimate, at a 90% confidence level, the proportion of all households that receive welfare

Sample proportion = 17.5%

Let n be the sample size

Standard error of sample proportion= [tex]\sqrt{\frac{pq}{n} } =\sqrt{\frac{0.175*0.825}{n} }[/tex]

Z critical for 90% = 1.645

Margin of error = 1.645 * std error

Since margin of error<0.025 we have

[tex]1.645*\sqrt{\frac{0.175*0.825}{n} }<0.025\\0.625046/0.025 <\sqrt{n} \\n>625[/tex]

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