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An advertiser wishes to estimate the proportion of adults in Utah who already own a gym membership. He wishes to create a 90% confidence interval with a margin of error of 0.10. How many people must he sample?

Respuesta :

Answer:

Step-by-step explanation:

Given that an advertiser wishes to estimate the proportion of adults in Utah who already own a gym membership.

Confidence level wanted = 90%

Critical value Z for 90% = 1.645

p is not known hence for maximum std error we know p =0.5

Let us take p =0.5

Margin of error = [tex]1.645*\sqrt{\frac{pq}{n} } =\\1.645*0.5(\frac{1}{\sqrt{n} } ) 0.10\\n=67.65\\n~68[/tex]

Sample size can be 68

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