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lana [24]
3 years ago
12

A researcher wants to construct a 90% confidence interval for the proportion of elementary school students in Seward County who

receive free or reduced-price school lunches. A state-wide survey indicates that the proportion is 0.50. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.1?
Mathematics
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

The sample  size is  n =68

Step-by-step explanation:

    The population proportion is  \^ p =  0.50

     The margin of error is  E =  0.1

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the sample size is mathematically represented as  

    n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>  n = [\frac{1.645 }{0.1} ]^2 * 0.50  (1 - 0.50 )

=>  n =68

     

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