We need to find the homeownership proportion in Southern California. The homeownership proportion is believed to be around 50%. How many householders would you survey for a 93% CI where the maximum likely sampling error is 1.5%?

Answer :

Answer:

The value is  [tex]n = 3648 [/tex]

Step-by-step explanation:

From the question we are told that

     The sample proportion is  [tex]\^ p = 0.5[/tex]

     The margin of error is  [tex]E = 0.015[/tex]

From the question we are told the confidence level is  93% , hence the level of significance is    

      [tex]\alpha = (100 - 93 ) \%[/tex]

=>   [tex]\alpha = 0.07[/tex]

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

   [tex]Z_{\frac{\alpha }{2} } =  1.812[/tex]

Generally the sample size is mathematically represented as  

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

=>  [tex]n = [\frac{ 1.812 }{ 0.015} ]^2 *0.5 (1 - 0.5 ) [/tex]  

=>  [tex]n = 3648 [/tex]

 

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