According to a report in USA Today, more and more parents are helping their young adult children buy homes. You would like to construct a 90% confidence interval of the proportion of all young adults in Louisville, Kentucky, who received help from their parents in buying a home. How large a sample should you take so that the margin of error is no more than 0.06?

Respuesta :

Answer:

The value is [tex]n = 188 [/tex]

Step-by-step explanation:

From the question we are told that

    The margin of error is  [tex]E = 0.06[/tex]

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

      [tex]\alpha = (100 - 90 ) \%[/tex]

=>   [tex]\alpha = 0.10[/tex]

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

   [tex]Z_{\frac{\alpha }{2} } =  1.645[/tex]

  Generally will assume that the sample proportion young adults in Louisville, Kentucky, who received help from their parents in buying a home is  [tex]\^ p = 0.5[/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{1.645}{0.06 } ]^2 * 0.5 (1 - 0.5 ) [/tex]

=>  [tex]n = 188 [/tex]