A survey of 73 students found that 37% were in favor of raising tuition to pave new parking lots. The standard deviation of the sample proportion is 9.8%. How large a sample (to the nearest person) would be required to reduce this standard deviation to 4.7%?

Respuesta :

Answer:

The value  is   [tex]n_2  = 317[/tex]

Step-by-step explanation:

From the question we are told that

  The sample size is  n  =  73

   The proportion that in favor of raising the tuition is  [tex]\^{p} =  0.37[/tex]

   The standard deviation is  [tex]s_1  = 0.098[/tex]

    The required standard deviation  [tex]s_2  = 0.047[/tex]

Generally the requires standard deviation is mathematically represented as

      [tex]s_2  = s_1 *  \sqrt{\frac{n_1}{n_2} }[/tex]

=>    [tex]\frac{s_2}{s_1}  =  \sqrt{\frac{n_1}{n_2} }[/tex]

=>    [tex]\frac{n_1}{n_2} =[\frac{s_2}{s_1} ]^2[/tex]

=>     [tex]\frac{73}{n_2} =[\frac{0.047}{0.098} ]^2[/tex]

=>     [tex]\frac{73}{n_2} =0.2300[/tex]

=>     [tex]n_2 =  \frac{73}{0.2300}[/tex]

=>   [tex]n_2  = 317[/tex]