Question

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%?

Answer 1

Answer:

The value  is   $n_2 = 317$

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  $\^{p} = 0.37$

   The standard deviation is  $s_1 = 0.098$

    The required standard deviation  $s_2 = 0.047$

Generally the requires standard deviation is mathematically represented as

      $s_2 = s_1 * \sqrt{\frac{n_1}{n_2} }$

=>    $\frac{s_2}{s_1} = \sqrt{\frac{n_1}{n_2} }$

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

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

=>     $\frac{73}{n_2} =0.2300$

=>     $n_2 = \frac{73}{0.2300}$

=>   $n_2 = 317$