Question

The amount of money spent on textbooks per year for students is approximately normal.

Answer 1

Answer:a

a

   $336.04 < \mu < 443.96$

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are  told that

   The sample size  $n = 19$

    The sample mean is  $\= x = \ \ 390$

    The  standard deviation is  $\sigma = \ \ 120$

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           $\alpha = 100 - 95$

          $\alpha = 5 \%$

          $\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

    So  

         $Z_{\frac{\alpha }{2} } = 1.96$

The  margin of error is mathematically represented as

      $E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=>    $E = 1.96 * \frac{120}{\sqrt{19} }$

=>   $E = 53.96$

The 95% confidence interval is  

     $\= x - E < \mu < \= x + E$

=>   $390 - 53.96 < \mu < 390 - 53.96$

=>  $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         $n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

       $\r p = \frac{210}{500}$

       $\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is  $\alpha = 0.01$

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

      $Z_{\frac{\alpha }{2} } = 2.58$

  Generally the margin of error is mathematically represented as

       $E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=>   $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=>     $E = 0.0093$

The 99% confidence interval  is

     $\r p - E < p < \r p + E$

     $0.42 - 0.0093 < p < 0.42 + 0.0093$

     $0.4107 < p < 0.4293$