Question

A survey of 800 women shoppers found that 17% of them shop on impulse. What is the 98% confidence interval for the true proportion of women shoppers who shop on impulse

Answer 1

Answer:

The 98% confidence interval for the true proportion of women shoppers who shop on impulse is  

              $0.1391 < p < 0.2009$

Step-by-step explanation:

From the question we are  told that

     The  sample size is n = 800

      The sample  proportion is  $\r p = 0.17$

 

Given that the confidence level is  98%  

The level of significance is evaluated as

      $\alpha = 100 -98$

     $\alpha = 2$%

      $\alpha = 0.02$

given that this is a two tailed test

    $\frac{\alpha }{2} = \frac{0.02}{2} = 0.01$

The critical values obtained from the normal distribution table is  

   $z_{\frac{\alpha }{2} } = 2.33$  

Now the the margin of error is mathematically evaluated as

         $MOE = 2.33 * \sqrt{\frac{0.17 (1-0.17)}{800} }$

          $MOE = 0.0309$

the 98% confidence interval for the true proportion of women shoppers who shop on impulse is mathematically evaluated as

      $0.17 - 0.0309 < p < 0.17 + 0.0309$

       $0.1391 < p < 0.2009$